Consider two vectors and , . The angle between them is given by . Let , where is parallel to and is perpendicular to . Then the value is equal to [2025]

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Solution

Q.
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]

1 14

2 10

3 $\frac{23}{2}$

4 $\frac{25}{2}$

Ans.
(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, $\vec{v} = {\vec{v}}_{1} + {\vec{v}}_{2}$

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

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

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

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