Let the line L_{1} be parallel to the vector | vedclass

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(B) The equation of line $L_{1}$ is $\frac{x-2}{-3} = \frac{y-6}{2} = \frac{z-7}{4} = \lambda_{1}$. Thus, any point $C$ on $L_{1}$ is $(-3\lambda_{1}+

Vedclass · Accepted Answer

$290$

Vedclass · Answer

(B) The equation of line $L_{1}$ is $\frac{x-2}{-3} = \frac{y-6}{2} = \frac{z-7}{4} = \lambda_{1}$. Thus, any point $C$ on $L_{1}$ is $(-3\lambda_{1}+2, 2\lambda_{1}+6, 4\lambda_{1}+7)$.The equation of line $L_{2}$ is $\frac{x-4}{2} = \frac{y-3}{1} = \frac{z-5}{3} = \lambda_{2}$. Thus, any point $D$ on $L_{2}$ is $(2\lambda_{2}+4, \lambda_{2}+3, 3\lambda_{2}+5)$.The vector $\overrightarrow{CD} = (2\lambda_{2}+3\lambda_{1}+2)\hat{i} + (\lambda_{2}-2\lambda_{1}-3)\hat{j} + (3\lambda_{2}-4\lambda_{1}-2)\hat{k}$.Since $L_{3}$ is parallel to $-3\hat{i}+5\hat{j}+16\hat{k}$, the components of $\overrightarrow{CD}$ must be proportional to $(-3, 5, 16)$:$\frac{2\lambda_{2}+3\lambda_{1}+2}{-3} = \frac{\lambda_{2}-2\lambda_{1}-3}{5} = \frac{3\lambda_{2}-4\lambda_{1}-2}{16} = k$.Solving these equations, we get $\lambda_{1} = -3$ and $\lambda_{2} = 2$.Substituting these values, we get $C = (11, 0, -5)$ and $D = (8, 5, 11)$.Then $\overrightarrow{CD} = (8-11)\hat{i} + (5-0)\hat{j} + (11-(-5))\hat{k} = -3\hat{i} + 5\hat{j} + 16\hat{k}$.Therefore, $|\overrightarrow{CD}|^2 = (-3)^2 + 5^2 + 16^2 = 9 + 25 + 256 = 290$.

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