Let ( , , ) be the image of the point (8,5,7) in the line (x-1)/(2)=(y+1)/(3)=(z-2)/(5).

Question

Let $(\alpha ,\beta ,\gamma )$ be the image of the point $(8,5,7)$ in the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{5}$. Then $\alpha +\beta +\gamma $ is equal to :

Accepted Answer

$$(x, y, z) \equiv(2 \lambda+1,3 \lambda+1,5 \lambda+2)$$

DR of $P Q$ :

$$(2 \lambda-7,3 \lambda-6,5 \lambda-5)$$

DR of line : $(2,3,5)$

$$\[\begin{gathered} \therefore \quad  2(2 \lambda-7)+3(3 \lambda-6)+5(5 \lambda-5)=0 \ \Rightarrow \quad  \lambda=\frac{3}{2} \ \therefore \quad \quad  Q\left(4, \frac{7}{2}, \frac{19}{2}\right) \ \therefore \quad \quad  (\alpha, \beta, y) \equiv(0,2,12) \quad\left(Q \text { is mid point of } P \quad  P^{\prime}\right) \ \alpha+\beta+y \equiv 14 \end{gathered}$$\]

Solutions · 1