Consider the lines L_1: frac{x+1}{3}=frac{y+2 | vedclass

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(B, D, C) $1.$ The direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{v_2} = \hat{i} + 2\hat{j} + 3\hat{k}

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$(B, D, C)$

Vedclass · Answer

(B, D, C) $1.$ The direction vectors of $L_1$ and $L_2$ are $\vec{v_1} = 3\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{v_2} = \hat{i} + 2\hat{j} + 3\hat{k}$.The vector perpendicular to both is $\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & 2 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(3-4) - \hat{j}(9-2) + \hat{k}(6-1) = -\hat{i} - 7\hat{j} + 5\hat{k}$.The magnitude is $|\vec{n}| = \sqrt{(-1)^2 + (-7)^2 + 5^2} = \sqrt{1 + 49 + 25} = \sqrt{75} = 5\sqrt{3}$.The unit vector is $\frac{-\hat{i} - 7\hat{j} + 5\hat{k}}{5\sqrt{3}}$. Thus,option $(B)$ is correct.$2.$ The shortest distance between lines passing through $A(-1, -2, -1)$ and $B(2, -2, 3)$ with directions $\vec{v_1}, \vec{v_2}$ is $d = \frac{|(\vec{AB}) \cdot (\vec{v_1} 	imes \vec{v_2})|}{|\vec{v_1} 	imes \vec{v_2}|}$.$\vec{AB} = (2 - (-1))\hat{i} + (-2 - (-2))\hat{j} + (3 - (-1))\hat{k} = 3\hat{i} + 0\hat{j} + 4\hat{k}$.$d = \frac{|(3\hat{i} + 0\hat{j} + 4\hat{k}) \cdot (-\hat{i} - 7\hat{j} + 5\hat{k})|}{5\sqrt{3}} = \frac{|-3 + 0 + 20|}{5\sqrt{3}} = \frac{17}{5\sqrt{3}}$. Thus,option $(D)$ is correct.$3.$ The plane passes through $(-1, -2, -1)$ with normal $\vec{n} = -\hat{i} - 7\hat{j} + 5\hat{k}$.Equation: $-1(x+1) - 7(y+2) + 5(z+1) = 0 \Rightarrow -x - 1 - 7y - 14 + 5z + 5 = 0 \Rightarrow x + 7y - 5z + 10 = 0$.The distance from $(1, 1, 1)$ is $d = \frac{|1 + 7(1) - 5(1) + 10|}{\sqrt{1^2 + 7^2 + (-5)^2}} = \frac{|1 + 7 - 5 + 10|}{\sqrt{1 + 49 + 25}} = \frac{13}{\sqrt{75}}$. Thus,option $(C)$ is correct.

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