Let $(\alpha, \beta, \gamma)$ be the mirror image of the point $(2,3,5)$ in the line $\frac{x-1}{2}=

Question

Let $(\alpha, \beta, \gamma)$ be the mirror image of the point $(2,3,5)$ in the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$. Then $2 \alpha+3 \beta+4 \gamma$ is equal to\

Professor PC Thomas Classes · Answer

(d) : Given equation of the line is\ $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$\ Let $A(2,3,5)$ be the given point and let $B$ be the foot of the perpendicular drawn from the point $A$ on the line.\ $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\lambda, \lambda \in R$\ $\Rightarrow \quad x=2 \lambda+1, y=3 \lambda+2, z=4 \lambda+3$\ $	herefore \quad B \equiv(2 \lambda+1,3 \lambda+2,4 \lambda+3)$\ Direction ratios of $A B$ are $2 \lambda-1,3 \lambda-1$, 4λ-2.\ Now, $A B$ is perpendicular to the given line.\ $	herefore \quad 2(2 \lambda-1)+3(3 \lambda-1)+4(4 \lambda-2)=0$\ $\Rightarrow \quad 4 \lambda-2+9 \lambda-3+16 \lambda-8=0$\ $\Rightarrow \quad 29 \lambda-13=0 \Rightarrow \lambda=\frac{13}{29}$\ $	herefore \quad$ Coordinates of $B$ are $\left(\frac{55}{29}, \frac{97}{29}, \frac{139}{29}ight)$\ Now, using midpoint formula,\ $\frac{55}{29}=\frac{2+\alpha}{2} ; \frac{97}{29}=\frac{3+\beta}{2} ; \frac{139}{29}=\frac{5+\gamma}{2}$\ $\Rightarrow \quad \alpha=\frac{52}{29}, \beta=\frac{107}{29}, \gamma=\frac{133}{29}$\ $	herefore \quad 2 \alpha+3 \beta+4 \gamma$\ $=2 	imes \frac{52}{29}+\frac{3 	imes 107}{29}+\frac{4 	imes 133}{29}=\frac{957}{29}=33$\