Let a→ be a non-zero vector parallel to the line of intersection of the two planes described by i^+j^, i^+k^ and i^−j^, j^−k^. If θ is the angle between the vector a→ and the vector b→=2i^−2j^+k^ and a→ . b→=6, then the ordered pair (θ, |a→ × b→|) is equal to

D.r’s perpendicular to plane containingi^ +j^ and i^ +k^=|i^j^k^110101|=(1, −1, −1) D.r’s perpendicular to plane containing i^−j^, j^−k^ is i^j^k^1−1001−1=i^+j^+k^D.r’s a→ , parallel to the line of intersection of the above two planes is =|i^j^k^1111−1−1| =(0, 1, −1)∴ a→ = λ (j^−k^) a→ . b→ = 6 ⇒ −3λ=6 λ=−2 ∴ a→=−2j^+2k^ Angle between a→ & b→⇒ cosθ=a→ . b→|a→| |b→|=622× 3=12∴ θ=π4a→ × b→=|i^j^k^0−222−21| = 2i^+4j^+4k^∴ |a→ × b→ | = 6

Let a→ be a non-zero vector parallel to the line of intersection of the two planes described by i^+j^, i^+k^ and i^−j^, j^−k^. If θ is the angle between the vector a→ and the vector b→=2i^−2j^+k^ and a→ . b→=6, then the ordered pair (θ, |a→ × b→|) is equal to

D.r’s perpendicular to plane containingi^ +j^ and i^ +k^=|i^j^k^110101|=(1, −1, −1) D.r’s perpendicular to plane containing i^−j^, j^−k^ is i^j^k^1−1001−1=i^+j^+k^D.r’s a→ , parallel to the line of intersection of the above two planes is =|i^j^k^1111−1−1| =(0, 1, −1)∴ a→ = λ (j^−k^) a→ . b→ = 6 ⇒ −3λ=6 λ=−2 ∴ a→=−2j^+2k^ Angle between a→ & b→⇒ cosθ=a→ . b→|a→| |b→|=622× 3=12∴ θ=π4a→ × b→=|i^j^k^0−222−21| = 2i^+4j^+4k^∴ |a→ × b→ | = 6

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Let $\vec{a}$ be a non-zero vector parallel to the line of intersection of the two planes described by $\hat{i} + \hat{j}, \hat{i} + \hat{k}$ and $\hat{i} - \hat{j}, \hat{j} - \hat{k}$ . If $θ$ is the angle between the vector $\vec{a}$ and the vector $\vec{b} = 2 \hat{i} - 2 \hat{j} + \hat{k}$ and $\vec{a} . \vec{b} = 6$ , then the ordered pair $(θ, | \vec{a} \times \vec{b} |)$ is equal to

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$(\frac{π}{4}, 6)$

$(\frac{π}{4}, 3 \sqrt{6})$

$(\frac{π}{3}, 3 \sqrt{6})$

$(\frac{π}{3}, 6)$

answer is B.

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Detailed Solution

D.r’s perpendicular to plane containing
$\hat{i} + \hat{j} and \hat{i} + \hat{k} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{matrix} |$ $= (1, - 1, - 1)$

D.r’s perpendicular to plane containing $\hat{i} - \hat{j}, \hat{j} - \hat{k}$ is

$|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & - 1 & 0 \\ 0 & 1 & - 1 \end{array}| = \hat{i} + \hat{j} + \hat{k}$
D.r’s $\vec{a}$ , parallel to the line of intersection of the above two planes is
$= | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & - 1 & - 1 \end{matrix} | = (0, 1, - 1) ∴ \vec{a} = λ (\hat{j} - \hat{k})$
$\vec{a} . \vec{b} = 6 \Rightarrow - 3 λ = 6 λ = - 2$
$∴ \vec{a} = - 2 \hat{j} + 2 \hat{k}$

Angle between $\vec{a} & \vec{b} \Rightarrow cosθ = \frac{\vec{a} . \vec{b}}{| \vec{a} | | \vec{b} |}$
$= \frac{6}{2 \sqrt{2} \times 3} = \frac{1}{\sqrt{2}}$

$∴ θ = \frac{π}{4}$

$\vec{a} \times \vec{b} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & - 2 & 2 \\ 2 & - 2 & 1 \end{matrix} | = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$
$∴ | \vec{a} \times \vec{b} | = 6$