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 $\mathrm{x}$-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
$\mathrm{L}_{1}=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda$
$\mathrm{M}(\lambda, 1+2 \lambda, 2+3 \lambda)$
$\overrightarrow{\mathrm{PM}}=(\lambda-1) \hat{\mathrm{i}}+(1+2 \lambda) \hat{\mathrm{j}}+(3 \lambda-5) \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{PM}}$ is perpendicular to line $\mathrm{L}_{1}$
$\overrightarrow{\mathrm{PM}} \overrightarrow{\mathrm{b}}=0 n(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})$
$\Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0$
$14 \lambda=14 \Rightarrow \lambda=1$
$\therefore \mathrm{M}=(1,3,5)$
$\overrightarrow{\mathrm{Q}}=2 \overrightarrow{\mathrm{M}}-\overrightarrow{\mathrm{P}}[\mathrm{M}$ is midpoint of $\overrightarrow{\mathrm{P}} & \overrightarrow{\mathrm{Q}}]$
$\overrightarrow{\mathrm{Q}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}-\hat{\mathrm{i}}-7 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{Q}}=\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$
$\therefore(\alpha, \beta, \gamma)=(1,6,3)$
Required line having direction $\operatorname{cosine}(l, \mathrm{~m}, \mathrm{n})$
$l^{2}+m^{2}+n^{2}=1$
$\Rightarrow l^{2}+\left(-\frac{1}{2}\right)^{2}+\left(-\frac{1}{\sqrt{2}}\right)^{2}=1$
$l^{2}=\frac{1}{4}$
$\therefore l=\frac{1}{2}$ [Line make acute angle with $\mathrm{x}$-axis]
Equation of line passing through $(1,6,3)$ will be
$\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mu\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}\right)$
Option 3 satisfying for $\mu=4$
