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]

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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]

0%

(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 $α = 5$ .

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]

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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 (14 cos θ)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]

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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]