Let Q(a, b, c) be the image of the point P(3, 2, 1) in the line (x-1)/(1) = (y)/(2) = (z-1…

Question

Let Q(a, b, c) be the image of the point P(3, 2, 1) in the line $\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}$. Then the distance of Q from the line $\frac{x-9}{3} = \frac{y-9}{2} = \frac{z-5}{-2}$ is

Accepted Answer

Point $P(3,2,1)$

Let Line ${L}_{1}:\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ and ${L}_{2}:\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$

Direction vector of ${L}_{1}:(1,2,1)$

Let the first line be ${L}_{1}:\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}=\lambda $.

Any general point $M$ on this line is $M(\lambda +1,2\lambda ,\lambda +1)$.

If $M$ is the foot of the perpendicular from $P(3,2,1)$ to ${L}_{1}$, the direction ratio of $PM$ is $(\lambda -2,2\lambda -2,\lambda )$.

Since $PM\perp {L}_{1}$ : Since $PM\perp {L}_{1}$ :

$1(\lambda -2)+2(2\lambda -2)+1(\lambda )=0$

$\Rightarrow$ $\lambda -2+4\lambda -4+\lambda =0\implies 6\lambda =6\implies \lambda =1$

The foot of the perpendicular $M$ is $(2,2,2)$. Since $M$ is the midpoint of $PQ$, where $Q$ is $(a,b,c)$ :

$\frac{a+3}{2}=2\implies a=1$

$\frac{b+2}{2}=2\implies b=2$

$\frac{c+1}{2}=2\implies c=3$

So, $Q=(1,2,3)$.

Find the distance of $Q(1,2,3)$ from the second line

The second line is ${L}_{2}:\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}=\mu $.

Let $A(9,9,5)$ be a point on ${L}_{2}$ and $d=3\hat{i}+2\hat{j}-2\hat{k}$ be its direction vector.

The vector $AQ=(1-9)\hat{i}+(2-9)\hat{j}+(3-5)\hat{k}=-8\hat{i}-7\hat{j}-2\hat{k}$.

The distance $D$ of a point from a line is given by:

Distance of point from line :

$D=\frac{|\overset{	o \leftarrow }{AQ}	imes \vec{d}|}{|\vec{d}|}$

Cross product

$\begin{gathered} \overset{	o \leftarrow }{AQ}	imes \vec{d}=|\begin{array}{ccc}\hat{i} \quad  \hat{j} \quad  \hat{k} \ -8 \quad  -7 \quad  -2 \ 3 \quad  2 \quad  -2\end{array}| \ =\hat{i}[(-7)(-2)-(-2)(2)]-\hat{j}[(-8)(-2)-(-2)(3)]+\hat{k}[(-8)(2)-(-7)(3)] \ =\hat{i}(14+4)-\hat{j}(16+6)+\hat{k}(-16+21) \ \overset{	o \leftarrow }{AQ}	imes \vec{d}=18\hat{i}-22\hat{j}+5\hat{k} \end{gathered}$

Magnitudes

$\begin{gathered} |\overset{	o \leftarrow }{AQ}	imes \vec{d}|=\sqrt{{18}^{2}+(-22{)}^{2}+{5}^{2}}=\sqrt{833} \ |\vec{d}|=\sqrt{{3}^{2}+{2}^{2}+(-2{)}^{2}}=\sqrt{17} \end{gathered}$

Distance of point from line

$D=\frac{|\overset{	o \leftarrow }{AQ}	imes \vec{d}|}{|\vec{d}|}=\sqrt{\frac{833}{17}}=\sqrt{49}=7$

Solutions · 1