Let the image of the point (1,0,7) in the line [(x)/(1)=(y-1)/(2)=(z-2)/(3)] be the point …

Question

Let the image of the point $(1,0,7)$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$ be the point $$(\alpha, \beta, \gamma)$$. Then which one of the following points lies on the line passing through $$(\alpha, \beta, \gamma)$$ and...

Accepted Answer

${L}_{1}=\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda $

$\begin{gathered} M(\lambda ,1+2\lambda ,2+3\lambda ) \ \overset{	o \leftarrow }{PM}=(\lambda -1)\hat{i}+(1+2\lambda )\hat{j}+(3\lambda -5)\hat{k} \end{gathered}$

$\overset{	o \leftarrow }{PM}$ is perpendicular to line ${L}_{1}$

$\begin{gathered} \overset{	o \leftarrow }{PM}\overset{	o }{b}=0\, (\overset{	o }{b}=\hat{i}+2\hat{j}+3\hat{k}) \ \Rightarrow \lambda -1+4\lambda +2+9\lambda -15=0 \ 14\lambda =14\Rightarrow \lambda =1 \ 	herefore M=(1,3,5) \ \overset{	o }{Q}=2\overset{	o }{M}-\overset{	o }{P}[M\overset{	o }{P}\overset{	o }{Q}] \end{gathered}$

$\begin{gathered} \overset{	o }{Q}=2\hat{i}+6\hat{j}+10\hat{k}-\hat{i}-7\hat{k} \ \overset{	o }{Q}=\hat{i}+6\hat{j}+3\hat{k} \ 	herefore (\alpha ,\beta ,\gamma )=(1,6,3) \end{gathered}$

Required line having direction cosine $(l,m,n)$

$\begin{gathered} {l}^{2}+{m}^{2}+{n}^{2}=1 \ \Rightarrow {l}^{2}+{(-\frac{1}{2})}^{2}+{(-\frac{1}{\sqrt{2}})}^{2}=1 \end{gathered}$

${l}^{2}=\frac{1}{4}$

$	herefore l=\frac{1}{2}$ [Line make acute angle with $x$-axis]

Equation of line passing through $(1,6,3)$ will be

$\overset{	o }{r}=(\hat{i}+6\hat{j}+3\hat{k})+\mu (\frac{1}{2}\hat{i}-\frac{1}{2}\hat{j}-\frac{1}{\sqrt{2}}\hat{k})$

Option (3) satisfying for $\mu =4$

Solutions · 1