JEE Previous Year Question Bank on Vector Algebra for free Scalar or Dot Product

Q 1 :

Let a unit vector which makes an angle of 60° with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and an angle of 45° with $\hat{i} - \hat{k}$ be $\vec{C}$ . Then $\vec{C} + (- \frac{1}{2} \hat{i} + \frac{1}{3 \sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k})$ is [2024]

$\frac{\sqrt{2}}{3} \hat{i} + \frac{1}{3 \sqrt{2}} \hat{j} - \frac{1}{2} \hat{k}$
$- \frac{\sqrt{2}}{3} \hat{i} + \frac{\sqrt{2}}{3} \hat{j} + (\frac{1}{2} + \frac{2 \sqrt{2}}{3}) \hat{k}$
$\frac{\sqrt{2}}{3} \hat{i} - \frac{1}{2} \hat{k}$
$(\frac{1}{\sqrt{3}} + \frac{1}{2}) \hat{i} + (\frac{1}{\sqrt{3}} - \frac{1}{3 \sqrt{2}}) \hat{j} + (\frac{1}{\sqrt{3}} + \frac{\sqrt{2}}{3}) \hat{k}$

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(3)

Let $\vec{C} = a \hat{i} + b \hat{j} + c \hat{k}$ , then we have $| \vec{C} |$ $= 1$

$\Rightarrow a^{2} + b^{2} + c^{2} = 1$ ... (i)

Angle between $\vec{C}$ and $2 \hat{i} + 2 \hat{j} - \hat{k}$ is 60°.

$∴ (a \hat{i} + b \hat{j} + c \hat{k}) \cdot (2 \hat{i} + 2 \hat{j} - \hat{k}) = (\sqrt{4 + 4 + 1}) \cos 60 °$ $(∵ a^{2} + b^{2} + c^{2} = 1)$

$\Rightarrow 2 a + 2 b - c = \frac{3}{2}$ ... (ii)

Angle between $\vec{C}$ and $\hat{i} - \hat{k}$ is 45°.

$∴ (a \hat{i} + b \hat{j} + c \hat{k}) \cdot (\hat{i} - \hat{k}) = \sqrt{1 + 0 + 1} \cos 45 °$

$\Rightarrow a - c = 1$ ... (iii)

Solving (i), (ii) and (iii), we get $a + 2 b = \frac{1}{2}$ and $a^{2} + b^{2} + {(a - 1)}^{2} = 1$

$\Rightarrow 2 a^{2} - 2 a + b^{2} = 0 \Rightarrow 2 a^{2} - 2 a + {(\frac{2 a - 1}{4})}^{2} = 0$

$\Rightarrow 32 a^{2} - 32 a + 4 a^{2} - 4 a + 1 = 0 \Rightarrow 36 a^{2} - 36 a + 1 = 0$

$\Rightarrow a = \frac{36 \pm \sqrt{{(36)}^{2} - 4 (36)}}{2 \times 36} = \frac{1}{2} \pm \frac{\sqrt{2}}{3}$

$∴ b = \frac{1 - 2 a}{4} \Rightarrow b = \frac{1 - 1 \mp \frac{2 \sqrt{2}}{3}}{4} = \mp \frac{1}{3 \sqrt{2}}$

$∴ c = - \frac{1}{2} \pm \frac{\sqrt{2}}{3}$

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

Q 2 :

For $λ > 0$ , let $θ$ be the angle between the vectors $\vec{a} = \hat{i} + λ \hat{j} - 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$ . If the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are mutually perpendicular, then the value of ${(14 \cos θ)}^{2}$ is equal to [2024]

20
25
50
40

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(2)

$\vec{a} + \vec{b} = 4 \hat{i} + (λ - 1) \hat{j} - \hat{k}$ and $\vec{a} - \vec{b} = - 2 \hat{i} + (λ + 1) \hat{j} - 5 \hat{k}$

Now, given that $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are mutually perpendicular.

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

$\Rightarrow - 8 + λ^{2} - 1 + 5 = 0 \Rightarrow λ = \pm 2$

$∴ λ = 2 (∵ λ > 0)$

Now, $\cos θ \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | \cdot | \vec{b} |} = \frac{(\hat{i} + 2 \hat{j} - 3 \hat{k}) \cdot (3 \hat{i} - \hat{j} + 2 \hat{k})}{\sqrt{{(1)}^{2} + {(2)}^{2} + {(3)}^{2}} \sqrt{{(3)}^{2} + {(- 1)}^{2} + {(2)}^{2}}}$

$\Rightarrow \cos θ = \frac{- 5}{14}$

$∴ {(14 \cos θ)}^{2} = {(14 \times \frac{- 5}{14})}^{2} = 25$ .

Q 3 :

Let P(x, y, z) be a point in the first octant, whose projection in the xy-plane is the point Q. Let OP = $γ$ ; the angle between OQ and the positive x-axis be $θ$ ; and the angle between OP and the positive z-axis be $ϕ$ , where O is the origin. Then the distance of P from the x-axis is [2024]

$γ \sqrt{1 - \sin^{2} θ \cos^{2} ϕ}$
$γ \sqrt{1 - \sin^{2} ϕ \cos^{2} θ}$
$γ \sqrt{1 + \cos^{2} θ \sin^{2} ϕ}$
$γ \sqrt{1 + \cos^{2} ϕ \sin^{2} θ}$

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(2)

Projection of P in xy-plane is given by Q(x, y, 0)

Now, OP = $γ$

$\Rightarrow \sqrt{x^{2} + y^{2} + z^{2}} = γ \Rightarrow x^{2} + y^{2} + z^{2} = γ^{2}$ ... (i)

Now, angle between OQ and positive x-axis is $θ$ .

$∴ \frac{(x \hat{i} + y \hat{j}) \cdot (\hat{i})}{\sqrt{x^{2} + y^{2}}} = \cos θ \Rightarrow \cos θ = \frac{x}{\sqrt{x^{2} + y^{2}}}$ ... (ii)

Similarly, angle between OP and positive z-axis is $ϕ$ .

$\Rightarrow \cos ϕ = \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \Rightarrow \sin^{2} ϕ = \frac{x^{2} + y^{2}}{x^{2} + y^{2} + z^{2}}$ ... (iii)

Now, distance of P from x-axis $= \sqrt{y^{2} + z^{2}}$

$= \sqrt{γ^{2} - x^{2}}$ [Using (i)]

$= \sqrt{γ^{2} - \cos^{2} θ (x^{2} + y^{2})}$ [Using (ii)]

$= \sqrt{γ^{2} - \cos^{2} θ \sin^{2} ϕ \cdot γ^{2}}$ [Using (iii)]

$= γ \sqrt{1 - \cos^{2} θ \sin^{2} ϕ}$ .

Q 4 :

Let $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $\vec{b} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ and $\vec{c} = 3 \hat{i} - \hat{j} + λ \hat{k}$ be three vectors. Let $\vec{r}$ be a unit vector along $\vec{b} + \vec{c}$ . If $\vec{r} \cdot \vec{a} = 3$ , then $3 λ$ is equal to : [2024]

21
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25
27

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(3)

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

Now, $\vec{b} + \vec{c} = 5 \hat{i} + 2 \hat{j} + (λ - 5) \hat{k}$

$\vec{r}$ is a unit vector along $\vec{b} + \vec{c} \Rightarrow \vec{r} = \frac{\vec{b} + \vec{c}}{| \vec{b} + \vec{c} |}$

$∴ \vec{r} = \frac{5 \hat{i} + 2 \hat{j} + (λ - 5) \hat{k}}{\sqrt{25 + 4 + {(λ - 5)}^{2}}}$

Now, $\vec{r} \cdot \vec{a} = 3$

$\frac{1}{\sqrt{29 + {(λ - 5)}^{2}}} [5 + 4 + 3 (λ - 5)] = 3$

$\Rightarrow \frac{1}{29 + {(λ - 5)}^{2}} {[9 + 3 (λ - 5)]}^{2} = 9$

$\Rightarrow {[3 + (λ - 5)]}^{2} = 29 + {(λ - 5)}^{2}$

$\Rightarrow 9 + {(λ - 5)}^{2} + 6 (λ - 5) = 29 + {(λ - 5)}^{2}$

$\Rightarrow 9 + 6 (λ - 5) = 29 \Rightarrow λ = \frac{25}{3}$

Hence, $3 λ = 3 \times \frac{25}{3} = 25$ .

Q 5 :

Consider a $△ A B C$ where A(1, 3, 2), B(–2, 8, 0) and C(3, 6, 7). If the angle bisector of $∠ B A C$ meets the line BC at D, then the length of the projection of the vector $\vec{A D}$ on the vector $\vec{A C}$ IS : [2024]

$\frac{37}{2 \sqrt{38}}$
$\sqrt{19}$
$\frac{39}{2 \sqrt{38}}$
$\frac{\sqrt{38}}{2}$

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(1)

We have, $A B = \sqrt{3^{2} + 5^{2} + 2^{2}} = \sqrt{38}$

$A C = \sqrt{2^{2} + 3^{2} + 5^{2}} = \sqrt{38}$

$∴ \frac{B D}{D C} = \frac{A B}{A C} = 1 : 1$

i.e., D is the mid point of BC.

So. coordinates of $D = (\frac{1}{2}, 7, \frac{7}{2})$

Now, $\vec{A D} = \frac{- 1}{2} \hat{i} + 4 \hat{j} + \frac{3}{2} \hat{k}$ and $\vec{A C} = 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$

$∴$ Projection of $\vec{A D}$ on $\vec{A C} = \frac{\vec{A D} \cdot \vec{A C}}{| \vec{A C} |}$

$= \frac{- 1 + 12 + 15 / 2}{\sqrt{4 + 9 + 25}} = \frac{37}{2 \sqrt{38}}$ .

Q 6 :

Let a unit vector $\hat{u} = x \hat{i} + y \hat{j} + z \hat{k}$ make angles $\frac{π}{2}, \frac{π}{3}$ and $\frac{2 π}{3}$ with the vectors $\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k}$ , $\frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$ and $\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}$ respectively. If $\vec{v} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$ , then $| \hat{u} - \vec{v} |^{2}$ is equal to [2024]

9
$\frac{11}{2}$
7
$\frac{5}{2}$

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(4)

Let $\vec{a} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k}$ , $\vec{b} = \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{\sqrt{2}} \hat{k}$ and $\vec{c} = \frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j}$

Now, $\hat{u} \cdot \vec{a} = | \hat{u} | \cdot | \vec{a} | \cos θ$ ,

$\Rightarrow \frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 1 \cdot 1 \cdot \cos \frac{π}{2} \Rightarrow x + z = 0$ ... (i)

Again $\hat{u} \cdot \vec{b} = \frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 1 \cdot 1 \cdot \cos \frac{π}{3} \Rightarrow y + z = \frac{1}{\sqrt{2}}$ ... (ii)

and $\hat{u} \cdot \vec{c} = \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 1 \cdot 1 \cdot \cos \frac{2 π}{3} \Rightarrow x + y = \frac{- 1}{\sqrt{2}}$ ... (iii)

Solving (i), (ii) and (iii), we get

$x = \frac{- 1}{\sqrt{2}}, y = 0, z = \frac{1}{\sqrt{2}} ∴ \hat{u} = \frac{- 1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{k}$

So, $| \hat{u} - \vec{v} |^{2} =$ $| - \sqrt{2} \hat{i} - \frac{1}{\sqrt{2}} \hat{j} |^{2}$ $= 2 + \frac{1}{2} = \frac{5}{2}$ .

Q 7 :

The Least positive integral value of $α$ , for which the angle between the vectors $α \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $α \hat{i} + 2 α \hat{j} - 2 \hat{k}$ is acute, is __________. [2024]

(5)

$\cos θ = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |}$

As $θ$ is acute, so $\cos θ > 0$ .

$\frac{(α \hat{i} - 2 \hat{j} + 2 \hat{k}) \cdot (α \hat{i} + 2 α \hat{j} - 2 \hat{k})}{\sqrt{α^{2} + {(- 2)}^{2} + {(2)}^{2}} \times \sqrt{α^{2} + {(2 α)}^{2} + {(- 2)}^{2}}} > 0$

$\Rightarrow \frac{α^{2} - 4 α - 4}{\sqrt{α^{2} + 8} \times \sqrt{5 α^{2} + 4}} > 0$

$\Rightarrow α^{2} - 4 α - 4 > 0 \Rightarrow α^{2} - 4 α + 4 - 4 - 4 > 0$

$\Rightarrow {(α - 2)}^{2} - 8 > 0 \Rightarrow {(α - 2)}^{2} > 8$

$\Rightarrow α - 2 > 2 \sqrt{2} or α - 2 < - 2 \sqrt{2}$ (Neglect)

$\Rightarrow α > 2 + 2 \sqrt{2} = 4.83$

$\Rightarrow$ Least positive integral value of $α = 5$ .

Q 8 :

If $\vec{a}$ is a non-zero vector such its projections on the vectors $2 \hat{i} - \hat{j} + 2 \hat{k}$ , $\hat{i} + 2 \hat{j} - 2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $\vec{a}$ is: [2025]

$\frac{1}{\sqrt{155}} (7 \hat{i} + 9 \hat{j} + 5 \hat{k})$
$\frac{1}{\sqrt{155}} (7 \hat{i} + 9 \hat{j} - 5 \hat{k})$
$\frac{1}{\sqrt{155}} (- 7 \hat{i} + 9 \hat{j} - 5 \hat{k})$
$\frac{1}{\sqrt{155}} (- 7 \hat{i} + 9 \hat{j} + 5 \hat{k})$

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(1)

Let $\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}, \vec{u} = 2 \hat{i} - \hat{j} + 2 \hat{k}, \vec{v} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ and $\vec{w} = \hat{k}$

When, $\frac{\vec{a} \cdot \vec{u}}{| \vec{u} |} = \frac{\vec{a} \cdot \vec{v}}{| \vec{v} |} = \frac{\vec{a} \cdot \vec{w}}{| \vec{w} |}$

Now, $\frac{\vec{a} \cdot \vec{u}}{| \vec{u} |} = \frac{\vec{a} \cdot \vec{w}}{| \vec{w} |}$

$\Rightarrow \frac{(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (2 \hat{i} - \hat{j} + 2 \hat{k})}{3} = \frac{(x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{k})}{1}$

$\Rightarrow 2 x - y + 2 z = 3 z$

$\Rightarrow 2 x - y - z = 0$ ... (i)

Also, $\frac{\vec{a} \cdot \vec{v}}{| \vec{v} |} = \frac{\vec{a} \cdot \vec{w}}{| \vec{w} |}$

$\Rightarrow \frac{x + 2 y - 2 z}{3} = z$

$\Rightarrow x + 2 y - 5 z = 0$ ... (ii)

and $\frac{\vec{a} \cdot \vec{u}}{| \vec{u} |} = \frac{\vec{a} \cdot \vec{v}}{| \vec{v} |}$

$\Rightarrow 2 x - y + 2 z = x + 2 y - 2 z$

$\Rightarrow x - 3 y + 4 z = 0$ ... (iii)

From (i), (ii) and (iii), we get x = 7, y = 9 and z = 5

$\vec{a} = \frac{7 \hat{i} + 9 \hat{j} + 5 \hat{k}}{\sqrt{{(7)}^{2} + {(9)}^{2} + {(5)}^{2}}} = \frac{7 \hat{i} + 9 \hat{j} + 5 \hat{k}}{\sqrt{155}}$

Q 9 :

Consider two vectors $\vec{u} = 3 \hat{i} - \hat{j}$ and $\vec{v} = 2 \hat{i} + \hat{j} - λ \hat{k}$ , $λ > 0$ . The angle between them is given by $\cos^{- 1} (\frac{\sqrt{5}}{2 \sqrt{7}})$ . Let $\vec{v} = {\vec{v}}_{1} + {\vec{v}}_{2}$ , where ${\vec{v}}_{1}$ is parallel to $\vec{u}$ and ${\vec{v}}_{2}$ is perpendicular to $\vec{u}$ . Then the value $| {\vec{v}}_{1} |^{2} + | {\vec{v}}_{2} |^{2}$ is equal to [2025]

14
10
$\frac{23}{2}$
$\frac{25}{2}$

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(1)

We have, $\vec{u} = 3 \hat{i} - \hat{j}$ , $\vec{v} = 2 \hat{i} + \hat{j} - λ \hat{k}$

Now, the angle between $\vec{u}$ and $\vec{v}$ is

$\Rightarrow \cos θ = \frac{\vec{u} \cdot \vec{v}}{| u | | v |} \Rightarrow \frac{5}{\sqrt{10} \sqrt{5 + λ^{2}}} = \frac{\sqrt{5}}{2 \sqrt{7}}$

$\Rightarrow λ^{2} = 9 \Rightarrow λ = 3$ ( $∵ λ > 0$ )

Now, $\vec{v} = {\vec{v}}_{1} + {\vec{v}}_{2}$

$\Rightarrow | \vec{v} |^{2} = {\vec{v}}_{1}^{2} + {\vec{v}}_{2}^{2} + 2 {\vec{v}}_{1} \cdot {\vec{v}}_{2}$

$\Rightarrow 14 = {\vec{v}}_{1}^{2} + {\vec{v}}_{2}^{2} + 0$ ( $∵ {\vec{v}}_{1} ⊥ {\vec{v}}_{2}$ )

$\Rightarrow {\vec{v}}_{1}^{2} + {\vec{v}}_{2}^{2} = 14$ .

Q 10 :

Let the angle $θ, 0 < θ < \frac{π}{2}$ between two unit vectors $\hat{a}$ and $\hat{b}$ be $\sin^{- 1} (\frac{\sqrt{65}}{9})$ . If the vector $\vec{c} = 3 \hat{a} + 6 \hat{b} + 9 (\hat{a} \times \hat{b})$ , then the value of $9 (\vec{c} \cdot \hat{a}) - 3 (\vec{c} \cdot \hat{b})$ is [2025]

24
29
27
31

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(2)

Given, $\vec{c} = 3 \hat{a} + 6 \hat{b} + 9 (\hat{a} \times \hat{b})$ and

$θ = \sin^{- 1} (\frac{\sqrt{65}}{9}) \Rightarrow \sin θ = \frac{\sqrt{65}}{9} \Rightarrow \cos θ = \frac{4}{9}$ $[∵ 0 < θ < \frac{π}{2}]$

Now, $\vec{c} \cdot \hat{a} = 3 | \hat{a} |^{2} + 6 \hat{a} \cdot \hat{b} + 9 (\hat{a} \times \hat{b}) \cdot \hat{a}$

$= 3 + \frac{6 \times 4}{9} + 0$ [ $∵ \hat{a} \cdot \hat{b} = \cos θ$ ]

$= \frac{51}{9}$

and $\vec{c} \cdot \hat{b} = 3 \hat{a} \cdot \hat{b} + 6 | \hat{b} |^{2} + 9 (\hat{a} \times \hat{b}) \cdot \hat{b}$

$= \frac{3 \times 4}{9} + 6 = \frac{22}{3}$

Now, $9 (\vec{c} \cdot \hat{a}) - 3 (\vec{c} \cdot \hat{b}) = 51 - 22 = 29$ .

Q 11 :

Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{| \vec{a} + \vec{b} | + | \vec{a} - \vec{b} |}{| \vec{a} + \vec{b} | - | \vec{a} - \vec{b} |} = \sqrt{2} + 1$ . Then $\frac{| \vec{a} + \vec{b} |^{2}}{| \vec{a} |^{2}}$ is : [2025]

$4 + 2 \sqrt{2}$
$2 + 4 \sqrt{2}$
$1 + \sqrt{2}$
$2 + \sqrt{2}$

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(4)

We have, $\frac{| \vec{a} + \vec{b} | + | \vec{a} - \vec{b} |}{| \vec{a} + \vec{b} | - | \vec{a} - \vec{b} |} = \sqrt{2} + 1$

Apply componendo and dividendo, we get

$\Rightarrow \frac{2 | \vec{a} + \vec{b} |}{2 | \vec{a} - \vec{b} |} = \frac{\sqrt{2} + 2}{\sqrt{2}}$

$\Rightarrow | \vec{a} + \vec{b} | = (1 + \sqrt{2}) | \vec{a} - \vec{b} |$

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

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

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

$\Rightarrow \frac{\vec{a} \cdot \vec{b}}{| \vec{a} |^{2}} = \frac{2 + 2 \sqrt{2}}{4 + 2 \sqrt{2}} = \frac{1}{\sqrt{2}}$

Now, we have

$\frac{| \vec{a} + \vec{b} |^{2}}{| \vec{a} |^{2}} = 1 + \frac{| \vec{b} |^{2}}{| \vec{a} |^{2}} + \frac{2 \vec{a} \cdot \vec{b}}{| \vec{a} |^{2}} = 1 + 1 + 2 (\frac{1}{\sqrt{2}}) = 2 + \sqrt{2}$

Q 12 :

Let $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$ . Let $\hat{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$ . Then such a vector $\hat{c}$ is : [2025]

$\frac{1}{\sqrt{2}} (- \hat{i} + \hat{k})$
$\frac{1}{\sqrt{3}} (\hat{i} - \hat{j} + \hat{k})$
$\frac{1}{\sqrt{3}} (- \hat{i} + \hat{j} - \hat{k})$
$\frac{1}{\sqrt{5}} (\hat{j} - 2 \hat{k})$

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(1)

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

Let $\vec{c} = λ \vec{a} + μ \vec{b}$

Now, $\vec{c} \cdot \vec{a} = 0$

$\Rightarrow (λ \vec{a} + μ \vec{b}) \cdot \vec{a} = 0 \Rightarrow λ \vec{a} \cdot \vec{a} + μ \vec{b} \cdot \vec{a} = 0$

$\Rightarrow λ (6) + μ (3) = 0 \Rightarrow μ = - 2 λ$

$∴ \vec{c} = λ \vec{a} - 2 λ \vec{b}$

$= λ (\hat{i} + 2 \hat{j} + \hat{k} - 2 (2 \hat{i} + \hat{j} - \hat{k}))$

$= λ (- 3 \hat{i} + 3 \hat{k})$

Since, $\vec{c}$ is a unit vector.

$\Rightarrow$ $| λ (- 3 \hat{i} + 3 \hat{k}) |$ $= 1 \Rightarrow$ $| λ |$ $\sqrt{9 + 9} = 1$

$\Rightarrow λ = \pm \frac{1}{3 \sqrt{2}} \Rightarrow \hat{c} = \pm \frac{1}{\sqrt{2}} (- \hat{i} + \hat{k})$ .

Q 13 :

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{π}{3}$ . If $λ \vec{a} + 2 \vec{b}$ and $3 \vec{a} - λ \vec{b}$ are perpendicular to each other, then the number of values of $λ$ in [–1, 3] is : [2025]

1
0
2
3

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(2)

$\vec{a} \cdot \vec{b} =$ $| \vec{a} |$ $| \vec{b} |$ $\cos \frac{π}{3} = \frac{1}{2}$ [ $∵ | \vec{a} | = | \vec{b} | = 1$ ]

Now, $(λ \vec{a} + 2 \vec{b}) \cdot (3 \vec{a} - λ \vec{b}) = 0$

$\Rightarrow 3 λ \vec{a} \cdot \vec{a} - λ^{2} \vec{a} \cdot \vec{b} + 6 \vec{b} \cdot \vec{a} - 2 λ \vec{b} \cdot \vec{b} = 0$

$\Rightarrow 3 λ - \frac{λ^{2}}{2} + 3 - 2 λ = 0$ $[∵ | \vec{a} | = | \vec{b} | = 1 and \vec{a} \cdot \vec{b} = \frac{1}{2}]$

$\Rightarrow λ^{2} - 2 λ - 6 = 0 \Rightarrow λ = 1 \pm \sqrt{7} \notin [- 1, 3]$

$\Rightarrow$ Number of values of $λ$ = 0

Q 14 :

Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the are AC such that $\frac{length of arc A B}{length of arc B C} = \frac{1}{5}$ , and $\vec{O C} = α \vec{O A} + β \vec{O B}$ , then $α + \sqrt{2} (\sqrt{3} - 1) β$ is equal to [2025]

$2 - \sqrt{3}$
$5 \sqrt{3}$
$2 \sqrt{3}$
$2 + \sqrt{3}$

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(1)

Here, $∠ A O B = 15 °$ and $∠ B O C = 75 °$

Given $\vec{O C} = α \vec{O A} + β \vec{O B}$ ... (i)

$\vec{O A} \cdot \vec{O C} = α \vec{O A} \cdot \vec{O A} + β \vec{O B} \cdot \vec{O A}$

$0 = α | \vec{O A} |^{2} + β$ $| \vec{O B} |$ $| \vec{O A} |$ $\cdot \cos 15 °$ [ $∵ \vec{O A} ⊥ \vec{O C}$ ]

$\Rightarrow α + β \cos 15 ° = 0$ ... (ii)

[ $∵$ $| \vec{O A} |$ $=$ $| \vec{O B} |$ $=$ $| \vec{O C} |$ ]

$\vec{O B} \cdot \vec{O C} = α \vec{O A} \cdot \vec{O B} + β \vec{O B} \cdot \vec{O B}$ [Form (i)]

$| \vec{O B} |$ $| \vec{O C} |$ $\cos 75 ° = α$ $| \vec{O A} |$ $| \vec{O B} |$ $\cos 15 ° + β | \vec{O B} |^{2}$

$\Rightarrow \cos 75 ° = α \cos 15 ° + β$ ... (iii)

From (ii) and (iii), we get

$\cos 75 ° = - β \cos^{2} 15 + β$

$\Rightarrow \cos 75 ° = β (1 - \cos^{2} 15 °) = β \sin^{2} 15 °$

$β = \frac{\sin 15 °}{\sin^{2} 15 °} = \frac{1}{\sin 15 °} = \frac{2 \sqrt{2}}{\sqrt{3} - 1}$ [ $∵ \cos (90 ° - θ) = \sin θ$ ]

From (ii), $α = - β \cos 15 ° = \frac{- 2 \sqrt{2}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

$\Rightarrow α = \frac{- (\sqrt{3} + 1)}{(\sqrt{3} - 1)}$

$∴ α + \sqrt{2} (\sqrt{3} - 1) β = \frac{- (\sqrt{3} + 1)}{(\sqrt{3} - 1)} + \sqrt{2} \frac{(\sqrt{3} - 1) \times 2 \sqrt{2}}{(\sqrt{3} - 1)} = (2 - \sqrt{3})$

Q 15 :

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = 3 \hat{i} + 2 \hat{j} - \hat{k}$ , $\vec{c} = μ \hat{j} + u \hat{k}$ and $\hat{d}$ be a unit vector such that $\vec{a} \times \hat{d} = \vec{b} \times \hat{d}$ and $\vec{c} \cdot \hat{d} = 1$ . If $\vec{c}$ is perpendicular to $\vec{a}$ , then $| 3 λ \hat{d} + μ \vec{c} |^{2}$ is equal to __________. [2025]

(5)

We have, $\vec{a} \times \hat{d} - \vec{b} \times \hat{d} = 0$

$\Rightarrow (\vec{a} - \vec{b}) \times \hat{d} = 0$

$∴ \hat{d} = t (\vec{a} - \vec{b})$ , for any scalar t.

$\Rightarrow \hat{d} = t (- 2 \hat{i} - \hat{j} + 2 \hat{k})$

Since, $| \hat{d} |$ $= 1$

$∴$ $| t |$ $= \frac{1}{3}$

Now, $\vec{c} \cdot \vec{a} = 0$ [ $∵ \vec{c}$ is perpendicular to $\vec{a}$ ]

$\Rightarrow λ + μ = 0 \Rightarrow μ = - λ$

$∴ \vec{c} = λ (\hat{j} - \hat{k}) \Rightarrow | \vec{c} |^{2} = 2 λ^{2}$

Also, $\vec{c} \cdot \hat{d} = 1$

$\Rightarrow λ (\hat{j} - \hat{k}) \cdot t (- 2 \hat{i} - \hat{j} + 2 \hat{k}) = 1$

$\Rightarrow λ t = \frac{- 1}{3} \Rightarrow λ^{2} = 1$ $[∵ | t | = \frac{1}{3}]$

$∴ | 3 λ \hat{d} + μ \vec{c} |^{2} = 9 λ^{2} | \hat{d} |^{2} + μ^{2} | \vec{c} |^{2} + 6 λ μ (\hat{d} \cdot \vec{c})$

$= 3 λ^{2} + 2 λ^{4} = 5$ [ $∵ λ^{2} = 1$ ]

Q 16 :

An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If $\vec{O P} = \vec{u}$ , $\vec{O R} = \vec{v}$ and $\vec{O Q} = α \vec{u} + β \vec{v}$ , then $α, β^{2}$ are the roots of the equation [2023]

$x^{2} + x - 2 = 0$
$3 x^{2} + 2 x - 1 = 0$
$x^{2} - x - 2 = 0$
$3 x^{2} - 2 x - 1 = 0$

1

2

3

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(3)

$We have, | \vec{u} | = | \vec{v} | = | α \vec{u} + β \vec{v} | = r$

$Also, (\vec{u}) \cdot (α \vec{u} + β \vec{v}) = 0$

$Now, \vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos 45 ° = \frac{r^{2}}{\sqrt{2}}$

$∴$ $α | \vec{u} |^{2} + β \vec{u} \cdot \vec{v} = 0$

$\Rightarrow r^{2} α + β \frac{r^{2}}{\sqrt{2}} = 0$

$\Rightarrow α + β \times \frac{1}{\sqrt{2}} = 0 \Rightarrow α = - \frac{β}{\sqrt{2}} [∵ r \neq 0]$

$Also, | α \vec{u} + β \vec{v} |^{2} = r^{2}$ $\Rightarrow α^{2} | \vec{u} |^{2} + β^{2} | \vec{v} |^{2} + 2 α β \vec{u} \cdot \vec{v} = r^{2}$

$\Rightarrow α^{2} + β^{2} + \sqrt{2} α β = 1$ $\Rightarrow β^{2} = 2 and α = - 1$

$∴$ $α, β^{2} are roots of equation x^{2} - x - 2 = 0$

Q 17 :

Let $\vec{α} = 4 \hat{i} + 3 \hat{j} + 5 \hat{k}$ and $\vec{β} = \hat{i} + 2 \hat{j} - 4 \hat{k}$ . Let ${\vec{β}}_{1}$ be parallel to $\vec{α}$ and ${\vec{β}}_{2}$ be perpendicular to $\vec{α}$ . If $\vec{β} = {\vec{β}}_{1} + {\vec{β}}_{2}$ , then the value of $5 {\vec{β}}_{2} \cdot (\hat{i} + \hat{j} + \hat{k})$ is [2023]

9
11
7
6

1

2

3

4

(3)

$Given \vec{α} = 4 \hat{i} + 3 \hat{j} + 5 \hat{k} and \vec{β} = \hat{i} + 2 \hat{j} - 4 \hat{k}$

$As, {\vec{β}}_{1} ∥ \vec{α} \Rightarrow {\vec{β}}_{1} = λ \vec{α}$

$\Rightarrow {\vec{β}}_{1} = λ (4 \hat{i} + 3 \hat{j} + 5 \hat{k})$ ...(i)

${\vec{β}}_{2} ⊥ \vec{α} \Rightarrow {\vec{β}}_{2} \cdot \vec{α} = 0$

$Let {\vec{β}}_{2} = x \hat{i} + y \hat{j} + z \hat{k}$

$\Rightarrow 4 x + 3 y + 5 z = 0$ ...(ii) $[{\vec{β}}_{2} \cdot \vec{α} = 0]$

$Now, \vec{β} = {\vec{β}}_{1} + {\vec{β}}_{2}$

$\Rightarrow \hat{i} + 2 \hat{j} - 4 \hat{k} = (4 λ + x) \hat{i} + (3 λ + y) \hat{j} + (5 λ + z) \hat{k}$

$On comparing, we get$

$4 λ + x = 1 \Rightarrow x = 1 - 4 λ$ ...(iii)

$3 λ + y = 2 \Rightarrow y = 2 - 3 λ$ ...(iv)

$5 λ + z = - 4 \Rightarrow z = - 4 - 5 λ$ ...(v)

$Putting the value of x, y, z in (ii), we get$

$4 (1 - 4 λ) + 3 (2 - 3 λ) + 5 (- 4 - 5 λ) = 0$

$\Rightarrow 4 - 16 λ + 6 - 9 λ - 20 - 25 λ = 0$

$\Rightarrow - 10 - 50 λ = 0 \Rightarrow - 50 λ = 10 \Rightarrow λ = - \frac{1}{5}$

$Put the value of λ in (iii), (iv), (v), we get$

$∴$ $x = \frac{9}{5}, y = \frac{13}{5}, z = - 3$ $∴$ ${\vec{β}}_{2} = \frac{9}{5} \hat{i} + \frac{13}{5} \hat{j} - 3 \hat{k}$

$∴$ $5 {\vec{β}}_{2} \cdot (\hat{i} + \hat{j} + \hat{k}) = 9 + 13 - 15 = 7$

Q 18 :

The vector $\vec{a} = - \hat{i} + 2 \hat{j} + \hat{k}$ is rotated through a right angle, passing through the $y$ -axis in its way and the resulting vector is $\vec{b}$ . Then the projection of $3 \vec{a} + \sqrt{2} \vec{b}$ on $\vec{c} = 5 \hat{i} + 4 \hat{j} + 3 \hat{k}$ is [2023]

$3 \sqrt{2}$
$1$
$2 \sqrt{3}$
$\sqrt{6}$

1

2

3

4

(1)

$\vec{b}, \vec{a} and \hat{j} are coplanar.$

$\vec{b} = λ \vec{a} + μ \hat{j} = λ (- \hat{i} + 2 \hat{j} + \hat{k}) + μ \hat{j}$

$= - λ \hat{i} + (2 λ + μ) \hat{j} + λ \hat{k},$ $\vec{b} \cdot \vec{a} = 0$

$(- \hat{i} + 2 \hat{j} + \hat{k}) \cdot (- λ \hat{i} + (2 λ + μ) \hat{j} + λ \hat{k}) = 0$

$λ + 2 (2 λ + μ) + λ = 0 \Rightarrow 6 λ + 2 μ = 0 \Rightarrow μ + 3 λ = 0$

$\vec{b} = λ \vec{a} - 3 λ \hat{j} = λ (- \hat{i} + 2 \hat{j} + \hat{k}) - 3 λ \hat{j}$ $= λ (- \hat{i} - \hat{j} + \hat{k})$

$| \vec{b} | = \sqrt{3} | λ | = \sqrt{6} [∵ | \vec{a} | = \sqrt{6}] \Rightarrow | λ | = \sqrt{2} \Rightarrow λ \neq \sqrt{2}$

As for this value of $λ$ angle between b and y-axis is not acute.

$Therefore λ = - \sqrt{2};$

$3 \vec{a} + \sqrt{2} \vec{b} = - \hat{i} + 8 \hat{j} + \hat{k};$ $| \vec{c} | = 5 \sqrt{2}$

$\frac{(3 \vec{a} + \sqrt{2} \vec{b}) \cdot \vec{c}}{| \vec{c} |} = \frac{(- \hat{i} + 8 \hat{j} + \hat{k}) \cdot (5 \hat{i} + 4 \hat{j} + 3 \hat{k})}{5 \sqrt{2}}$

$= \frac{- 5 + 32 + 3}{5 \sqrt{2}} = \frac{30}{5 \sqrt{2}} = \frac{6}{\sqrt{2}} = 3 \sqrt{2}$

Q 19 :

Let $\vec{a} = 2 \hat{i} + \hat{j} + \hat{k}$ , and $\vec{b}$ and $\vec{c}$ be two non-zero vectors such that $| \vec{a} + \vec{b} + \vec{c} |$ $=$ $| \vec{a} + \vec{b} - \vec{c} |$ and $\vec{b} \cdot \vec{c} = 0$ .

Consider the following two statements:

(A) $| \vec{a} + λ \vec{c} | \geq | \vec{a} |$ for all $λ \in R$ .

(B) $\vec{a}$ and $\vec{c}$ are always parallel.

Then, [2023]

both (A) and (B) are correct
only (A) is correct
only (B) is correct
neither (A) nor (B) is correct

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2

3

4

(2)

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

$| \vec{a} + \vec{b} + \vec{c} |^{2} = | \vec{a} + \vec{b} - \vec{c} |^{2}$

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

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

$\Rightarrow 0 + 2 \vec{c} \cdot \vec{a} = - 0 - 2 \vec{c} \cdot \vec{a}$ $[∵ \vec{b} \cdot \vec{c} = 0]$

$\Rightarrow \vec{c} \cdot \vec{a} = 0 \Rightarrow \vec{a} and \vec{c} are perpendicular.$

$Hence (B) is incorrect.$

$Now, | \vec{a} + λ \vec{c} |^{2} \geq | \vec{a} |^{2}$

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

$\Rightarrow λ^{2} | \vec{c} |^{2} \geq 0 (∵ \vec{a} \cdot \vec{c} = 0)$

$We have \vec{c} \neq 0 \Rightarrow λ^{2} | \vec{c} |^{2} > 0 is true for λ \in R$ , $∴ λ^{2} | \vec{c} |^{2} \geq 0 is also true.$

$Hence (A) is correct.$

Q 20 :

Let $\vec{c} and \vec{d}$ be vectors such that $| \vec{c} + \vec{d} |$ $= \sqrt{29}$ and $\vec{c} \times (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times \vec{d} .$ If $λ_{1}, λ_{2} (λ_{1} > λ_{2})$ are the possible values of $(\vec{c} + \vec{d}) \cdot (- 7 \hat{i} + 2 \hat{j} + 3 \hat{k}),$ then the equation $K^{2} x^{2} + (K^{2} - 5 K + λ_{1}) x y + (3 K + \frac{λ_{2}}{2}) y^{2} - 8 x + 12 y + λ_{2} = 0$ represents a circle, for k equal to: [2026]

-1
4
1
2

1

2

3

4

(3)

$| \vec{c} + \vec{d} |$ $= \sqrt{29}$

$\vec{c} + \vec{d} = λ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$

$λ = \pm 1$

$λ (- 14 + 6 + 12) = 4 λ, λ_{1} = 4, λ_{2} = - 4$

$k^{2} x^{2} + (k^{2} - 5 k + 4) x y + (3 k - 2) y^{2} - 8 x + 12 y - 4 = 0$

$is circle$

$k^{2} - 5 k + 4 = 0 \Rightarrow k = 1, 4$

$k^{2} = 3 k - 2 \Rightarrow k = 1, 2$

$k = 1$

Q 21 :

Let P be a point in the plane of the vectors $\vec{A B} = 3 \hat{i} + \hat{j} - \hat{k} and \vec{A C} = \hat{i} - \hat{j} + 3 \hat{k}$ such that P is equidistant from the lines AB and AC. If $| \vec{A P} |$ $= \frac{\sqrt{5}}{2},$ then the area of the triangle ABP is: [2026]

$\frac{\sqrt{26}}{4}$
2
$\frac{3}{2}$
$\frac{\sqrt{30}}{4}$

1

2

3

4

(4)

$\cos 2 θ = \frac{3 - 1 - 3}{\sqrt{11} \cdot \sqrt{11}} = - \frac{1}{11}$

$1 - 2 \sin^{2} θ = - \frac{1}{11} \Rightarrow 2 \sin^{2} θ = \frac{12}{11} \Rightarrow \sin θ = \sqrt{\frac{6}{11}}$

$∴$ $Area (∆ A P B) = \frac{1}{2} \times \sqrt{11} \cdot \frac{\sqrt{5}}{2} \cdot \sqrt{\frac{6}{11}} = \frac{\sqrt{30}}{4}$

Q 22 :

Let $\vec{a} = 2 \hat{i} - \hat{j} - \hat{k}, \vec{b} = \hat{i} + 3 \hat{j} - \hat{k}$ and $\vec{c} = 2 \hat{i} + \hat{j} + 3 \hat{k} .$ Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ , such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}} .$ Then $| \vec{v} |$ is equal to [2026]

$\frac{\sqrt{21}}{2}$
$\frac{\sqrt{35}}{2}$
$13$
7

1

2

3

4

(2)

$Let \vec{v} = \vec{a} + λ \vec{b}$

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

$\frac{\vec{v} \cdot \vec{c}}{| \vec{c} |} = \frac{1}{\sqrt{14}}$

$\frac{2 (2 + λ) + 3 λ - 1 - 3 - 3 λ}{\sqrt{14}} = \frac{1}{\sqrt{14}}$

$4 + 2 λ - 4 = 1$

$2 λ = 1 \Rightarrow λ = \frac{1}{2}$

$\vec{v} = \frac{5}{2} \hat{i} + \frac{1}{2} \hat{j} - \frac{3}{2} \hat{k}$

$| \vec{v} |$ $= \sqrt{\frac{25 + 1 + 9}{4}} = \frac{\sqrt{35}}{2}$

Q 23 :

Let $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} - \hat{k}, \vec{c} = λ \hat{i} + \hat{j} + \hat{k}$ and $\vec{v} = \vec{a} \times \vec{b} . If \vec{v} \cdot \vec{c} = 11$ and the length of the projection of $\vec{b} on \vec{c}$ , is p, then $9 p^{2}$ is equal to. [2026]

4
9
6
12

1

2

3

4

(4)

$\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k},$ $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k},$ $\vec{c} = λ \hat{i} + \hat{j} + \hat{k}, and \vec{v} = \vec{a} \times \vec{b} .$ $If \vec{v} \cdot \vec{c} = 11$

$\vec{v} = (\vec{a} \times \vec{b}) = (- \hat{i} + 7 \hat{j} + 5 \hat{k})$

$\vec{v} \cdot \vec{c} = 11 = (- \hat{i} + 7 \hat{j} + 5 \hat{k}) \cdot (λ \hat{i} + \hat{j} + \hat{k}) = 11$

$\Rightarrow - λ + 7 + 5 = 11$

$\Rightarrow λ = 1$

$Length of projection of \vec{b} on \vec{c} = \vec{b} \cdot \hat{c}$

$\Rightarrow$ $| (2 \hat{i} + \hat{j} - \hat{k}) \cdot \frac{(\hat{i} + \hat{j} + \hat{k})}{\sqrt{3}} |$ $= \frac{2 + 1 - 1}{\sqrt{3}} = p = \frac{2}{\sqrt{3}}$

$\Rightarrow 9 p^{2} = 9 (\frac{4}{3}) = 12$

Q 24 :

Let $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = λ \hat{j} + 2 \hat{k}, λ \in Z$ be two vectors. Let $\vec{c} = \vec{a} \times \vec{b} and \vec{d}$ be a vector of magnitude 2 in yz-plane. If $|| \vec{c} || = \sqrt{53},$ then the maximum possible value of ${(\vec{c} \cdot \vec{d})}^{2}$ is equal to: [2026]

104
52
26
208

1

2

3

4

(4)

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

$\vec{b} = λ \hat{j} + 2 \hat{k}, λ \in ℤ$

$\vec{c} = \vec{a} \times \vec{b} = (- 2 - λ) \hat{i} - 4 \hat{j} + 2 λ \hat{k}$

$| \vec{c} |$ $= \sqrt{53}$

$\Rightarrow 5 λ^{2} + 4 λ - 33 = 0$

$λ = 2.2 or - 3$

$\Rightarrow λ = - 3$

$\vec{c} = \hat{i} - 4 \hat{j} - 6 \hat{k}$

$let \vec{d} = y \hat{j} + z \hat{k}$

$| \vec{d} |$ $= 2$

$\Rightarrow y^{2} + z^{2} = 4$

${(\vec{c} \cdot \vec{d})}^{2} = {(4 y + 6 z)}^{2}$ $\leq {(\sqrt{4^{2} + 6^{2}} \times \sqrt{y^{2} + z^{2}})}^{2} \leq 208$

Q 25 :

Let a vector $\vec{a} = \sqrt{2} \hat{i} - \hat{j} + λ \hat{k}, λ > 0,$ make an obtuse angle with the vector $\vec{b} = - λ^{2} \hat{i} + 4 \sqrt{2} \hat{j} + 4 \sqrt{2} \hat{k}$ and an angle $θ, \frac{π}{6} < θ < \frac{π}{2}$ , with the positive z-axis. If the set of all possible values of $λ is (α, β) - {γ}$ , then $α + β + γ$ is equal to _______. [2026]

(5)

$\frac{\vec{a} \cdot \hat{k}}{| \vec{a} |} = \cos θ = \frac{λ}{\sqrt{3 + λ^{2}}} = \cos θ$

$\Rightarrow 0 < \frac{λ}{\sqrt{3 + λ^{2}}} < \frac{\sqrt{3}}{2}$

$\Rightarrow λ > 0 & 4 λ^{2} < 9 + 3 λ^{2} \Rightarrow λ^{2} < 9$

$\Rightarrow λ \in (0, 3) . . . (1)$

$\Rightarrow \vec{a} \cdot \vec{b} < 0$ $\Rightarrow - \sqrt{2} λ^{2} - 4 \sqrt{2} + 4 \sqrt{2} λ < 0$

$\Rightarrow λ^{2} - 4 λ + 4 > 0$ $\Rightarrow {(λ - 2)}^{2} > 0$

$\Rightarrow λ \neq 2 . . . (2)$

From (1) and (2)

$λ \in (0, 3) - {2}$

$∴$ $α = 0, β = 3, γ = 2$

$\Rightarrow α + β + γ = 5$

Topic Question Set

Q 1 :

Let a unit vector which makes an angle of 60° with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and an angle of 45° with $\hat{i} - \hat{k}$ be $\vec{C}$ . Then $\vec{C} + (- \frac{1}{2} \hat{i} + \frac{1}{3 \sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k})$ is [2024]

Q 4 :

Let $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $\vec{b} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ and $\vec{c} = 3 \hat{i} - \hat{j} + λ \hat{k}$ be three vectors. Let $\vec{r}$ be a unit vector along $\vec{b} + \vec{c}$ . If $\vec{r} \cdot \vec{a} = 3$ , then $3 λ$ is equal to : [2024]

Q 5 :

Consider a $△ A B C$ where A(1, 3, 2), B(–2, 8, 0) and C(3, 6, 7). If the angle bisector of $∠ B A C$ meets the line BC at D, then the length of the projection of the vector $\vec{A D}$ on the vector $\vec{A C}$ IS : [2024]

Q 7 :

The Least positive integral value of $α$ , for which the angle between the vectors $α \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $α \hat{i} + 2 α \hat{j} - 2 \hat{k}$ is acute, is __________. [2024]

Q 8 :

If $\vec{a}$ is a non-zero vector such its projections on the vectors $2 \hat{i} - \hat{j} + 2 \hat{k}$ , $\hat{i} + 2 \hat{j} - 2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $\vec{a}$ is: [2025]

Q 11 :

Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{| \vec{a} + \vec{b} | + | \vec{a} - \vec{b} |}{| \vec{a} + \vec{b} | - | \vec{a} - \vec{b} |} = \sqrt{2} + 1$ . Then $\frac{| \vec{a} + \vec{b} |^{2}}{| \vec{a} |^{2}}$ is : [2025]

Q 12 :

Let $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$ . Let $\hat{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$ . Then such a vector $\hat{c}$ is : [2025]

Q 13 :

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{π}{3}$ . If $λ \vec{a} + 2 \vec{b}$ and $3 \vec{a} - λ \vec{b}$ are perpendicular to each other, then the number of values of $λ$ in [–1, 3] is : [2025]

Q 14 :

Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the are AC such that $\frac{length of arc A B}{length of arc B C} = \frac{1}{5}$ , and $\vec{O C} = α \vec{O A} + β \vec{O B}$ , then $α + \sqrt{2} (\sqrt{3} - 1) β$ is equal to [2025]

Q 16 :

An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If $\vec{O P} = \vec{u}$ , $\vec{O R} = \vec{v}$ and $\vec{O Q} = α \vec{u} + β \vec{v}$ , then $α, β^{2}$ are the roots of the equation [2023]

Q 18 :

The vector $\vec{a} = - \hat{i} + 2 \hat{j} + \hat{k}$ is rotated through a right angle, passing through the $y$ -axis in its way and the resulting vector is $\vec{b}$ . Then the projection of $3 \vec{a} + \sqrt{2} \vec{b}$ on $\vec{c} = 5 \hat{i} + 4 \hat{j} + 3 \hat{k}$ is [2023]

Q 21 :

Let P be a point in the plane of the vectors $\vec{A B} = 3 \hat{i} + \hat{j} - \hat{k} and \vec{A C} = \hat{i} - \hat{j} + 3 \hat{k}$ such that P is equidistant from the lines AB and AC. If $| \vec{A P} |$ $= \frac{\sqrt{5}}{2},$ then the area of the triangle ABP is: [2026]

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Topic Question Set

Q 1 : Let a unit vector which makes an angle of 60° with 2i^+2j^–k^ and an angle of 45° with i^–k^ be C→. Then C→+(–12i^+132j^–23k^) is [2024]

Q 2 : For λ>0, let θ be the angle between the vectors a→=i^+λj^–3k^ and b→=3i^–j^+2k^. If the vectors a→+b→ and a→–b→ are mutually perpendicular, then the value of (14cosθ)2 is equal to [2024]

Q 3 : Let P(x, y, z) be a point in the first octant, whose projection in the xy-plane is the point Q. Let OP = γ; the angle between OQ and the positive x-axis be θ; and the angle between OP and the positive z-axis be ϕ, where O is the origin. Then the distance of P from the x-axis is [2024]

Q 4 : Let a→=i^+2j^+3k^, b→=2i^+3j^–5k^ and c→=3i^–j^+λk^ be three vectors. Let r→ be a unit vector along b→+c→. If r→·a→=3, then 3λ is equal to : [2024]

Q 5 : Consider a △ABC where A(1, 3, 2), B(–2, 8, 0) and C(3, 6, 7). If the angle bisector of ∠BAC meets the line BC at D, then the length of the projection of the vector AD→ on the vector AC→ IS : [2024]

Q 6 : Let a unit vector u^=xi^+yj^+zk^ make angles π2,π3 and 2π3 with the vectors 12i^+12k^, 12j^+12k^ and 12i^+12j^ respectively. If v→=12i^+12j^+12k^, then |u^–v→|2 is equal to [2024]

Q 7 : The Least positive integral value of α, for which the angle between the vectors αi^–2j^+2k^ and αi^+2αj^–2k^ is acute, is __________. [2024]

Q 8 : If a→ is a non-zero vector such its projections on the vectors 2i^–j^+2k^, i^+2j^–2k^ and k^ are equal, then a unit vector along a→ is: [2025]

Q 9 : Consider two vectors u→=3i^–j^ and v→=2i^+j^–λk^, λ>0. The angle between them is given by cos–1(527). Let v→=v→1+v→2, where v→1 is parallel to u→ and v→2 is perpendicular to u→. Then the value |v→1|2+|v→2|2 is equal to [2025]

Q 10 : Let the angle θ,0<θ<π2 between two unit vectors a^ and b^ be sin–1(659). If the vector c→=3a^+6b^+9(a^×b^), then the value of 9(c→·a^)–3(c→·b^) is [2025]

Q 11 : Let a→ and b→ be the vectors of the same magnitude such that |a→+b→|+|a→–b→||a→+b→|–|a→–b→|=2+1. Then |a→+b→|2|a→|2 is : [2025]

Q 12 : Let a→=i^+2j^+k^ and b→=2i^+j^–k^. Let c^ be a unit vector in the plane of the vectors a→ and b→ and be perpendicular to a→. Then such a vector c^ is : [2025]

Q 13 : Let a→ and b→ be two unit vectors such that the angle between them is π3. If λa→+2b→ and 3a→–λb→ are perpendicular to each other, then the number of values of λ in [–1, 3] is : [2025]

Q 14 : Let the arc AC of a circle subtend a right angle at the centre O. If the point B on the arc AC, divides the are AC such that length of arc ABlength of arc BC=15, and OC→=αOA→+βOB→, then α+2(3–1)β is equal to [2025]

Q 15 : Let a→=i^+j^+k^, b→=3i^+2j^–k^, c→=μj^+uk^ and d^ be a unit vector such that a→×d^=b→×d^ and c→·d^=1. If c→ is perpendicular to a→, then |3λd^+μc→|2 is equal to __________. [2025]

Q 16 : An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If OP→=u→, OR→=v→ and OQ→=αu→+βv→, then α,β2 are the roots of the equation [2023]

Q 17 : Let α→=4i^+3j^+5k^ and β→=i^+2j^-4k^. Let β→1 be parallel to α→ and β→2 be perpendicular to α→. If β→=β→1+β→2, then the value of 5β→2·(i^+j^+k^) is [2023]

Q 18 : The vector a→=-i^+2j^+k^ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is b→. Then the projection of 3a→+2 b→ on c→=5i^+4j^+3k^ is [2023]

Q 19 : Let a→=2i^+j^+k^, and b→ and c→ be two non-zero vectors such that |a→+b→+c→|=|a→+b→-c→| and b→·c→=0. Consider the following two statements: (A) |a→+λc→|≥|a→| for all λ∈R. (B) a→ and c→ are always parallel. Then, [2023]

Q 20 : Let c→ and d→ be vectors such that |c→+d→|=29 and c→×(2i^+3j^+4k^)=(2i^+3j^+4k^)×d→. If λ1,λ2(λ1>λ2) are the possible values of (c→+d→)·(-7i^+2j^+3k^), then the equation K2x2+(K2-5K+λ1)xy+(3K+λ22)y2-8x+12y+λ2=0 represents a circle, for k equal to: [2026]

Q 21 : Let P be a point in the plane of the vectors AB→=3i^+j^-k^ and AC→=i^-j^+3k^ such that P is equidistant from the lines AB and AC. If |AP→|=52, then the area of the triangle ABP is: [2026]

Q 22 : Let a→=2i^-j^-k^, b→=i^+3j^-k^ and c→=2i^+j^+3k^. Let v→ be the vector in the plane of the vectors a→ and b→, such that the length of its projection on the vector c→ is 114. Then |v→| is equal to [2026]

Q 23 : Let a→=i^-2j^+3k^, b→=2i^+j^-k^, c→=λi^+j^+k^ and v→=a→×b→. If v→·c→=11 and the length of the projection of b→ on c→, is p, then 9p2 is equal to. [2026]

Q 24 : Let a→=2i^-j^+k^ and b→=λj^+2k^ , λ∈Z be two vectors. Let c→=a→×b→ and d→ be a vector of magnitude 2 in yz-plane. If |c→|=53, then the maximum possible value of (c→·d→)2 is equal to: [2026]

Q 25 : Let a vector a→=2i^-j^+λk^, λ>0, make an obtuse angle with the vector b→=-λ2i^+42j^+42k^ and an angle θ, π6<θ<π2, with the positive z-axis. If the set of all possible values of λ is (α,β)-{γ}, then α+β+γ is equal to _______. [2026]

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Q 1 :

Let a unit vector which makes an angle of 60° with $2 \hat{i} + 2 \hat{j} - \hat{k}$ and an angle of 45° with $\hat{i} - \hat{k}$ be $\vec{C}$ . Then $\vec{C} + (- \frac{1}{2} \hat{i} + \frac{1}{3 \sqrt{2}} \hat{j} - \frac{\sqrt{2}}{3} \hat{k})$ is [2024]

Q 4 :

Let $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ , $\vec{b} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ and $\vec{c} = 3 \hat{i} - \hat{j} + λ \hat{k}$ be three vectors. Let $\vec{r}$ be a unit vector along $\vec{b} + \vec{c}$ . If $\vec{r} \cdot \vec{a} = 3$ , then $3 λ$ is equal to : [2024]

Q 5 :

Consider a $△ A B C$ where A(1, 3, 2), B(–2, 8, 0) and C(3, 6, 7). If the angle bisector of $∠ B A C$ meets the line BC at D, then the length of the projection of the vector $\vec{A D}$ on the vector $\vec{A C}$ IS : [2024]

Q 7 :

The Least positive integral value of $α$ , for which the angle between the vectors $α \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $α \hat{i} + 2 α \hat{j} - 2 \hat{k}$ is acute, is __________. [2024]

Q 8 :

If $\vec{a}$ is a non-zero vector such its projections on the vectors $2 \hat{i} - \hat{j} + 2 \hat{k}$ , $\hat{i} + 2 \hat{j} - 2 \hat{k}$ and $\hat{k}$ are equal, then a unit vector along $\vec{a}$ is: [2025]

Q 11 :

Let $\vec{a}$ and $\vec{b}$ be the vectors of the same magnitude such that $\frac{| \vec{a} + \vec{b} | + | \vec{a} - \vec{b} |}{| \vec{a} + \vec{b} | - | \vec{a} - \vec{b} |} = \sqrt{2} + 1$ . Then $\frac{| \vec{a} + \vec{b} |^{2}}{| \vec{a} |^{2}}$ is : [2025]

Q 12 :

Let $\vec{a} = \hat{i} + 2 \hat{j} + \hat{k}$ and $\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}$ . Let $\hat{c}$ be a unit vector in the plane of the vectors $\vec{a}$ and $\vec{b}$ and be perpendicular to $\vec{a}$ . Then such a vector $\hat{c}$ is : [2025]

Q 13 :

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{π}{3}$ . If $λ \vec{a} + 2 \vec{b}$ and $3 \vec{a} - λ \vec{b}$ are perpendicular to each other, then the number of values of $λ$ in [–1, 3] is : [2025]

Q 16 :

An arc PQ of a circle subtends a right angle at its centre O. The mid point of the arc PQ is R. If $\vec{O P} = \vec{u}$ , $\vec{O R} = \vec{v}$ and $\vec{O Q} = α \vec{u} + β \vec{v}$ , then $α, β^{2}$ are the roots of the equation [2023]

Q 18 :

The vector $\vec{a} = - \hat{i} + 2 \hat{j} + \hat{k}$ is rotated through a right angle, passing through the $y$ -axis in its way and the resulting vector is $\vec{b}$ . Then the projection of $3 \vec{a} + \sqrt{2} \vec{b}$ on $\vec{c} = 5 \hat{i} + 4 \hat{j} + 3 \hat{k}$ is [2023]

Q 21 :

Let P be a point in the plane of the vectors $\vec{A B} = 3 \hat{i} + \hat{j} - \hat{k} and \vec{A C} = \hat{i} - \hat{j} + 3 \hat{k}$ such that P is equidistant from the lines AB and AC. If $| \vec{A P} |$ $= \frac{\sqrt{5}}{2},$ then the area of the triangle ABP is: [2026]