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]

Question

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]

Smart Achievers · Accepted Answer

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

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

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

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

Solution

1 14

2 10

3 $\frac{23}{2}$

4 $\frac{25}{2}$

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Solution

Q. 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]

1 14

2 10

3 232

4 252

Ans. (1) We have, u→=3i^–j^, v→=2i^+j^–λk^ Now, the angle between u→ and v→ is ⇒ cos θ=u→·v→|u||v| ⇒ 5105+λ2=527 ⇒ λ2=9 ⇒ λ=3 (∵ λ>0) Now, v→=v→1+v→2 ⇒ |v→|2=v→12+v→22+2v→1·v→2 ⇒ 14=v→12+v→22+0 (∵ v→1⊥v→2) ⇒ v→12+v→22=14.

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3 $\frac{23}{2}$

4 $\frac{25}{2}$