The square of the distance of the point P(5,6,7) from the line (x-2)/(2)=(y-5)/(3)=(z-2)/(…

Question

The square of the distance of the point $P(5,6,7)$ from the line $\frac{x-2}{2}=\frac{y-5}{3}=\frac{z-2}{4}$ is equal to:

Accepted Answer

Write the given line in vector form.

From

$\frac{x-2}{2}=\frac{y-5}{3}=\frac{z-2}{4},$

a point on the line is

$A(2,5,2)$

and its direction vector is

$\vec{b}=2\hat{i}+3\hat{j}+4\hat{k}.$

Given point is

$P(5,6,7).$

Now,

$\overset{	o \leftarrow }{AP}=(5-2)\hat{i}+(6-5)\hat{j}+(7-2)\hat{k}=3\hat{i}+\hat{j}+5\hat{k}.$

The distance $d$ of point $P$ from the line is given by

$d=\frac{|\overset{	o \leftarrow }{AP}	imes \vec{b}|}{|\vec{b}|}.$

Since we need the square of the distance,

${d}^{2}=\frac{|\overset{	o \leftarrow }{AP}	imes \vec{b}{|}^{2}}{|\vec{b}{|}^{2}}.$

Now calculate the cross product:

$\overset{	o \leftarrow }{AP}	imes \vec{b}=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 3 \quad  1 \quad  5 \ 2 \quad  3 \quad  4 \end{vmatrix}$

$=\hat{i}(1\cdot 4-5\cdot 3)-\hat{j}(3\cdot 4-5\cdot 2)+\hat{k}(3\cdot 3-1\cdot 2)$

$=\hat{i}(4-15)-\hat{j}(12-10)+\hat{k}(9-2)$

$=-11\hat{i}-2\hat{j}+7\hat{k}.$

So,

$|\overset{	o \leftarrow }{AP}	imes \vec{b}{|}^{2}=(-11{)}^{2}+(-2{)}^{2}+{7}^{2}=121+4+49=174.$

Also,

$|\vec{b}{|}^{2}={2}^{2}+{3}^{2}+{4}^{2}=4+9+16=29.$

Therefore,

${d}^{2}=\frac{174}{29}=6.$

So, the square of the distance is

$6$

Hence, the correct option is Option C.

Solutions · 1