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(A) The line $L$ passes through points $A(\hat{i}-9 \hat{k})$ and $B(7 \hat{j}+\hat{k})$. The direction vector of the line is $\vec{b} = B - A = (7 \h

Vedclass · Accepted Answer

$\frac{8 \sqrt{2}}{15}$

Vedclass · Answer

(A) The line $L$ passes through points $A(\hat{i}-9 \hat{k})$ and $B(7 \hat{j}+\hat{k})$. The direction vector of the line is $\vec{b} = B - A = (7 \hat{j} + \hat{k}) - (\hat{i} - 9 \hat{k}) = -\hat{i} + 7 \hat{j} + 10 \hat{k}$.The plane $\pi$ passes through $6 \hat{i} + \hat{j}$ and is perpendicular to $\vec{n} = \hat{i} + \hat{j} + \hat{k}$.The angle $	heta$ between a line with direction vector $\vec{b}$ and a plane with normal vector $\vec{n}$ is given by $\sin 	heta = \left| \frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|} ight|$.Calculate the dot product: $\vec{b} \cdot \vec{n} = (-1)(1) + (7)(1) + (10)(1) = -1 + 7 + 10 = 16$.Calculate the magnitudes: $|\vec{b}| = \sqrt{(-1)^2 + 7^2 + 10^2} = \sqrt{1 + 49 + 100} = \sqrt{150} = 5 \sqrt{6}$.$|\vec{n}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$.Substitute into the formula: $\sin 	heta = \left| \frac{16}{(5 \sqrt{6})(\sqrt{3})} ight| = \frac{16}{5 \sqrt{18}} = \frac{16}{5 	imes 3 \sqrt{2}} = \frac{16}{15 \sqrt{2}}$.Rationalizing the denominator: $\sin 	heta = \frac{16 \sqrt{2}}{15 	imes 2} = \frac{8 \sqrt{2}}{15}$.

Let $L$ be the line passing through the points $\hat{i}-9 \hat{k}$ and $7 \hat{j}+\hat{k}$ and $\pi$ be the plane passing through the point $6 \hat{i}+\hat{j}$ and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$. If $\theta$ is the angle between $L$ and $\pi$,then $\sin \theta=$

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