Question

Line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ intersects the plane $3x+6y+5z=0$ in the point $A.$ Then the d.R.s of the line $OA$ are

• A. $\text{1,6, - 8}$
• B. $\text{8,6, - 1}$
• C. $\text{6,8,1}$
• D. $\text{8,1, - 6}$

Answer 1

The correct option is A $\text{1,6, - 8}$

$L:\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=r$

Plane:$3x+6y+5z=0$

$\therefore$ Let $2r+1,3r+2,4r+3$ intersect the plane at $a$

$\therefore 3(2r+1)+6(3r+2)+5(4r+3)=0$

$44r+30=0$

$r=\frac{-15}{22}$

$\overrightarrow{A}=(-\frac{4}{11},-\frac{1}{22},\frac{3}{11})$

$\overrightarrow{OA}=(-\frac{4}{11}\hat{i},-\frac{1}{22}\hat{j},\frac{3}{11}\hat{k})$

direction ratios of $\overrightarrow{OA}$ are $(8,1,-6)$