Question

Find the ratio in which (the plane) $2x+3y+5z=1$ divides the line joining the points $(1,0,-3)$ and $(1,-5,7)$.

• A. $1:2$
• B. $2:3$
• C. $3:1$
• D. None of these

Answer 1

Answer: (B)

$2:3$

Here, $2x+3y+5z=1$ divides $(1,0,-3)$ and $(1,-5,7)$ in the ratio of $k:1$ at point $P.$

Then, $P=(\frac{k+1}{k+1},\frac{-5k}{k+1},\frac{7k-3}{k+1})$ which must satisfy $2x+3y+5x=1$

$\Rightarrow 2\left(\frac{k+1}{k+1}\right)+3\left(\frac{-5k}{k+1}\right)+5\left(\frac{7k-3}{k+1}\right)=1$

$\Rightarrow 2k+2-15k+35k-15=k+1\Rightarrow 21k=14$

$\Rightarrow k=\frac{2}{3}$

$\therefore 2x+3y+5z=1$

It divides the joint of $(1,0-3)$ and $(1,-5,7)$ in the radio of $2:3.$