Question

The line $\frac {x-2}{3}=\frac {y-3}{4}=\frac {z-4}{5}$ is parallel to the plane

• A. $3x + 4y + 5z = 7$
• B. $x + y + z = 2$
• C. $2x + 3y + 4z = 0$
• D. $2x + y - 2z = 0$

Answer 1

Answer: (D)

Given equation of line is
$\frac{x-2}{3}=\frac{Y-3}{4}=\frac{z-4}{5}$
So, DR's of a line are $(3,4,5)$.
Since, line is parallel to the plane, therefore normal to the plane is perpendicular to the line. $\therefore a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
Consider equation of plane is
$2 x+y-2 z =0$
$\therefore 3 \times 2+4 \times 1-2 \times 5 =6+4-10$
$=0$
$\therefore $ Required equation of plane is $2 x+y-2 z=0$