JEE Previous Year Question Bank on Vector or Cross Product for free

Q 1 :

If A(1, –1, 2), B(5, 7, –6), C(3, 4, –10) and D(–1, –4, –2) are the vertices of a quadrilateral ABCD, then its area is : [2024]

$24 \sqrt{7}$
$12 \sqrt{29}$
$48 \sqrt{7}$
$24 \sqrt{29}$

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(2)

We have, A(1, –1, 2), B(5, 7, –6), C(3, 4, –10) and D(–1, –4, –2)

$\vec{A C}$ and $\vec{B D}$ are diagonals of ABCD.

Now, $\vec{A C} = 2 \hat{i} + 5 \hat{j} - 12 \hat{k}$

and $\vec{B D} = 6 \hat{i} + 11 \hat{j} - 4 \hat{k}$

Area, of ABCD = $\frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & - 12 \\ 6 & 11 & - 4 \end{matrix} | |$

$= \frac{1}{2} | 112 \hat{i} - 64 \hat{j} - 8 \hat{k} | = \frac{1}{2} \times \sqrt{16704}$

$= \frac{1}{2} \times 24 \sqrt{29} = 12 \sqrt{29}$ .

Q 2 :

Let $\vec{a} = 2 \hat{i} + 5 \hat{j} - \hat{k}$ , $\vec{b} = 2 \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $\vec{c}$ be three vectors such that $(\vec{c} + \hat{i}) \times (\vec{a} + \vec{b} + \hat{i}) = \vec{a} \times (\vec{c} + \hat{i})$ . If $\vec{a} \cdot \vec{c} = - 29$ , then $\vec{c} \cdot (- 2 \hat{i} + \hat{j} + \hat{k})$ is equal to : [2024]

15
5
12
10

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(2)

Consider $\vec{a} + \vec{b} + \hat{i}$

$= 2 \hat{i} + 5 \hat{j} - \hat{k} + 2 \hat{i} - 2 \hat{j} + 2 \hat{k} + \hat{i}$

$= 5 \hat{i} + 3 \hat{j} + \hat{k}$

Now, $(\vec{c} + \hat{i}) \times (\vec{a} + \vec{b} + \hat{i}) = \vec{a} \times (\vec{c} + \hat{i})$

$\Rightarrow (\vec{c} + \hat{i}) \times (5 \hat{i} + 3 \hat{j} + \hat{k}) = (2 \hat{i} + 5 \hat{j} - \hat{k}) \times (\vec{c} + \hat{i})$

$\Rightarrow (\vec{c} + \hat{i}) \times (5 \hat{i} + 3 \hat{j} + \hat{k} + 2 \hat{i} + 5 \hat{j} - \hat{k}) = 0$

$\Rightarrow (\vec{c} + \hat{i}) \times (7 \hat{i} + 8 \hat{j}) = 0$

$\Rightarrow \vec{c} + \hat{i} = λ (7 \hat{i} + 8 \hat{j})$

$\vec{a} \cdot \vec{c} + \vec{a} \cdot \hat{i} = λ \vec{a} \cdot (7 \hat{i} + 8 \hat{j})$

$\Rightarrow - 29 + 2 = λ (14 + 40) \Rightarrow λ = \frac{- 1}{2}$

$∴ \vec{c} + \hat{i} = \frac{- 7}{2} \hat{i} - 4 \hat{j}$

$\Rightarrow \vec{c} = \frac{- 9}{2} \hat{i} - 4 \hat{j}$

$∴ \vec{c} \cdot (- 2 \hat{i} + \hat{j} + \hat{k}) = 9 - 4 = 5$ .

Q 3 :

Consider three vectors $\vec{a}, \vec{b}, \vec{c}$ . Let $| \vec{a} | = 2, | \vec{b} | = 3$ and $\vec{a} = \vec{b} \times \vec{c}$ . If $α \in [0, \frac{π}{3}]$ is the angle between the vectors $\vec{b}$ and $\vec{c}$ , then the minimum value of $27 | \vec{c} - \vec{a} |^{2}$ is equal to : [2024]

105
124
121
110

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(2)

Consider $| \vec{c} - \vec{a} |^{2}$

$= | \vec{c} |^{2} + | \vec{a} |^{2} - 2 \vec{a} \cdot \vec{c} = | \vec{c} |^{2} + 4 - 2 (\vec{b} \times \vec{c}) \cdot \vec{c} = | \vec{c} |^{2} + 4 - 0$

$∴ | \vec{c} - \vec{a} |^{2} = | \vec{c} |^{2} + 4$ ... (i)

Now, $| \vec{a} | = | \vec{b} \times \vec{c} | \Rightarrow 2 = | \vec{b} | | \vec{c} | \sin α, α \in [0, \frac{π}{3}]$

$= 3 | \vec{c} | \sin α$

$\Rightarrow | \vec{c} | = \frac{2}{3} cosec α$

$\Rightarrow | \vec{c} |_{\min} = \frac{2}{3} \times \frac{2}{\sqrt{3}}$ $(∵ α \in [0, \frac{π}{3}])$

$\Rightarrow 27 | \vec{c} - \vec{a} |_{\min}^{2} = 27 (\frac{16}{27} + 4)$

=16 + 108 = 124.

Q 4 :

If A(3, 1, –1), $B (\frac{5}{3}, \frac{7}{3}, \frac{1}{3})$ , C(2, 2, 1) and $D (\frac{10}{3}, \frac{2}{3}, \frac{- 1}{3})$ are the vertices of a quadrilateral ABCD, then its area is [2024]

$\frac{5 \sqrt{2}}{3}$
$\frac{4 \sqrt{2}}{3}$
$2 \sqrt{2}$
$\frac{2 \sqrt{2}}{3}$

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(2)

Gicen, A(3, 1, –1), $B (\frac{5}{3}, \frac{7}{3}, \frac{1}{3})$ , C(2, 2, 1) and $D (\frac{10}{3}, \frac{2}{3}, \frac{- 1}{3})$ are vertices of quadrilateral ABCD

$\vec{A C} = (2 - 3) \hat{i} + (2 - 1) \hat{j} + (1 + 1) \hat{k} = - \hat{i} + \hat{j} + 2 \hat{k}$

$\vec{B D} = (\frac{10}{3} - \frac{5}{3}) \hat{i} + (\frac{2}{3} - \frac{7}{3}) \hat{j} + (\frac{- 1}{3} - \frac{1}{3}) \hat{k} = \frac{5}{3} \hat{i} - \frac{5}{3} \hat{j} - \frac{2}{3} \hat{k}$

Area of quadrilateral ABCD = $\frac{1}{2}$ $| \vec{A C} \times \vec{B D} |$

$= \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 1 & 2 \\ 5 / 3 & - 5 / 3 & - 2 / 3 \end{matrix} | |$ $= \frac{1}{2}$ $| \hat{i} (\frac{8}{3}) - \hat{j} (\frac{- 8}{3}) + \hat{k} (0) |$

$= \frac{1}{2} \sqrt{\frac{64}{9} + \frac{64}{9}} = \frac{4 \sqrt{2}}{3} sq . units$ .

Q 5 :

Let $\vec{a} = 2 \hat{i} + \hat{j} - \hat{k}$ , $\vec{b} = ((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i}$ . Then the square of the projection of $\vec{a}$ on $\vec{b}$ is : [2024]

$\frac{1}{3}$
$\frac{1}{5}$
$\frac{2}{3}$
2

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(4)

We have, $\vec{a} = 2 \hat{i} + \hat{j} - \hat{k}$

and $\vec{b} = ((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i}$

$=$ $(| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & - 1 \\ 1 & 1 & 0 \end{matrix} | \times \hat{i})$ $\times \hat{i} = ((\hat{i} - \hat{j} + \hat{k}) \times \hat{i})$

$=$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 1 \\ 1 & 0 & 0 \end{matrix} |$ $\times \hat{i} = (\hat{j} + \hat{k}) \times \hat{i} = - \hat{k} + \hat{j}$

$\Rightarrow \vec{b} = \hat{j} - \hat{k}$

So, projection of $\vec{a}$ on $\vec{b}$ = $\frac{\vec{a} \cdot \vec{b}}{| \vec{b} |} = \frac{2}{\sqrt{2}} = \sqrt{2}$

$∴$ Square of projection = 2.

Q 6 :

Let $\vec{a} = 6 \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + \hat{j}$ . If $\vec{c}$ is a vector such that $| \vec{c} | \geq 6, \vec{a} \cdot \vec{c} = 6 | \vec{c} |, | \vec{c} - \vec{a} | = 2 \sqrt{2}$ and the angle between $\vec{a} \times \vec{b}$ and $\vec{c}$ is 60°, then $| (\vec{a} \times \vec{b}) \times \vec{c} |$ is equal to : [2024]

$\frac{3}{2} \sqrt{3}$
$\frac{9}{2} (6 - \sqrt{6})$
$\frac{9}{2} (6 + \sqrt{6})$
$\frac{3}{2} \sqrt{6}$

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(3)

We have, $| (\vec{a} \times \vec{b}) \times \vec{c} |$ $= | \vec{a} \times \vec{b} | | \vec{c} | \sin 60 °$

Now, $\vec{a} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 1 & - 1 \\ 1 & 1 & 0 \end{matrix} |$ $= \hat{i} - \hat{j} + 5 \hat{k}$

So, $| \vec{a} \times \vec{b} |$ $= \sqrt{1 + 1 + 25} = \sqrt{27} = 3 \sqrt{3}$

Also, $| \vec{c} - \vec{a} |$ $= 2 \sqrt{2}$ [Given]

$\Rightarrow | \vec{c} |^{2} + | \vec{a} |^{2} - 2 | \vec{c} \cdot \vec{a} | = 8$

$\Rightarrow | \vec{c} |^{2} + 38 - 12 | \vec{c} | = 8$ $[∵ \vec{c} \cdot \vec{a} = 6 | \vec{c} |]$

$\Rightarrow | \vec{c} |^{2} - 12 | \vec{c} | + 30 = 0$

$\Rightarrow | \vec{c} | = \frac{12 + \sqrt{24}}{2}$ $[∵ | \vec{c} | \geq 6]$

$\Rightarrow | \vec{c} | = 6 + \sqrt{6}$

$∴$ $| (\vec{a} \times \vec{b}) \times \vec{c} |$ $= \sqrt{27} \times (6 + \sqrt{6}) \times \frac{\sqrt{3}}{2} = \frac{9}{2} (6 + \sqrt{6})$ .

Q 7 :

The set of all $α$ , for which the vectors $\vec{a} = α t \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = t \hat{i} - 2 \hat{j} - 2 α t \hat{k}$ are inclined at an obtuse angle for all $t \in R$ , is [2024]

[0, 1)
(–2, 0]
$(- \frac{4}{3}, 0]$
$(- \frac{4}{3}, 1)$

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(3)

Given, $\vec{a} = α t \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = t \hat{i} - 2 \hat{j} - 2 α t \hat{k}$

So, $\vec{a} \cdot \vec{b} < 0 \forall t \in R$ [ $∴$ $\vec{a}$ and $\vec{b}$ are inclined at an obtuse angle]

$\Rightarrow α t^{2} - 12 + 6 α t < 0 \forall t \in R \Rightarrow α < 0 and D < 0$

$\Rightarrow 36 α^{2} + 48 α < 0 \Rightarrow 12 α (3 α + 4) < 0$

$\Rightarrow \frac{- 4}{3} < α < 0$

Also, for $α = 0, \vec{a} \cdot \vec{b} < 0 \Rightarrow α \in (- \frac{4}{3}, 0]$ .

Q 8 :

Let $\vec{a} = 4 \hat{i} - \hat{j} + \hat{k}$ , $\vec{b} = 11 \hat{i} - \hat{j} + \hat{k}$ and $\vec{c}$ be a vector that $(\vec{a} + \vec{b}) \times \vec{c} = \vec{c} \times (- 2 \vec{a} + 3 \vec{b})$ . If $(2 \vec{a} + 3 \vec{b}) \cdot \vec{c} = 1670$ , then $| \vec{c} |^{2}$ is equal to : [2024]

1600
1618
1627
1609

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(2)

We have, $\vec{a} = 4 \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = 11 \hat{i} - \hat{j} + \hat{k}$

$\vec{a} \cdot \vec{b} = 44 + 1 + 1 = 46, | \vec{a} |^{2} = 18, | \vec{b} |^{2} = 123$

$(\vec{a} + \vec{b}) \times \vec{c} = \vec{c} \times (- 2 \vec{a} + 3 \vec{b})$ ... (i) [Given]

and $(2 \vec{a} + 3 \vec{b}) \cdot \vec{c} = 1670$ ... (ii)

$∴ (\vec{a} + \vec{b}) \times \vec{c} = (2 \vec{a} - 3 \vec{b}) \times \vec{c}$ [From (i)]

$\Rightarrow (- \vec{a} + 4 \vec{b}) \times \vec{c} = 0 \Rightarrow \vec{c} = λ (4 \vec{b} - \vec{a})$

$\Rightarrow (2 \vec{a} + 3 \vec{b}) \cdot λ (4 \vec{b} - \vec{a}) = 1670$ [From (ii)]

$\Rightarrow λ (5 \vec{a} \cdot \vec{b} - 2 | \vec{a} |^{2} + 12 | \vec{b} |^{2}) = 1670$

$\Rightarrow λ = \frac{1670}{5 \times 46 - 2 \times 18 + 12 \times 123} = 1$

Now, $\vec{c} = 4 \vec{b} - \vec{a} = 4 (11 \hat{i} - \hat{j} + \hat{k}) - (4 \hat{i} - \hat{j} + \hat{k})$

$= 40 \hat{i} - 3 \hat{j} + 3 \hat{k}$

$∴ | \vec{c} |^{2} = 1600 + 9 + 9 = 1618$ .

Q 9 :

Let three vectors $\vec{a} = α \hat{i} + 4 \hat{j} + 2 \hat{k}$ , $\vec{b} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k},$ $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}$ form a triangle such that $\vec{c} = \vec{a} - \vec{b}$ and the area of the triangle is $5 \sqrt{6}$ . If $α$ is a positive real number, then $| \vec{c} |^{2}$ is equal to: [2024]

14
16
12
10

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(1)

We have, $\vec{c} = \vec{a} - \vec{b}$

$\Rightarrow \vec{c} = (α - 5) \hat{i} + \hat{j} - 2 \hat{k}$

Let ABC be the given triangle.

Area of $△ A B C = \frac{1}{2} | \vec{a} \times \vec{b} | = \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ α & 4 & 2 \\ 5 & 3 & 4 \end{matrix} | |$

$\Rightarrow \frac{1}{2}$ $| 10 \hat{i} - (4 α - 10) \hat{j} + (3 α - 20) \hat{k} |$ $= 5 \sqrt{6}$ [Given]

$= 100 + {(4 α - 10)}^{2} + {(3 α - 20)}^{2} = 600$

$\Rightarrow 25 α^{2} - 200 α = 0 \Rightarrow α = 8$ $[∵ α .]$

$\Rightarrow | \vec{c} |^{2} = 9 + 1 + 4 = 14$ .

Q 10 :

Let $\vec{O A} = 2 a$ and $\vec{O B} = 6 \vec{a} + 5 \vec{b}$ and $\vec{O C} = 3 \vec{b}$ where O is the origin. If the area of the parallelogram with adjacent sides $\vec{O A}$ and $\vec{O C}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to: [2024]

35
40
38
32

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(1)

We have, $| \vec{O A} \times \vec{O C} |$ $= 15$

$\Rightarrow$ $| 2 \vec{a} \times 3 \vec{b} |$ $= 15 \Rightarrow$ $| \vec{a} \times \vec{b} |$ $= \frac{5}{2}$ ... (i)

Area of quadrilateral $O A B C = \frac{1}{2}$ $| \vec{O B} \times \vec{A C} |$

$= \frac{1}{2}$ $| (6 \vec{a} + 5 \vec{b}) \times (3 \vec{b} - 2 \vec{a}) |$ $= \frac{1}{2}$ $| 18 (\vec{a} \times \vec{b}) - 10 (\vec{b} \times \vec{a}) |$

$= \frac{1}{2}$ $| 18 (\vec{a} \times \vec{b}) + 10 (\vec{a} \times \vec{b}) |$ $= 14$ $| \vec{a} \times \vec{b} |$

$= 14 \times \frac{5}{2} [usin g (i)]$

= 35 sq. units.

Q 11 :

Let $\vec{a} = 2 \hat{i} + α \hat{j} + \hat{k}$ , $\vec{b} = - \hat{i} + \hat{k}$ , $\vec{c} = β \hat{j} - \hat{k}$ , where $α$ and $β$ are integers and $α β = - 6$ . Let the values of the ordered pair $(α, β)$ , for which the area of the parallelogram of diagonals $\vec{a} + \vec{b}$ and $\vec{b} + \vec{c}$ is $\frac{\sqrt{21}}{2}$ , be $(α_{1}, β_{1})$ and $(α_{2}, β_{2})$ . Then $α_{1}^{2} + β_{1}^{2} - α_{2} β_{2}$ is equal to [2024]

21
17
24
19

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(4)

Area of parallelogram $= \frac{1}{2} \times$ $| (\vec{a} + \vec{b}) \times (\vec{b} + \vec{c}) |$

$\Rightarrow \frac{\sqrt{21}}{2} = \frac{1}{2}$ $| (\vec{a} + \vec{b}) \times (\vec{b} + \vec{c}) |$

$\Rightarrow \sqrt{21} =$ $| (\hat{i} + α \hat{j} + 2 \hat{k}) \times (- \hat{i} + β \hat{j}) |$

$\Rightarrow \sqrt{21} =$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & α & 2 \\ - 1 & β & 0 \end{matrix} | |$

$\Rightarrow \sqrt{21} =$ $| - 2 β \hat{i} - 2 \hat{j} + (α + β) \hat{k} |$

$\Rightarrow α^{2} + 5 β^{2} + 2 α β = 17$

$\Rightarrow α^{2} + 5 β^{2} = 17 + 12$ $[∵ α β = - 6]$

$\Rightarrow α^{2} + 5 β^{2} = 29$

and $α β = - 6, α, β$ are integers.

$\Rightarrow α = - 3, β = 2 or α = 3, β = - 2$

$α_{1}^{2} + β_{1}^{2} - α_{2} β_{2} = 9 + 4 + 6 = 19$ .

Q 12 :

Between the following two statements:

Statement-I: Let $\vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$ . Then the vector $\vec{r}$ satisfying $\vec{a} \times \vec{r} = \vec{a} \times \vec{b}$ and $\vec{a} \cdot \vec{r} = 0$ is of magnitude $\sqrt{10}$ .

Statement-II: In a triangle ABC, cos 2A + cos 2B + cos 2C $\geq - \frac{3}{2}$ . [2024]

Both Statement-I and Statement-II are incorrect.
Statement-I is incorrect but Statement-II is correct.
Statement-I is correct but Statement-II is incorrect.
Both Statement-I and Statement-II are correct.

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(2)

We have, $\vec{a} \times \vec{r} = \vec{a} \times \vec{b}$

$\Rightarrow \vec{a} \times (\vec{r} - \vec{b}) = 0 \Rightarrow \vec{a} = λ (\vec{r} - \vec{b})$

$\Rightarrow \vec{a} \cdot \vec{a} = λ (\vec{a} \cdot \vec{r} - \vec{a} \cdot \vec{b})$

$\Rightarrow | \vec{a} |^{2} = λ (0 - 7) \Rightarrow 14 = - 7 λ$

$\Rightarrow λ = - 2 \Rightarrow \vec{r} = \vec{b} - \frac{\vec{a}}{2} = \frac{2 \vec{b} - \vec{a}}{2} = \frac{3 \hat{i} + \hat{k}}{2}$

So, $| \vec{r} | = \frac{\sqrt{10}}{2}$

So, Statement-I is incorrect.

Let $△$ ABC be given triangle and O be its circumcentre

Now, OA = $\vec{a}$ , OB = $\vec{b}$ and OC = $\vec{c}$

Now, |OA| = |OB| = |OC|

(distance from circumcentre will be same for all vertices.)

Consider (OA + OB + OC)

$= {(\vec{a} + \vec{b} + \vec{c})}^{2} \geq 0$

$\Rightarrow | \vec{a} |^{2} + | \vec{b} |^{2} + | \vec{c} |^{2} + 2 \vec{a} \cdot \vec{b} + 2 \vec{b} \cdot \vec{c} + 2 \vec{c} \cdot \vec{a} \geq 0$

$\Rightarrow 3 | \vec{a} |^{2} + 2 | \vec{a} | | \vec{b} | \cos 2 C + 2 | \vec{b} | | \vec{c} | \cos 2 A + 2 | \vec{c} | | \vec{a} | \cos 2 B \geq 0$

$\Rightarrow 3 | \vec{a} |^{2} + 2 | \vec{a} |^{2} [\cos 2 A + \cos 2 B + \cos 2 C] \geq 0$

$\Rightarrow 2 [\cos 2 A + \cos 2 B + \cos 2 C] \geq - 3$ $[∵ | \vec{a} | > 0]$

$\Rightarrow$ cos 2A + cos 2B + cos 2C $\geq - \frac{3}{2}$

Hence, Statement-II is correct.

Q 13 :

Let $\vec{O A} = \vec{a}$ , $\vec{O B} = 12 \vec{a} + 4 \vec{b}$ and $\vec{O C} = b$ , where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then $\frac{area of the quadrilateral O A B C}{area of S}$ is equal to __________. [2024]

8
10
7
6

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(1)

We have,

Area of parallelogram, $S =$ $| \vec{O A} \times \vec{O C} |$ $=$ $| \vec{a} \times \vec{b} |$ ... (i)

Area of quad. OABC = Area of $△$ OAB + Area of $△$ OBC

$= \frac{1}{2}$ $| \vec{O A} \times \vec{O B} |$ $+ \frac{1}{2}$ $| \vec{O B} \times \vec{O C} |$

$= \frac{1}{2}$ $| \vec{a} \times (12 \vec{a} + 4 \vec{b}) |$ $+ \frac{1}{2}$ $| (12 \vec{a} + 4 \vec{b}) \times \vec{b} |$

$= \frac{1}{2} \times (4 | \vec{a} \times \vec{b} | + 12 | \vec{a} \times \vec{b} |) = 8$ $| \vec{a} \times \vec{b} |$

$∴ \frac{Area of quad . O A B C}{Area of parallelogram} = \frac{8 | \vec{a} \times \vec{b} |}{| \vec{a} \times \vec{b} |} = 8$ .

Q 14 :

Let A(2, 3, 5) and C(–3, 4, –2) be opposite vertices of a parallelogram ABCD. If the diagonal $\vec{B D} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , then the area of the parallelogram is equal to [2024]

$\frac{1}{2} \sqrt{410}$
$\frac{1}{2} \sqrt{474}$
$\frac{1}{2} \sqrt{586}$
$\frac{1}{2} \sqrt{306}$

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(2)

Area of parallelogram ABCD

$= \frac{1}{2}$ $| \vec{A C} \times \vec{B D} |$

$= \frac{1}{2}$ $| (- 5 \hat{i} + \hat{j} - 7 \hat{j}) \times (\hat{i} + 2 \hat{j} + 3 \hat{k}) |$

$= \frac{1}{2}$ $| | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 5 & 1 & - 7 \\ 1 & 2 & 3 \end{matrix} | |$

$= \frac{1}{2}$ $| (3 + 14) \hat{i} - (- 15 + 7) \hat{j} + (- 10 - 1) \hat{k} |$

$= \frac{1}{2}$ $| (17 \hat{i} + 8 \hat{j} - 11 \hat{k}) |$

$= \frac{1}{2} \sqrt{289 + 64 + 121} = \frac{1}{2} \sqrt{474}$ sq. units.

Q 15 :

Let $\vec{a} = \hat{i} + α \hat{j} + β \hat{k}$ , $α, β \in R$ . Let a vector $\vec{b}$ be such that the angle between $\vec{a}$ and $\vec{b}$ is $\frac{π}{4}$ and $| \vec{b} |^{2} = 6$ . If $\vec{a} \cdot \vec{b} = 3 \sqrt{2}$ , then the value of $(α^{2} + β^{2}) | \vec{a} \times \vec{b} |^{2}$ is equal to [2024]

95
85
90
75

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(3)

Given, $\vec{a} \cdot \vec{b} = 3 \sqrt{2}$ and angle between $\vec{a}$ and $\vec{b}$ is $\frac{π}{4}$ .

$\Rightarrow | \vec{a} | | \vec{b} | \cos \frac{π}{4} = 3 \sqrt{2} \Rightarrow | \vec{a} | (\sqrt{6}) (\frac{1}{\sqrt{2}}) = 3 \sqrt{2}$

$\Rightarrow | \vec{a} | = \sqrt{6}$

Since, $\vec{a} = \hat{i} + α \hat{j} + β \hat{k}$ $∴ | \vec{a} | = \sqrt{1 + α^{2} + β^{2}} = \sqrt{6}$

$\Rightarrow 1 + α^{2} + β^{2} = 6 \Rightarrow α^{2} + β^{2} = 5$

also, $| \vec{a} \times \vec{b} |^{2} = | \vec{a} |^{2} | \vec{b} |^{2} \sin^{2} \frac{π}{4} = 6 \times 6 \times \frac{1}{2} = 18$

$∴ (α^{2} + β^{2}) | \vec{a} \times \vec{b} |^{2} = 5 \times 18 = 90$ .

Q 16 :

Let $\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ , $\vec{b} = 4 \hat{i} + \hat{j} + 7 \hat{k}$ and $\vec{c} = \hat{i} - 3 \hat{j} + 4 \hat{k}$ be three vectors. If a vectors $\vec{p}$ satisfies $\vec{p} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{p} \cdot \vec{a} = 0$ , then $\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k})$ is equal to [2024]

28
32
36
24

1

2

3

4

(2)

Given, $\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ , $\vec{b} = 4 \hat{i} + \hat{j} + 7 \hat{k}$ and $\vec{c} = \hat{i} - 3 \hat{j} + 4 \hat{k}$

Also, $\vec{p} \times \vec{b} = \vec{c} \times \vec{b}$

$(\vec{p} - \vec{c}) \times \vec{b} = 0 \Rightarrow \vec{b} | | \vec{p} - \vec{c} \Rightarrow (\vec{p} - \vec{c}) = λ \vec{b}$

let $λ \vec{b} + \vec{c} = \vec{p}$

$\Rightarrow \vec{p} = \hat{i} (4 λ + 1) + \hat{j} (λ - 3) + \hat{k} (7 λ + 4) \Rightarrow \vec{p} \cdot \vec{a} = 0$

$\Rightarrow (4 λ + 1) 3 + (λ - 3) (1) + (7 λ + 4) (- 2) = 0$

$\Rightarrow 12 λ + 3 + λ - 3 - 14 λ - 8 \Rightarrow λ = - 8$

$∴ \vec{p} = - 31 \hat{i} - 11 \hat{j} - 52 \hat{k}$

$\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k}) = - 31 + 11 + 52$

$\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k}) = 32$ .

Q 17 :

Let ABC be a triangle of area $15 \sqrt{2}$ and the vectors $\vec{A B} = \hat{i} + 2 \hat{j} - 7 \hat{k}$ , $\vec{B C} = a \hat{i} + b \hat{j} + c \hat{k}$ and $\vec{A C} = 6 \hat{i} + d \hat{j} - 2 \hat{k}$ , d > 0. Then the square of the length of the largest side of the triangle ABC is __________. [2024]

(54)

Area of triangle, ABC = $15 \sqrt{2}$

$\frac{1}{2}$ $| \vec{A B} \times \vec{A C} |$ $= 15 \sqrt{2}$ ... (i)

Now, $\vec{A B} \times \vec{A C} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 7 \\ 6 & d & - 2 \end{matrix} |$

$= (7 d - 4) \hat{i} - 40 \hat{j} + (d - 12) \hat{k}$ ... (ii)

From (i) and (ii), we get

$\frac{1}{2} \sqrt{{(7 d - 4)}^{2} + 1600 + {(d - 12)}^{2}} = 15 \sqrt{2}$

$\Rightarrow 5 d^{2} - 8 d - 4 = 0$

$\Rightarrow d = \frac{- 2}{5}$ (Rejected as d > 0) or d = 2

Also, $\vec{A B} + \vec{B C} = \vec{A C}$

$\Rightarrow (\hat{i} + 2 \hat{j} - 7 \hat{k}) + (a \hat{i} + b \hat{j} + c \hat{k}) = 6 \hat{i} + d \hat{j} - 2 \hat{k}$

$\Rightarrow (a + 1) \hat{i} + (2 + b) \hat{j} + (- 7 + c) \hat{k} = 6 \hat{i} + 2 \hat{j} - 2 \hat{k}$

$\Rightarrow a + 1 = 6 \Rightarrow a = 5$

and b + 2 = 2 $\Rightarrow$ b = 0 and c – 7 = –2 $\Rightarrow$ c = 5

Hence, $| \bar{A B} |$ $= \sqrt{54},$ $| \bar{A C} |$ $= \sqrt{44},$ $| \bar{B C} |$ $= \sqrt{50}$

Largest side has length of $\sqrt{54}$

$∴$ $| \vec{A B} |^{2}$ $= 54$ .

Q 18 :

Let $\vec{a} = \hat{i} - 3 \hat{j} + 7 \hat{k}$ , $\vec{b} = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a} + 2 \vec{b}) \times \vec{c} = 3 (\vec{c} \times \vec{a})$ . If $\vec{a} \cdot \vec{c} = 130$ , then $\vec{b} \cdot \vec{c}$ is equal to __________. [2024]

(30)

Let $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}$

Now, $\vec{a} + 2 \vec{b} = 5 \hat{i} - 5 \hat{j} + 9 \hat{k}$

Given, $(\vec{a} + 2 \vec{b}) \times \vec{c} = 3 (\vec{c} \times \vec{a})$

$\Rightarrow$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & - 5 & 9 \\ x & y & z \end{matrix} |$ $= 3$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 1 & - 3 & 7 \end{matrix} |$

$\Rightarrow \hat{i} (- 5 z - 9 y) - \hat{j} (5 z - 9 x) + \hat{k} (5 y + 5 x)$

$\Rightarrow 3$ $[\hat{i} (7 y + 3 z) - \hat{j} (7 x - z) + \hat{k} (- 3 x - y)]$

$\Rightarrow - 5 z - 9 y = 21 y + 9 z and 5 z - 9 x = 21 x - 3 z$

$\Rightarrow - 30 y = 14 z and 8 z = 30 x$

$\Rightarrow z = \frac{- 30}{14} y and z = \frac{30}{8} x$

$\Rightarrow z = \frac{- 15}{7} y = \frac{15}{4} x$

$\Rightarrow - 4 y = 7 x$

Now, $\vec{a} \cdot \vec{c} = 130$

$\Rightarrow x - 3 y + 7 z = 130$

$\Rightarrow x - 3 (- \frac{7}{4} x) + 7 (\frac{15}{4} x) = 130$

$\Rightarrow 130 x = 4 \times 130 \Rightarrow x = 4, y = - 7 and z = 15$

$∴ \vec{b} \cdot \vec{c} = (2 \hat{i} - \hat{j} + \hat{k}) (4 \hat{i} - 7 \hat{j} + 15 \hat{k})$

= 8 + 7 +15 = 30.

Q 19 :

Let $\vec{a} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ , $\vec{b} = 3 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and a vector $\vec{c}$ be such that $\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times \vec{c} = \hat{i} + 8 \hat{j} + 13 \hat{k}$ . If $\vec{a} \cdot \vec{c} = 13$ , then $(24 - \vec{b} \cdot \vec{c})$ is equal to __________. [2024]

(46)

We have, $\vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times \vec{c} = \hat{i} + 8 \hat{j} + 13 \hat{k}$

$\Rightarrow (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}) + (\vec{b} \times \vec{c}) = \hat{i} + 8 \hat{j} + 13 \hat{k}$

$\Rightarrow \vec{a} \times (\vec{a} \times \vec{b}) + \vec{a} \times (\vec{a} \times \vec{c}) + \vec{a} \times (\vec{b} \times \vec{c}) = \vec{a} \times (\hat{i} + 8 \hat{j} + 13 \hat{k})$

$\Rightarrow (\vec{a} \cdot \vec{b}) \vec{a} - a^{2} \vec{b} + (\vec{a} \cdot \vec{c}) \vec{a} - a^{2} \vec{c} + (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} = \vec{a} \times (\hat{i} + 8 \hat{j} + 13 \hat{k})$

$\Rightarrow - 26 \vec{a} - 29 \vec{b} + 13 \vec{a} - 29 \vec{c} + 13 \vec{b} + 26 \vec{c} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 3 & 4 \\ 1 & 8 & 13 \end{matrix} |$

$\Rightarrow - 13 \vec{a} - 16 \vec{b} - 3 \vec{c} = - 71 \hat{i} - 22 \hat{j} + 19 \hat{k}$

$\Rightarrow - 13 \vec{a} \cdot \vec{b} - 16 b^{2} - 3 \vec{b} \cdot \vec{c} = - 213 - 88 - 95$

$\Rightarrow 338 - 800 - 3 \vec{b} \cdot \vec{c} = - 396$

$\Rightarrow - 462 - 3 \vec{b} \cdot \vec{c} = - 396 \Rightarrow \vec{b} \cdot \vec{c} = - 22$

. $∴ 24 - \vec{b} \cdot \vec{c} = 24 + 22 = 46$

Q 20 :

Let $\vec{a} = 9 \hat{i} - 13 \hat{j} + 25 \hat{k}$ , $\vec{b} = 3 \hat{i} + 7 \hat{j} - 13 \hat{k}$ and $\vec{c} = 17 \hat{i} - 2 \hat{j} + \hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} = (\vec{b} + \vec{c}) \times \vec{a}$ and $\vec{r} \cdot (\vec{b} - \vec{c}) = 0$ , then $\frac{| 593 \vec{r} + 67 \vec{a} |^{2}}{{(593)}^{2}}$ is equal to __________. [2024]

(569)

We have, $\vec{r} \times \vec{a} = (\vec{b} + \vec{c}) \times \vec{a}$

$\Rightarrow [\vec{r} - (\vec{b} + \vec{c})] \times \vec{a} = 0 \Rightarrow \vec{r} - (\vec{b} + \vec{c}) = λ \vec{a}$ , for some scalar $λ$ .

$\Rightarrow \vec{r} = λ \vec{a} + \vec{b} + \vec{c}$

Also, $\vec{r} \cdot (\vec{b} - \vec{c}) = 0 \Rightarrow (λ \vec{a} + \vec{b} + \vec{c}) \cdot (\vec{b} - \vec{c}) = 0$

$\Rightarrow λ \vec{a} \cdot \vec{b} - λ \vec{a} \cdot \vec{c} + | \vec{b} |^{2} - \vec{b} \cdot \vec{c} + \vec{b} \cdot \vec{c} - | \vec{c} |^{2} = 0$

$\Rightarrow λ = \frac{| \vec{c} |^{2} - | \vec{b} |^{2}}{\vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}} = \frac{294 - 227}{- 389 - 204} = \frac{- 67}{593}$

$\Rightarrow \vec{r} = \vec{b} + \vec{c} - \frac{67}{593} \vec{a} ∴ \frac{| 593 \vec{r} + 67 \vec{a} |^{2}}{{(593)}^{2}} = \frac{(593 {(\vec{b} + \vec{c}))}^{2}}{{(593)}^{2}}$

$= | \vec{b} + \vec{c} |^{2} = | 20 \hat{i} + 5 \hat{j} - 12 \hat{k} |^{2} = 569$ .

Q 21 :

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = - \hat{i} - 8 \hat{j} + 2 \hat{k}$ and $\vec{c} = 4 \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ be three vectors such that $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$ . If the angle between the vector $\vec{c}$ and the vector $3 \hat{i} + 4 \hat{j} + \hat{k}$ is $θ$ , then the greatest integer less than or equal to $\tan^{2} θ$ is __________. [2024]

(38)

We have, $\vec{b} \times \vec{a} = \vec{c} \times \vec{a}$

$\Rightarrow - 10 \hat{i} + 3 \hat{j} + 7 \hat{k} = \hat{i} (c_{2} - c_{3}) - \hat{j} (4 - c_{3}) + \hat{k} (4 - c_{2})$

On comparing, we get $c_{2} = - 3, c_{3} = 7$

$∴ \vec{c} = 4 \hat{i} - 3 \hat{j} + 7 \hat{k}$

Now, $\frac{\vec{c} \cdot (3 \hat{i} + 4 \hat{j} + \hat{k})}{| \vec{c} | | 3 \hat{i} + 4 \hat{j} + \hat{k} |} \Rightarrow \cos θ$

$\Rightarrow \cos θ = \frac{12 - 12 + 7}{\sqrt{74} \sqrt{26}} \Rightarrow \cos θ = \frac{7}{2 \sqrt{481}}$

So, $\tan θ = \frac{25 \sqrt{3}}{7} \Rightarrow \tan^{2} θ = 38.26$

$\Rightarrow [\tan^{2} θ] = [38.26] = 38$

Q 22 :

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $| \vec{a} | = 1, | \vec{b} | = 4$ and $\vec{a} \cdot \vec{b} = 2$ . If $\vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $α$ , then 192 $\sin^{2} α$ is equal to __________. [2024]

(48)

Given $| \vec{a} |$ $= 1,$ $| \vec{b} |$ $= 4 and \vec{a} \cdot \vec{b} = 2$

$\Rightarrow | \vec{a} \times \vec{b} |^{2} = | \vec{a} |^{2} | \vec{b} |^{2} - {(\vec{a} \cdot \vec{b})}^{2}$

$| \vec{a} \times \vec{b} |^{2} = 16 - 4 = 12$

Now $\vec{c} = 2 \vec{a} \times \vec{b} - 3 \vec{b}$

$\Rightarrow | \vec{c} |^{2} = 4 | \vec{a} \times \vec{b} |^{2} + 9 | \vec{b} |^{2}$

$| \vec{c} |^{2} = 4 \times 12 + 9 \times 16 = 192$

$\vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b}$ [Taking dot product with $\vec{b}$ ]

$\vec{b} \cdot \vec{c} = 0 - 3 | \vec{b} |^{2} = - 48$

Let $α$ be the angle between $\vec{b}$ and $\vec{c}$ then

$\cos α = \frac{\vec{b} \cdot \vec{c}}{| \vec{b} | | \vec{c} |} = \frac{- 48}{4 \times 8 \sqrt{3}} = \frac{- \sqrt{3}}{2} \Rightarrow α = \frac{5 π}{6}$

then $192 \sin^{2} α = 192 \sin^{2} (150 °) = 192 \times \sin^{2} (180 ° - 30 °)$

$= 192 \times \sin^{2} 30 ° = 192 \times \frac{1}{4} = 48$ .

Q 23 :

Let $\vec{a} = 3 \hat{i} + 2 \hat{j} + \hat{k}$ , $\vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a} + \vec{b}) \times \vec{c} = 2 (\vec{a} \times \vec{b}) + 24 \hat{j} - 6 \hat{k}$ and $(\vec{a} - \vec{b} + \hat{i}) \cdot \vec{c} = - 3$ . Then $| \vec{c} |^{2}$ is equal to __________. [2024]

(38)

We have, $\vec{a} = 3 \hat{i} + 2 \hat{j} + \hat{k}$ , $\vec{b} = 2 \hat{i} - \hat{j} + 3 \hat{k}$

Let $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}$

$(\vec{a} + \vec{b}) \times \vec{c} = (3 \hat{i} + 2 \hat{j} + \hat{k} + 2 \hat{i} - \hat{j} + 3 \hat{k}) \times (x \hat{i} + y \hat{j} + z \hat{k})$

$= (5 \hat{i} + \hat{j} + 4 \hat{k}) \times (x \hat{i} + y \hat{j} + z \hat{k})$

$= (z - 4 y) \hat{i} - (5 z - 4 x) \hat{j} + (5 y - x) \hat{k}$

$\vec{a} \times \vec{b} = 7 \hat{i} - 7 \hat{j} - 7 \hat{k}$

$∴ (z - 4 y) \hat{i} - (5 z - 4 x) \hat{j} + (5 y - x) \hat{k} = 2 (7 \hat{i} - 7 \hat{j} - 7 \hat{k}) + 24 \hat{j} - 6 \hat{k}$

On comparing, we get

z – 4y = 14 ... (i), 5z – 4x = –10 ... (ii), 5y – x = –20 ... (iii)

From (i), z = 14 +4y

Now, $(\vec{a} - \vec{b} + \hat{i}) \cdot \vec{c} = - 3$

$\Rightarrow (3 \hat{i} + 2 \hat{j} + \hat{k} - 2 \hat{i} + \hat{j} - 3 \hat{k} + \hat{i}) \cdot (x \hat{i} + y \hat{j} + z \hat{k}) = - 3$

$\Rightarrow (2 \hat{i} + 3 \hat{j} - 2 \hat{k}) \cdot (x \hat{i} + y \hat{j} + z \hat{k}) = - 3$

2x + 3y – 2z = –3

$\Rightarrow 2 x + 3 (\frac{x - 20}{5}) - 2 (\frac{4 x - 10}{5}) = - 3$

$\Rightarrow 10 x + 3 x - 60 - 8 x + 20 = - 15$

$\Rightarrow 5 x - 25 = 0 \Rightarrow x = 5 \Rightarrow y = \frac{5 - 20}{5} = - 3$

z = 14 + 4y = 2

$∴ \vec{c} = 5 \hat{i} - 3 \hat{j} + 2 \hat{k}; | \vec{c} |^{2} = 25 + 9 + 4 = 38$ .

Q 24 :

Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the $△$ BCD is equal to : [2025]

$\sqrt{110}$
$7 \sqrt{3}$
$\sqrt{340}$
12

1

2

3

4

(1)

We have, ar( $△$ ABC) = 5

$\Rightarrow \frac{1}{2} \times b \times c = 5 \Rightarrow b c = 10$ ,

ar( $△$ ACD) = 6

$\Rightarrow \frac{1}{2} \times c \times d = 6 \Rightarrow c d = 12$ ,

$a r (△ A B D) = 7$

$\Rightarrow \frac{1}{2} \times b \times d = 7 \Rightarrow b d = 14$

$\vec{B C} = - b \hat{i} + c \hat{j} and \vec{B D} = - b \hat{i} + d \hat{k}$

$\vec{B C} \times \vec{B D} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - b & c & 0 \\ - b & 0 & d \end{matrix} |$ $= c d \hat{i} + b d \hat{j} + b c \hat{k}$

$| \vec{B C} \times \vec{B D} |$ $= \sqrt{c^{2} d^{2} + b^{2} d^{2} + b^{2} c^{2}}$

$= \sqrt{12^{2} + 14^{2} + 10^{2}} = \sqrt{440}$

$∴ A r e a (△ B C D) = \frac{1}{2}$ $| \overset{—}{B C} \times \overset{—}{B D} |$

$\Rightarrow \frac{1}{2}$ $| \sqrt{440} |$ $= \sqrt{110} sq . units$

Q 25 :

Let $\vec{a} = 2 \hat{i} - 3 \hat{j} + \hat{k}$ , $\vec{b} = 3 \hat{i} + 2 \hat{j} + 5 \hat{k}$ and a vector $\vec{c}$ be such that $(\vec{a} - \vec{c}) \times \vec{b} = - 18 \hat{i} - 3 \hat{j} + 12 \hat{k}$ and $\vec{a} \cdot \vec{c} = 3$ . If $\vec{b} \times \vec{c} = \vec{d}$ , then $| \vec{a} \cdot \vec{d} |$ is equal to : [2025]

12
18
15
9

1

2

3

4

(3)

$\vec{a} \times \vec{b} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & - 3 & 1 \\ 3 & 2 & 5 \end{matrix} |$

$= \hat{i} (- 15 - 2) - \hat{j} (10 - 3) + \hat{k} (4 + 9) = - 17 \hat{i} - 7 \hat{j} + 13 \hat{k}$

Consider, $(\vec{a} \times \vec{b}) - ((\vec{a} - \vec{c}) \times \vec{b})$

$= (\vec{a} \times \vec{b}) - (\vec{a} \times \vec{b} - \vec{c} \times \vec{b}) = - \vec{d}$ [ $∵ \vec{b} \times \vec{c} = \vec{d}$ ]

$∴ \vec{d} = (\vec{a} - \vec{c}) \times \vec{b} - (\vec{a} \times \vec{b})$

$= - 18 \hat{i} - 3 \hat{j} + 12 \hat{k} - (- 17 \hat{i} - 7 \hat{j} + 13 \hat{k}) = - \hat{i} + 4 \hat{j} - \hat{k}$

$∴ \vec{a} \cdot \vec{d} = - 2 - 12 - 1 = - 15 \Rightarrow$ $| \vec{a} \cdot \vec{d} |$ $=$ $| - 15 |$ $= 15$ .

Q 26 :

Let the position vectors of the vertices A, B and C of a tetrahedron ABCD be $\hat{i} + 2 \hat{j} + \hat{k}$ , $\hat{i} + 3 \hat{j} - 2 \hat{k}$ and $2 \hat{i} + \hat{j} - \hat{k}$ respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E. If the length of AD is $\frac{\sqrt{110}}{3}$ and the volume of the tetrahedron is $\frac{\sqrt{805}}{6 \sqrt{2}}$ , then the position vector of E is [2025]

$\frac{1}{2} (\hat{i} + 4 \hat{j} + 7 \hat{k})$
$\frac{1}{6} (7 \hat{i} + 12 \hat{j} + \hat{k})$
$\frac{1}{6} (12 \hat{i} + 12 \hat{j} + \hat{k})$
$\frac{1}{12} (7 \hat{i} + 4 \hat{j} + 3 \hat{k})$

1

2

3

4

(2)

Coordinates of F are $(\frac{3}{2}, 2, \frac{- 3}{2})$ .

Area of $△$ ABC $= \frac{1}{2}$ $| \vec{A B} \times \vec{A C} |$

$= \frac{1}{2}$ $| - 5 \hat{i} - 3 \hat{j} - \hat{k} |$ $= \frac{\sqrt{35}}{2} sq . units$

Volume of Tetrahedron $= \frac{1}{3} \times Base Area \times Height$

$= \frac{1}{3} \times \frac{\sqrt{35}}{2} \times D E$

$∴ \frac{\sqrt{35}}{6} \times D E = \frac{\sqrt{805}}{6 \sqrt{2}} \Rightarrow D E = \sqrt{\frac{23}{2}}$

$∴ In △ A D E, A E = \sqrt{A D^{2} - D E^{2}} = \sqrt{\frac{13}{18}}$ $[∵ A D = \frac{\sqrt{110}}{3}]$

$∴ \vec{A E} =$ $| \vec{A E} |$ $(\frac{\hat{i} - 5 \hat{k}}{\sqrt{26}}) = \sqrt{\frac{13}{18}} (\frac{\hat{i} - 5 \hat{k}}{\sqrt{26}}) = \frac{\hat{i} - 5 \hat{k}}{6}$

$∴$ Position vector of 'E' $= (\frac{i - 5 k}{6}) + (\hat{i} + 2 \hat{j} + \hat{k})$

$= \frac{1}{6} (7 \hat{i} + 12 \hat{j} + \hat{k})$ .

Q 27 :

Let the point A divide the line segment joining the points P(–1, –1, 2) and Q(5, 5, 10) internally in the ratio r : 1 (r > 0). If O is the origin and $(\vec{O Q} \cdot \vec{O A}) - \frac{1}{5} | \vec{O P} \times \vec{O A} |^{2} = 10$ , then the value of r is : [2025]

7
$\sqrt{7}$
3
14

1

2

3

4

(1)

The point A divides line segment PQ internally in the ratio r : 1

$∴ A \equiv (\frac{5 r - 1}{r + 1}, \frac{5 r - 1}{r + 1}, \frac{10 r + 2}{r + 1})$

Now, $\vec{O Q} \cdot \vec{O A} = \frac{10}{r + 1} (15 r + 1)$

and $| \vec{O P} \times \vec{O A} |^{2} = \frac{r^{2}}{{(r + 1)}^{2}} (800)$

$(\vec{O Q} \cdot \vec{O A}) - \frac{| \vec{O P} \times \vec{O A} |^{2}}{5} = 10$

$\Rightarrow \frac{10}{r + 1} (15 r + 1) - \frac{1}{5} \frac{r^{2} (800)}{{(r + 1)}^{2}} = 10$

$\Rightarrow 2 r^{2} - 14 r = 0 \Rightarrow r = 7, r \neq 0$ .

Q 28 :

Let $\vec{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ , $\vec{b} = \vec{a} \times (\hat{i} - 2 \hat{k})$ and $\vec{c} = \vec{b} \times \hat{k}$ . Then the projection of $\vec{c} - 2 \hat{j}$ on $\vec{a}$ is : [2025]

$2 \sqrt{7}$
$3 \sqrt{7}$
$2 \sqrt{14}$
$\sqrt{14}$

1

2

3

4

(3)

Given: $\vec{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}$

and $\vec{b} = \vec{a} \times (\hat{i} - 2 \hat{k}) =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 2 \\ 1 & 0 & - 2 \end{matrix} |$ $= 2 \hat{i} + 8 \hat{j} + \hat{k}$

and $\vec{c} = \hat{b} \times \hat{k} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 8 & 1 \\ 0 & 0 & 1 \end{matrix} |$ $= 8 \hat{i} - 2 \hat{j}$

$\Rightarrow \vec{c} - 2 \hat{j} = 8 \hat{i} - 4 \hat{j}$

Now, Projection of $\vec{c} - 2 \hat{j}$ on $\vec{a}$ is given by

$= (\vec{c} - 2 \hat{j}) \cdot \hat{a} = (8 i - 4 j) \cdot \frac{(3 i - j + 2 k)}{\sqrt{3^{2} + 1^{2} + 2^{2}}}$

$= \frac{24 + 4 + 0}{\sqrt{14}} = \frac{28}{\sqrt{14}} = 2 \sqrt{14}$ .

Q 29 :

Let $\vec{a} = 2 \hat{i} - \hat{j} + 3 \hat{k}$ , $\vec{b} = 3 \hat{i} - 5 \hat{j} + \hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} \times \vec{c} = \vec{c} \times \vec{b}$ and $(\vec{a} + \vec{c}) \cdot (\vec{b} + \vec{c}) = 168$ . Then the maximum value of $| \vec{c} |^{2}$ is : [2025]

462
308
77
154

1

2

3

4

(2)

We have, $\vec{a} \times \vec{c} = \vec{c} \times \vec{b}$

$\Rightarrow \vec{a} \times \vec{c} = - \vec{b} \times \vec{c} \Rightarrow (\vec{a} + \vec{b}) \times \vec{c} = 0$

$\Rightarrow \vec{c} = λ (\vec{a} + \vec{b}) = λ (5 \hat{i} - 6 \hat{j} + 4 \hat{k})$

Also, $(\vec{a} + \vec{c}) \cdot (\vec{b} + \vec{c}) = 168$

$\Rightarrow \vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{c} \cdot \vec{c} = 168$

$\Rightarrow 14 + \vec{c} \cdot (\vec{a} + \vec{b}) + | \vec{c} |^{2} = 168$

$\Rightarrow 14 + λ \cdot 77 + λ^{2} \cdot 77 = 168$

$\Rightarrow 77 λ^{2} + 77 λ - 154 = 0$

$\Rightarrow λ^{2} + λ - 2 = 0 \Rightarrow λ = - 2, 1$

Maximum value of $| \vec{c} |^{2}$ occurs when $λ = - 2$

$| \vec{c} |^{2} = 77 λ^{2} = 77 \times 4 = 308$ .

Q 30 :

Let $\hat{a}$ be a unit vector perpendicular to the vectors $\vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ and $\vec{c} = 2 \hat{i} + 3 \hat{j} - \hat{k}$ , and makes an angle of $\cos^{- 1} (- \frac{1}{3})$ with the vector $\hat{i} + \hat{j} + \hat{k}$ . If $\hat{a}$ makes an angle of $\frac{π}{3}$ with the vector $\hat{i} + α \hat{j} + \hat{k}$ , then the value of $α$ is : [2025]

$\sqrt{3}$
$\sqrt{6}$
$- \sqrt{6}$
$- \sqrt{3}$

1

2

3

4

(3)

Let $\vec{m} = \hat{i} + \hat{j} + \hat{k}$

Since, $\vec{b} \times \vec{c} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 2 & 3 \\ 2 & 3 & - 1 \end{matrix} |$ $= - 7 \hat{i} + 7 \hat{j} + 7 \hat{k} = - 7 (\hat{i} - \hat{j} - \hat{k})$

$∵ \hat{a} parallel to \hat{b} \times \hat{c}$

$∴ \hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}} or \frac{- \hat{i} + \hat{j} + \hat{k}}{\sqrt{3}}$

Now, $\cos θ = \frac{\hat{a} \cdot \vec{m}}{| \vec{m} |} = \frac{1 - 1 - 1}{\sqrt{3} \sqrt{3}} = \frac{- 1}{3}$ ${∵ \cos^{- 1} (\frac{- 1}{3}) = θ}$

or $\cos θ = \frac{- 1 + 1 + 1}{\sqrt{3} \sqrt{3}} = \frac{1}{3}$ (rejected)

Hence, $\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$

Now, $\cos (\frac{π}{3}) = \frac{\hat{a} \cdot (\hat{i} + α \hat{j} + \hat{k})}{\sqrt{1 + α^{2} + 1}} \Rightarrow \frac{1}{2} = \frac{1 - α - 1}{\sqrt{3} \sqrt{α^{2} + 2}}$

$\Rightarrow \sqrt{3} \sqrt{α^{2} + 2} = - 2 α \Rightarrow α = \frac{- \sqrt{3 (α^{2} + 2)}}{2}, i . e ., α < 0$

$\Rightarrow 3 α^{2} + 6 = 4 α^{2} \Rightarrow α = - \sqrt{6}$ .

Topic Question Set

Q 1 :

If A(1, –1, 2), B(5, 7, –6), C(3, 4, –10) and D(–1, –4, –2) are the vertices of a quadrilateral ABCD, then its area is : [2024]

Q 4 :

If A(3, 1, –1), $B (\frac{5}{3}, \frac{7}{3}, \frac{1}{3})$ , C(2, 2, 1) and $D (\frac{10}{3}, \frac{2}{3}, \frac{- 1}{3})$ are the vertices of a quadrilateral ABCD, then its area is [2024]

Q 5 :

Let $\vec{a} = 2 \hat{i} + \hat{j} - \hat{k}$ , $\vec{b} = ((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i}$ . Then the square of the projection of $\vec{a}$ on $\vec{b}$ is : [2024]

Q 7 :

The set of all $α$ , for which the vectors $\vec{a} = α t \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = t \hat{i} - 2 \hat{j} - 2 α t \hat{k}$ are inclined at an obtuse angle for all $t \in R$ , is [2024]

Q 10 :

Let $\vec{O A} = 2 a$ and $\vec{O B} = 6 \vec{a} + 5 \vec{b}$ and $\vec{O C} = 3 \vec{b}$ where O is the origin. If the area of the parallelogram with adjacent sides $\vec{O A}$ and $\vec{O C}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to: [2024]

Q 13 :

Let $\vec{O A} = \vec{a}$ , $\vec{O B} = 12 \vec{a} + 4 \vec{b}$ and $\vec{O C} = b$ , where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then $\frac{area of the quadrilateral O A B C}{area of S}$ is equal to __________. [2024]

Q 14 :

Let A(2, 3, 5) and C(–3, 4, –2) be opposite vertices of a parallelogram ABCD. If the diagonal $\vec{B D} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , then the area of the parallelogram is equal to [2024]

Q 18 :

Let $\vec{a} = \hat{i} - 3 \hat{j} + 7 \hat{k}$ , $\vec{b} = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{c}$ be a vector such that $(\vec{a} + 2 \vec{b}) \times \vec{c} = 3 (\vec{c} \times \vec{a})$ . If $\vec{a} \cdot \vec{c} = 130$ , then $\vec{b} \cdot \vec{c}$ is equal to __________. [2024]

Q 22 :

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $| \vec{a} | = 1, | \vec{b} | = 4$ and $\vec{a} \cdot \vec{b} = 2$ . If $\vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $α$ , then 192 $\sin^{2} α$ is equal to __________. [2024]

Q 24 :

Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the $△$ BCD is equal to : [2025]

Q 27 :

Let the point A divide the line segment joining the points P(–1, –1, 2) and Q(5, 5, 10) internally in the ratio r : 1 (r > 0). If O is the origin and $(\vec{O Q} \cdot \vec{O A}) - \frac{1}{5} | \vec{O P} \times \vec{O A} |^{2} = 10$ , then the value of r is : [2025]

Q 28 :

Let $\vec{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ , $\vec{b} = \vec{a} \times (\hat{i} - 2 \hat{k})$ and $\vec{c} = \vec{b} \times \hat{k}$ . Then the projection of $\vec{c} - 2 \hat{j}$ on $\vec{a}$ is : [2025]

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Topic Question Set

Q 1 : If A(1, –1, 2), B(5, 7, –6), C(3, 4, –10) and D(–1, –4, –2) are the vertices of a quadrilateral ABCD, then its area is : [2024]

Q 2 : Let a→=2i^+5j^–k^, b→=2i^–2j^+2k^ and c→ be three vectors such that (c→+i^)×(a→+b→+i^)=a→×(c→+i^). If a→·c→=–29, then c→·(–2i^+j^+k^) is equal to : [2024]

Q 3 : Consider three vectors a→,b→,c→. Let |a→|=2, |b→|=3 and a→=b→×c→. If α∈[0,π3] is the angle between the vectors b→ and c→, then the minimum value of 27|c→–a→|2 is equal to : [2024]

Q 4 : If A(3, 1, –1), B(53,73,13), C(2, 2, 1) and D(103,23,–13) are the vertices of a quadrilateral ABCD, then its area is [2024]

Q 5 : Let a→=2i^+j^–k^, b→=((a→×(i^+j^))×i^)×i^. Then the square of the projection of a→ on b→ is : [2024]

Q 6 : Let a→=6i^+j^–k^ and b→=i^+j^. If c→ is a vector such that |c→|≥6,a→·c→=6|c→|,|c→–a→|=22 and the angle between a→×b→ and c→ is 60°, then |(a→×b→)×c→| is equal to : [2024]

Q 7 : The set of all α, for which the vectors a→=αti^+6j^–3k^ and b→=ti^–2j^–2αtk^ are inclined at an obtuse angle for all t∈R, is [2024]

Q 8 : Let a→=4i^–j^+k^, b→=11i^–j^+k^ and c→ be a vector that (a→+b→)×c→=c→×(–2a→+3b→). If (2a→+3b→)·c→=1670, then |c→|2 is equal to : [2024]

Q 9 : Let three vectors a→=αi^+4j^+2k^, b→=5i^+3j^+4k^, c→=xi^+yj^+zk^ form a triangle such that c→=a→–b→ and the area of the triangle is 56. If α is a positive real number, then |c→|2 is equal to: [2024]

Q 10 : Let OA→=2a and OB→=6a→+5b→ and OC→=3b→ where O is the origin. If the area of the parallelogram with adjacent sides OA→ and OC→ is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to: [2024]

Q 12 : Between the following two statements: Statement-I: Let a→=i^+2j^–3k^ and b→=2i^+j^–k^. Then the vector r→ satisfying a→×r→=a→×b→ and a→·r→=0 is of magnitude 10. Statement-II: In a triangle ABC, cos 2A + cos 2B + cos 2C ≥–32. [2024]

Q 13 : Let OA→=a→, OB→=12a→+4b→ and OC→=b, where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then area of the quadrilateral OABCarea of S is equal to __________. [2024]

Q 14 : Let A(2, 3, 5) and C(–3, 4, –2) be opposite vertices of a parallelogram ABCD. If the diagonal BD→=i^+2j^+3k^, then the area of the parallelogram is equal to [2024]

Q 15 : Let a→=i^+αj^+βk^, α,β∈R. Let a vector b→ be such that the angle between a→ and b→ is π4 and |b→|2=6. If a→·b→=32, then the value of (α2+β2)|a→×b→|2 is equal to [2024]

Q 16 : Let a→=3i^+j^-2k^, b→=4i^+j^+7k^ and c→=i^–3j^+4k^ be three vectors. If a vectors p→ satisfies p→×b→=c→×b→ and p→·a→=0, then p→·(i^–j^–k^) is equal to [2024]

Q 17 : Let ABC be a triangle of area 152 and the vectors AB→=i^+2j^–7k^, BC→=ai^+bj^+ck^ and AC→=6i^+dj^–2k^, d > 0. Then the square of the length of the largest side of the triangle ABC is __________. [2024]

Q 18 : Let a→=i^–3j^+7k^, b→=2i^–j^+k^ and c→ be a vector such that (a→+2b→)×c→=3(c→×a→). If a→·c→=130, then b→·c→ is equal to __________. [2024]

Q 19 : Let a→=2i^–3j^+4k^, b→=3i^+4j^–5k^ and a vector c→ be such that a→×(b→+c→)+b→×c→=i^+8j^+13k^. If a→·c→=13, then (24–b→·c→) is equal to __________. [2024]

Q 20 : Let a→=9i^–13j^+25k^, b→=3i^+7j^–13k^ and c→=17i^–2j^+k^ be three given vectors. If r→ is a vector such that r→×a→=(b→+c→)×a→ and r→·(b→–c→)=0, then |593r→+67a→|2(593)2 is equal to __________. [2024]

Q 21 : Let a→=i^+j^+k^, b→=–i^–8j^+2k^ and c→=4i^+c2 j^+c3k^ be three vectors such that b→×a→=c→×a→. If the angle between the vector c→ and the vector 3i^+4j^+k^ is θ, then the greatest integer less than or equal to tan2θ is __________. [2024]

Q 22 : Let a→ and b→ be two vectors such that |a→|=1, |b→|=4 and a→·b→=2. If c→=(2a→×b→)–3b→ and the angle between b→ and c→ is α, then 192 sin2α is equal to __________. [2024]

Q 23 : Let a→=3i^+2j^+k^, b→=2i^–j^+3k^ and c→ be a vector such that (a→+b→)×c→=2(a→×b→)+24j^–6k^ and (a→–b→+i^)·c→=–3. Then |c→|2 is equal to __________. [2024]

Q 24 : Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the △BCD is equal to : [2025]

Q 25 : Let a→=2i^–3j^+k^, b→=3i^+2j^+5k^ and a vector c→ be such that (a→–c→)×b→=–18i^–3j^+12k^ and a→·c→=3. If b→×c→=d→, then |a→·d→| is equal to : [2025]

Q 27 : Let the point A divide the line segment joining the points P(–1, –1, 2) and Q(5, 5, 10) internally in the ratio r : 1 (r > 0). If O is the origin and (OQ→·OA→)–15|OP→×OA→|2=10, then the value of r is : [2025]

Q 28 : Let a→=3i^–j^+2k^, b→=a→×(i^–2k^) and c→=b→×k^. Then the projection of c→–2j^ on a→ is : [2025]

Q 29 : Let a→=2i^–j^+3k^, b→=3i^–5j^+k^ and c→ be a vector such that a→×c→=c→×b→ and (a→+c→)·(b→+c→)=168. Then the maximum value of |c→|2 is : [2025]

Q 30 : Let a^ be a unit vector perpendicular to the vectors b→=i^–2j^+3k^ and c→=2i^+3j^–k^, and makes an angle of cos–1(–13) with the vector i^+j^+k^. If a^ makes an angle of π3 with the vector i^+αj^+k^, then the value of α is : [2025]

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Q 1 :

If A(1, –1, 2), B(5, 7, –6), C(3, 4, –10) and D(–1, –4, –2) are the vertices of a quadrilateral ABCD, then its area is : [2024]

Q 4 :

If A(3, 1, –1), $B (\frac{5}{3}, \frac{7}{3}, \frac{1}{3})$ , C(2, 2, 1) and $D (\frac{10}{3}, \frac{2}{3}, \frac{- 1}{3})$ are the vertices of a quadrilateral ABCD, then its area is [2024]

Q 5 :

Let $\vec{a} = 2 \hat{i} + \hat{j} - \hat{k}$ , $\vec{b} = ((\vec{a} \times (\hat{i} + \hat{j})) \times \hat{i}) \times \hat{i}$ . Then the square of the projection of $\vec{a}$ on $\vec{b}$ is : [2024]

Q 7 :

The set of all $α$ , for which the vectors $\vec{a} = α t \hat{i} + 6 \hat{j} - 3 \hat{k}$ and $\vec{b} = t \hat{i} - 2 \hat{j} - 2 α t \hat{k}$ are inclined at an obtuse angle for all $t \in R$ , is [2024]

Q 10 :

Let $\vec{O A} = 2 a$ and $\vec{O B} = 6 \vec{a} + 5 \vec{b}$ and $\vec{O C} = 3 \vec{b}$ where O is the origin. If the area of the parallelogram with adjacent sides $\vec{O A}$ and $\vec{O C}$ is 15 sq. units, then the area (in sq. units) of the quadrilateral OABC is equal to: [2024]

Q 13 :

Let $\vec{O A} = \vec{a}$ , $\vec{O B} = 12 \vec{a} + 4 \vec{b}$ and $\vec{O C} = b$ , where O is the origin. If S is the parallelogram with adjacent sides OA and OC, then $\frac{area of the quadrilateral O A B C}{area of S}$ is equal to __________. [2024]

Q 14 :

Let A(2, 3, 5) and C(–3, 4, –2) be opposite vertices of a parallelogram ABCD. If the diagonal $\vec{B D} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , then the area of the parallelogram is equal to [2024]

Q 22 :

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $| \vec{a} | = 1, | \vec{b} | = 4$ and $\vec{a} \cdot \vec{b} = 2$ . If $\vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b}$ and the angle between $\vec{b}$ and $\vec{c}$ is $α$ , then 192 $\sin^{2} α$ is equal to __________. [2024]

Q 24 :

Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the $△$ BCD is equal to : [2025]

Q 27 :

Let the point A divide the line segment joining the points P(–1, –1, 2) and Q(5, 5, 10) internally in the ratio r : 1 (r > 0). If O is the origin and $(\vec{O Q} \cdot \vec{O A}) - \frac{1}{5} | \vec{O P} \times \vec{O A} |^{2} = 10$ , then the value of r is : [2025]

Q 28 :

Let $\vec{a} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ , $\vec{b} = \vec{a} \times (\hat{i} - 2 \hat{k})$ and $\vec{c} = \vec{b} \times \hat{k}$ . Then the projection of $\vec{c} - 2 \hat{j}$ on $\vec{a}$ is : [2025]