Let the line L_1 be parallel to the vector -3i+2j+4k and pass through the point (2,6,7), a…

Question

Let the line ${L}_{1}$ be parallel to the vector $-3\hat{i}+2\hat{j}+4\hat{k}$ and pass through the point $(2,6,7)$, and the line ${L}_{2}$ be parallel to the vector $2\hat{i}+\hat{j}+3\hat{k}$ and pass through the point $(4,3,5)$. If the line ${L}_{3}$ is parallel to the vector $-3\hat{i}+5\hat{j}+16\hat{k}$ and intersects the lines ${L}_{1}$ and ${L}_{2}$ at the points $C$ and $D$, respectively, then ${|\overset{	o \leftarrow }{CD}|}^{2}$ is equal to:

Accepted Answer

$\begin{gathered} {L}_{1}:\frac{x-2}{-3}=\frac{y-6}{2}=\frac{z-7}{4}=t \ {L}_{2}:\frac{x-4}{2}=\frac{y-3}{1}=\frac{z-5}{3}=s \end{gathered}$

Any point on ${L}_{1}$ is $C=(-3t+2,2t+6,4t+7)$

Any point on ${L}_{2}$ is $D=(2s+4,s+3,3s+5)$

d.r's of $\overset{	o \leftarrow }{CD}=(2s+3t+2,s-2t-3,3s-4t-2)$

CD is parallel to $-3i+5j+16k$

$\begin{gathered} \frac{2s+3t+2}{-3}=\frac{s-2t-3}{5}=\frac{3s-4t-2}{16} \ 13s+9t+1=0\ldots .(i) \ 41s+36t+26=0\ldots .(2) \end{gathered}$

Solve (i) \& (2) $s=2;t=-3$

$\begin{gathered} C=(11,0,-5) \ D=(8,5,11) \ C{D}^{2}=290 \end{gathered}$

Solutions · 1