JEE Previous Year Question Bank on Scalar and Vector Triple Product for free

Q 1 :

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{c} = x \hat{i} + 2 \hat{j} + 3 \hat{k}, x \in R$ . If $\vec{d}$ is the unit vector in the direction of $\vec{b} + \vec{c}$ such that $\vec{a} \cdot \vec{d} = 1$ , then $(\vec{a} \times \vec{b}) \cdot \vec{c}$ is equal to [2024]

11
3
6
9

1

2

3

4

(1)

$\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and $\vec{c} = x \hat{i} + 2 \hat{j} + 3 \hat{k}, x \in R$

$\vec{b} + \vec{c} = (x + 2) \hat{i} + 6 \hat{j} - 2 \hat{k} a n d$ $| (\vec{b} + \vec{c}) |$

$= \sqrt{{(x + 2)}^{2} + {(6)}^{2} + {(- 2)}^{2}} = \sqrt{{(x + 2)}^{2} + 40}$

Now, $\vec{d} = \frac{x + 2}{\sqrt{40 + {(x + 2)}^{2}}} \hat{i} + \frac{6}{\sqrt{40 + {(x + 2)}^{2}}} \hat{j} - \frac{2}{\sqrt{40 + {(x + 2)}^{2}}} \hat{k}$

Given that $\vec{a} \cdot \vec{d} = 1$

$\Rightarrow \frac{x + 2 + 6 - 2}{\sqrt{40 + {(x + 2)}^{2}}} = 1$

$\Rightarrow x + 6 = \sqrt{40 + {(x + 2)}^{2}}$

$\Rightarrow {(x + 6)}^{2} = 40 + {(x + 2)}^{2}$

$\Rightarrow x^{2} + 36 + 12 x = 40 + x^{2} + 4 + 4 x$

$\Rightarrow x = 1$

Now, $(\vec{a} \times \vec{b})$ $\cdot \vec{c} =$ $[\vec{a} \vec{b} \vec{c}]$ $=$ $| \begin{matrix} 1 & 1 & 1 \\ 2 & 4 & - 5 \\ 1 & 2 & 3 \end{matrix} |$

= 1(12 + 10) – 1(6 + 5) + 1(4 – 4) = 11.

Q 2 :

Let $\vec{a} = - 5 \hat{i} + \hat{j} - 3 \hat{k}$ , $\vec{b} = \hat{i} + 2 \hat{j} - 4 \hat{k}$ and $\vec{c} = (((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i}$ . Then $\vec{c} \cdot (- \hat{i} + \hat{j} + \hat{k})$ is equal to: [2024]

–12
–15
–13
–10

1

2

3

4

(1)

We have, $\vec{a} = - 5 \hat{i} + \hat{j} - 3 \hat{k}$ , $\vec{b} = \hat{i} + 2 \hat{j} - 4 \hat{k}$

$∴ (\vec{a} \times \vec{b}) \times \hat{i} = (\vec{a} \cdot \hat{i}) \vec{b} - (\vec{b} \cdot \hat{i}) \vec{a} = - 5 \vec{b} - \vec{a}$

Now, $((- 5 \vec{b} - \vec{a}) \times \hat{i}) \times \hat{i} = (11 \hat{k} - 23 \hat{j}) \times \hat{i} = 11 \hat{j} - 23 \hat{k}$

$∴ \vec{c} \cdot (- \hat{i} + \hat{j} + \hat{k}) = 11 - 23 = - 12$ .

Q 3 :

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

36
20
24
32

1

2

3

4

(3)

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

[Using distributive property of multiplication]

$= [\vec{a} \vec{c} \vec{b}] - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$ ... (i)

Now, $\vec{a} \times \vec{c} = \vec{b} \Rightarrow (\vec{a} \times \vec{c}) \cdot \vec{b} = \vec{b} \cdot \vec{b}$

$\Rightarrow [\vec{a} \vec{c} \vec{b}] = | \vec{b} |^{2} = {(\sqrt{{(3)}^{2} + {(- 3)}^{2} + {(3)}^{2}})}^{2} = 27$

$\Rightarrow [\vec{a} \vec{c} \vec{b}] = 27$ ... (ii)

$\vec{a} \cdot \vec{b} = 3 (\hat{i} + 2 \hat{j} + \hat{k}) (\hat{i} - \hat{j} + \hat{k}) = 3 (1 - 2 + 1) = 0$ (iii)

and $\vec{a} \cdot \vec{c} = 3$ ... (iv)

Using (ii), (iii) and (iv) in (i), we get

$\vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} - \vec{c}) = 27 - 0 - 3 = 24$

Q 4 :

Let $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ be two vectors such that, $| \vec{a} | = 1$ , $\vec{a} \cdot \vec{b} = 2$ and $| \vec{b} | = 4$ . If $\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b}$ then the angle between $\vec{b}$ and $\vec{c}$ is equal to [2024]

$\cos^{- 1} (- \frac{\sqrt{3}}{2})$
$\cos^{- 1} (\frac{2}{\sqrt{3}})$
$\cos^{- 1} (- \frac{1}{\sqrt{3}})$
$\cos^{- 1} (\frac{2}{3})$

1

2

3

4

(1)

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

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

$\Rightarrow \vec{b} \cdot \vec{c} = 2 \vec{b} \cdot (\vec{a} \times \vec{b}) - 3 \vec{b} \cdot \vec{b} = 2 (0) - 3 | \vec{b} |^{2}$

$\Rightarrow \vec{b} \cdot \vec{c} = - 48 \Rightarrow | \vec{b} | | \vec{c} | \cos θ = - 48$

$\Rightarrow | \vec{c} | \cos θ = - 12$ ... (i)

Consider, $\vec{c} = 2 (\vec{a} \times \vec{b}) - 3 \vec{b} \Rightarrow | \vec{c} |^{2} = 4 | \vec{a} \times \vec{b} |^{2} + 9 | \vec{b} |^{2}$

$\Rightarrow | \vec{c} |^{2} = 4 (| \vec{a} |^{2} | \vec{b} |^{2} - {(\vec{a} \cdot \vec{b})}^{2}) + 9 | \vec{b} |^{2}$

$\Rightarrow | \vec{c} |^{2} = 4 (16 - 4) + 144 = 48 + 144$

$\Rightarrow | \vec{c} |^{2} = 192 \Rightarrow \vec{c} = 8 \sqrt{3}$

From (i), $\cos θ = \frac{- 12}{8 \sqrt{3}}$ [Using (i)]

$\Rightarrow \cos θ = \frac{- 3}{2 \sqrt{3}} \Rightarrow \cos θ = \frac{- \sqrt{3}}{2} \Rightarrow θ = \cos^{- 1} (- \frac{\sqrt{3}}{2})$

Q 5 :

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $| \vec{b} |$ $= 1$ and $| \vec{b} \times \vec{a} |$ $= 2$ . Then $| (\vec{b} \times \vec{a}) - \vec{b} |^{2}$ is equal to [2024]

5
1
4
3

1

2

3

4

(1)

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

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

$= {(2)}^{2} + 1 - 2 (0) = 5$ .

Q 6 :

Let $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $\vec{b} = 3 \hat{i} + \hat{j} - \hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$ . If the vector $\vec{c}$ is perpendicular to $\vec{b}$ and $\vec{a} \cdot \vec{c} = 5$ , then $| \vec{c} |$ is equal to [2025]

$\sqrt{\frac{11}{6}}$
16
$\frac{1}{3 \sqrt{2}}$
18

1

2

3

4

(1)

$| \vec{b} | = \sqrt{9 + 1 + 1} = \sqrt{11}, \vec{a} \cdot \vec{b} = 3 + 2 - 3 = 2$

As $\vec{c}$ is perpendicular to $\vec{b}$ and coplanar with $\vec{a}$ and $\vec{b}$ .

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

$= λ ((\vec{b} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b}) = λ (11 \vec{a} - 2 \vec{b})$

$\Rightarrow \vec{c} = λ (11 \hat{i} + 22 \hat{j} + 33 \hat{k} - 6 \hat{i} - 2 \hat{j} + 2 \hat{k}) = 5 λ (\hat{i} + 4 \hat{j} + 7 \hat{k})$

As $\vec{a} \cdot \vec{c} = 5 \Rightarrow 5 λ (1 + 8 + 21) = 5$

$\Rightarrow λ = \frac{1}{30} ∴ \vec{c} = \frac{1}{6} (\hat{i} + 4 \hat{j} + 7 \hat{k})$

$\Rightarrow$ $| \vec{c} |$ $= \frac{1}{6} \sqrt{1 + 16 + 49} = \sqrt{\frac{11}{6}}$

Q 7 :

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{d} = \vec{a} \times \vec{b}$ . If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} =$ $| \vec{c} |$ , $| \vec{c} - 2 \vec{a} |^{2} = 8$ and the angle between $\vec{d}$ and $\vec{c}$ is $\frac{π}{4}$ , then $| 10 - 3 \vec{b} \cdot \vec{c} |$ $+$ $| \vec{d} \times \vec{c} |^{2}$ is equal to __________. [2025]

0%

(6)

Given, $\vec{d} = \vec{a} \times \vec{b}$ , where $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2 \hat{i} + 2 \hat{j} + \hat{k}$

$∴ \vec{d} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 1 \end{matrix} |$ $= - \hat{i} + \hat{j}$ ... (i)

Also, $| \vec{c} - 2 \vec{a} |^{2} = 8 \Rightarrow | \vec{c} |^{2} + 4 | \vec{a} |^{2} - 4 \vec{a} \cdot \vec{c} = 8$

$\Rightarrow | \vec{c} |^{2} + 4 (3) - 4 | \vec{c} | = 8$ [Given, $\vec{a} \cdot \vec{c} =$ $| \vec{c} |$ ]

$\Rightarrow | \vec{c} |^{2} - 4 | \vec{c} | + 4 = 0 \Rightarrow {(| \vec{c} | - 2)}^{2} = 0 \Rightarrow | \vec{c} | = 2$

Now, $\vec{d} = \vec{a} \times \vec{b} \Rightarrow$ $| \vec{d} \times \vec{c} |$ $=$ $| (\vec{a} \times \vec{b}) \times \vec{c} |$

Using triple vector product properties, we have

$| \vec{d} |$ $| \vec{c} |$ $\sin \frac{π}{4} =$ $| ((\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a}) |$

$\sqrt{2} \times 2 \times \frac{1}{\sqrt{2}}$ $| 2 \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a} |$

Squaring both sides, we get

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

$\Rightarrow 4 = 4 (9) + 3 {(\vec{b} \cdot \vec{c})}^{2} - 4 \times 5 (\vec{b} \cdot \vec{c})$

$\Rightarrow 3 {(\vec{b} \cdot \vec{c})}^{2} - 20 {(\vec{b} \cdot \vec{c})}^{2} + 32 = 0$

Let $\vec{b} \cdot \vec{c} = t$ , we have

$3 t^{2} - 20 t + 32 = 0 \Rightarrow t = \frac{8}{3}, 4 \Rightarrow \vec{b} \cdot \vec{c} = \frac{8}{3} or \vec{b} \cdot \vec{c} = 4$

$∴$ $| 10 - 3 \vec{b} \cdot \vec{c} |$ $+ | \vec{d} \times \vec{c} |^{2} =$ $| 10 - 3 (\frac{8}{3}) |$ $+ {(2)}^{2} = 6$

or $| 10 - 3 (4) |$ $+ {(2)}^{2} = 6$ .

Q 8 :

Let the position vectors of the points A, B, C and D be $5 \hat{i} + 5 \hat{j} + 2 λ \hat{k}, \hat{i} + 2 \hat{j} + 3 \hat{k}, - 2 \hat{i} + λ \hat{j} + 4 \hat{k}$ and $- \hat{i} + 5 \hat{j} + 6 \hat{k}$ . Let the set $S =$ { $λ \in ℝ :$ the points A, B, C and D are coplanar}. Then $\sum_{λ \in S} {(λ + 2)}^{2}$ is equal to [2023]

$\frac{37}{2}$
$25$
$13$
$41$

1

2

3

4

(4)

We have, $\vec{A B} = - 4 \hat{i} - 3 \hat{j} + (3 - 2 λ) \hat{k}$

$\vec{A C} = - 7 \hat{i} + (λ - 5) \hat{j} + (4 - 2 λ) \hat{k}$

$\vec{A D} = - 6 \hat{i} + (6 - 2 λ) \hat{k}$

Since A, B, C and D are coplanar

$∴$ $| \begin{matrix} - 4 & - 3 & 3 - 2 λ \\ - 7 & λ - 5 & 4 - 2 λ \\ - 6 & 0 & 6 - 2 λ \end{matrix} |$ $= 0$

$\Rightarrow - 4 [(λ - 5) (6 - 2 λ) - 0] + 3 [- 7 (6 - 2 λ) + 6 (4 - 2 λ)]$ $+ (3 - 2 λ) [0 + 6 (λ - 5)] = 0$

$\Rightarrow - 4 (16 λ - 2 λ^{2} - 30) + 3 (2 λ - 18) + (3 - 2 λ) (6 λ - 30) = 0$

$\Rightarrow λ^{2} - 5 λ + 6 = 0 \Rightarrow λ = 2 or λ = 3$ $∴$ $S = {2, 3}$

So, $\sum_{λ \in S} {(λ + 2)}^{2} = 4^{2} + 5^{2} = 41$

Q 9 :

Let the vectors $\vec{a}, \vec{b}, \vec{c}$ represent three coterminous edges of a parallelepiped of volume V. Then the volume of the parallelepiped whose coterminous edges are represented by $\vec{a}, \vec{b} + \vec{c}$ and $\vec{a} + 2 \vec{b} + 3 \vec{c}$ is equal to [2023]

$V$
$3 V$
$2 V$
$6 V$

1

2

3

4

(1)

We have,

$Volume of parallelepiped = [\vec{a} \vec{b} \vec{c}] = V$ ...(i)

New volume of parallelepiped $= [\vec{a} \vec{b} + \vec{c} \vec{a} + 2 \vec{b} + 3 \vec{c}]$

$=$ $| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 2 & 3 \end{matrix} |$ $[\vec{a} \vec{b} \vec{c}] = {1 (3 - 2) - 0 + 0} [\vec{a} \vec{b} \vec{c}]$

$= [\vec{a} \vec{b} \vec{c}] = V$ $(Using (i))$

Q 10 :

The sum of all values of $α,$ for which the points whose position vectors are $\hat{i} - 2 \hat{j} + 3 \hat{k},$ $2 \hat{i} - 3 \hat{j} + 4 \hat{k},$ $(α + 1) \hat{i} + 2 \hat{k}$ and $9 \hat{i} + (α - 8) \hat{j} + 6 \hat{k}$ are coplanar, is equal to [2023]

6
- 2
2
4

1

2

3

4

(3)

Let

$\vec{O A} = \hat{i} - 2 \hat{j} + 3 \hat{k},$ $\vec{O B} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k},$ $\vec{O C} = (α + 1) \hat{i} + 2 \hat{k},$

$\vec{O D} = 9 \hat{i} + (α - 8) \hat{j} + 6 \hat{k}$

$∴$ $\vec{A B} = \hat{i} - \hat{j} + \hat{k},$ $\vec{A C} = α \hat{i} + 2 \hat{j} - \hat{k}$

$\vec{A D} = 8 \hat{i} + (α - 6) \hat{j} + 3 \hat{k}$

Now, $(\vec{A B} \times \vec{A C}) \cdot \vec{A D} = 0$ $\Rightarrow$ $| \begin{matrix} 1 & - 1 & 1 \\ α & 2 & - 1 \\ 8 & α - 6 & 3 \end{matrix} |$ $= 0$

$\Rightarrow 1 (6 + α - 6) + 1 (3 α + 8) + 1 (α^{2} - 6 α - 16) = 0$

$\Rightarrow α^{2} - 2 α - 8 = 0 \Rightarrow (α - 4) (α + 2) = 0$

$\Rightarrow$ $α = 4 or α = - 2$

$∴$ Required sum $= 4 - 2 = 2$

Topic Question Set

Q 2 :

Let $\vec{a} = - 5 \hat{i} + \hat{j} - 3 \hat{k}$ , $\vec{b} = \hat{i} + 2 \hat{j} - 4 \hat{k}$ and $\vec{c} = (((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i}$ . Then $\vec{c} \cdot (- \hat{i} + \hat{j} + \hat{k})$ is equal to: [2024]

Q 3 :

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

Q 5 :

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $| \vec{b} |$ $= 1$ and $| \vec{b} \times \vec{a} |$ $= 2$ . Then $| (\vec{b} \times \vec{a}) - \vec{b} |^{2}$ is equal to [2024]

0%

Q 9 :

Let the vectors $\vec{a}, \vec{b}, \vec{c}$ represent three coterminous edges of a parallelepiped of volume V. Then the volume of the parallelepiped whose coterminous edges are represented by $\vec{a}, \vec{b} + \vec{c}$ and $\vec{a} + 2 \vec{b} + 3 \vec{c}$ is equal to [2023]

Q 10 :

The sum of all values of $α,$ for which the points whose position vectors are $\hat{i} - 2 \hat{j} + 3 \hat{k},$ $2 \hat{i} - 3 \hat{j} + 4 \hat{k},$ $(α + 1) \hat{i} + 2 \hat{k}$ and $9 \hat{i} + (α - 8) \hat{j} + 6 \hat{k}$ are coplanar, is equal to [2023]

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Topic Question Set

Q 1 : Let a→=i^+j^+k^, b→=2i^+4j^–5k^ and c→=xi^+2j^+3k^, x∈R. If d→ is the unit vector in the direction of b→+c→ such that a→·d→=1, then (a→×b→)·c→ is equal to [2024]

Q 2 : Let a→=–5i^+j^–3k^, b→=i^+2j^–4k^ and c→=(((a→×b→)×i^)×i^)×i^. Then c→·(–i^+j^+k^) is equal to: [2024]

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

Q 4 : Let a→=a1i^+a2j^+a3k^ and b→=b1i^+b2j^+b3k^ be two vectors such that, |a→|=1, a→·b→=2 and |b→|=4. If c→=2(a→×b→)–3b→ then the angle between b→ and c→ is equal to [2024]

Q 5 : Let a→ and b→ be two vectors such that |b→|=1 and |b→×a→|=2. Then |(b→×a→)–b→|2 is equal to [2024]

Q 6 : Let a→=i^+2j^+3k^, b→=3i^+j^–k^ and c→ be three vectors such that c→ is coplanar with a→ and b→. If the vector c→ is perpendicular to b→ and a→·c→=5, then |c→| is equal to [2025]

Q 7 : Let a→=i^+j^+k^, b→=2i^+2j^+k^ and d→=a→×b→. If c→ is a vector such that a→·c→=|c→|, |c→–2a→|2=8 and the angle between d→ and c→ is π4, then |10–3b→·c→|+|d→×c→|2 is equal to __________. [2025]

0%

Q 8 : Let the position vectors of the points A, B, C and D be 5i^+5j^+2λk^, i^+2j^+3k^,-2i^+λj^+4k^ and -i^+5j^+6k^. Let the set S={λ∈ℝ: the points A, B, C and D are coplanar}. Then ∑λ∈S(λ+2)2 is equal to [2023]

Q 9 : Let the vectors a→,b→,c→ represent three coterminous edges of a parallelepiped of volume V. Then the volume of the parallelepiped whose coterminous edges are represented by a→,b→+c→ and a→+2b→+3c→ is equal to [2023]

Q 10 : The sum of all values of α, for which the points whose position vectors are i^-2j^+3k^, 2i^-3j^+4k^, (α+1)i^+2k^ and 9i^+(α-8)j^+6k^ are coplanar, is equal to [2023]

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Q 2 :

Let $\vec{a} = - 5 \hat{i} + \hat{j} - 3 \hat{k}$ , $\vec{b} = \hat{i} + 2 \hat{j} - 4 \hat{k}$ and $\vec{c} = (((\vec{a} \times \vec{b}) \times \hat{i}) \times \hat{i}) \times \hat{i}$ . Then $\vec{c} \cdot (- \hat{i} + \hat{j} + \hat{k})$ is equal to: [2024]

Q 3 :

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

Q 5 :

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $| \vec{b} |$ $= 1$ and $| \vec{b} \times \vec{a} |$ $= 2$ . Then $| (\vec{b} \times \vec{a}) - \vec{b} |^{2}$ is equal to [2024]

Q 9 :

Let the vectors $\vec{a}, \vec{b}, \vec{c}$ represent three coterminous edges of a parallelepiped of volume V. Then the volume of the parallelepiped whose coterminous edges are represented by $\vec{a}, \vec{b} + \vec{c}$ and $\vec{a} + 2 \vec{b} + 3 \vec{c}$ is equal to [2023]

Q 10 :

The sum of all values of $α,$ for which the points whose position vectors are $\hat{i} - 2 \hat{j} + 3 \hat{k},$ $2 \hat{i} - 3 \hat{j} + 4 \hat{k},$ $(α + 1) \hat{i} + 2 \hat{k}$ and $9 \hat{i} + (α - 8) \hat{j} + 6 \hat{k}$ are coplanar, is equal to [2023]