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
$2x+y-2z=0$
$\therefore 3\times 2+4\times 1-2\times 5=6+4-10$
$=0$
$\therefore$ Required equation of plane is $2x+y-2z=0$