let α→=i^+2j^−k^,β→=2i^−j^+3k^,γ→=2i^+j^+6k^ . if α→ and β→ are perpendicular c¯ then the angle between a→ and c→ is

since α→ and β→ are both perpendicular to vector ∴|δ→|=75λ2=75×425=23δ→∴δ→=λ(α→×β→)=λi^j^k^12−12−13=5λi^−5j^−5λk^ Now, δ→⋅γ→=10⇒(5λi^−5λj^−5λk^)⋅(2i^+j^+6k^)=10⇒10λ−5λ−30λ=10⇒λ=−25∴|δ→|=75λ2=75×425=23

let α→=i^+2j^−k^,β→=2i^−j^+3k^,γ→=2i^+j^+6k^ . if α→ and β→ are perpendicular c¯ then the angle between a→ and c→ is

since α→ and β→ are both perpendicular to vector ∴|δ→|=75λ2=75×425=23δ→∴δ→=λ(α→×β→)=λi^j^k^12−12−13=5λi^−5j^−5λk^ Now, δ→⋅γ→=10⇒(5λi^−5λj^−5λk^)⋅(2i^+j^+6k^)=10⇒10λ−5λ−30λ=10⇒λ=−25∴|δ→|=75λ2=75×425=23

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let $\vec{α} = \hat{i} + 2 \hat{j} - \hat{k}, \vec{β} = 2 \hat{i} - \hat{j} + 3 \hat{k}, \vec{γ} = 2 \hat{i} + \hat{j} + 6 \hat{k}$ . if $\vec{α} and \vec{β}$ are perpendicular $\bar{c}$ then the angle between $\vec{a} and \vec{c}$ is

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$\frac{1}{\sqrt{3}}$

$\sqrt{3}$

$2 \sqrt{3}$

$\frac{\sqrt{3}}{2}$

answer is C.

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Detailed Solution

since $\vec{α} and \vec{β}$ are both perpendicular to vector $∴ | \vec{δ} | = \sqrt{75 λ^{2}} = \sqrt{75 \times \frac{4}{25}} = 2 \sqrt{3}$ $\vec{δ}$
$\begin{aligned} ∴ \vec{δ} = λ (\vec{α} \times \vec{β}) \\ = λ |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & - 1 \\ 2 & - 1 & 3 \end{array}| = 5 λ \hat{i} - 5 \hat{j} - 5 λ \hat{k} \\ Now, \vec{δ} \cdot \vec{γ} = 10 \\ \Rightarrow (5 λ \hat{i} - 5 λ \hat{j} - 5 λ \hat{k}) \cdot (2 \hat{i} + \hat{j} + 6 \hat{k}) = 10 \\ \Rightarrow 10 λ - 5 λ - 30 λ = 10 \Rightarrow λ = - \frac{2}{5} \end{aligned}$
$∴ | \vec{δ} | = \sqrt{75 λ^{2}} = \sqrt{75 \times \frac{4}{25}} = 2 \sqrt{3}$