Three Dimensional Geometry Question 7
Question 7 - 2024 (27 Jan Shift 2)
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 making angles $\frac{2 \pi}{3}$ and $\frac{3 \pi}{4}$ with $y$-axis and $z$-axis respectively and an acute angle with $x$-axis ?
(1) $(1,-2,1+\sqrt{2})$
(2) $(1,2,1-\sqrt{2})$
(3) $(3,4,3-2 \sqrt{2})$
(4) $(3,-4,3+2 \sqrt{2})$
Show Answer
Answer (3)
Solution
$L _1=\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$
$M(\lambda, 1+2 \lambda, 2+3 \lambda)$
$\overrightarrow{PM}=(\lambda-1) \hat{i}+(1+2 \lambda) \hat{j}+(3 \lambda-5) \hat{k}$
$\overrightarrow{PM}$ is perpendicular to line $L _1$
$\overrightarrow{PM} \overrightarrow{b}=0 n(\overrightarrow{b}=\hat{i}+2 \hat{j}+3 \hat{k})$
$\Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0$
$14 \lambda=14 \Rightarrow \lambda=1$
$\therefore M=(1,3,5)$
$\overrightarrow{Q}=2 \overrightarrow{M}-\overrightarrow{P}[M$ is midpoint of $\overrightarrow{P} \& \overrightarrow{Q}]$
$\overrightarrow{Q}=2 \hat{i}+6 \hat{j}+10 \hat{k}-\hat{i}-7 \hat{k}$
$\overrightarrow{Q}=\hat{i}+6 \hat{j}+3 \hat{k}$
$\therefore(\alpha, \beta, \gamma)=(1,6,3)$
Required line having direction $\operatorname{cosine}(l, m, n)$
$l^{2}+m^{2}+n^{2}=1$
$\Rightarrow l^{2}+\left(-\frac{1}{2}\right)^{2}+\left(-\frac{1}{\sqrt{2}}\right)^{2}=1$
$l^{2}=\frac{1}{4}$
$\therefore l=\frac{1}{2}$ [Line make acute angle with $x$-axis]
Equation of line passing through $(1,6,3)$ will be
$\overrightarrow{r}=(\hat{i}+6 \hat{j}+3 \hat{k})+\mu\left(\frac{1}{2} \hat{i}-\frac{1}{2} \hat{j}-\frac{1}{\sqrt{2}} \hat{k}\right)$
Option 3 satisfying for $\mu=4$
