Question

The value of $m$ for which straight line $3x-2y+z+3=0=4x-3y+4z+1$ is parallel to the plane $2x-y+mx-2=0$ is

• A. $-2$
• B. $8$
• C. $-18$
• D. $11$

Answer 1

The correct option is A $-2$

write the straight line in canonical form:

$3x-2y+z+3=0=4x-3y+4z+1$

the straight line is the solution of the following equation

$3x-2y+z+3=0----\left(1\right)$

$4x-3y+4z+1=0---\left(2\right)$

multipluing equation (1) by 4 and subtracting :

$12x-8y+4z+12=0$

$4x-3y+4z+1=0$

$8x-5y+11=0$

$\therefore x=\frac{x+\frac{11}{8}}{\frac{5}{8}}$

from the above equation substitute x in (1) equation

we get $15y-33-16y+8z+24=0$

$\therefore y=8z-9$

$\therefore y=\frac{z-\frac{9}{8}}{\frac{1}{8}}$

$\therefore$ DR's of the line is $(\frac{5}{8},1,\frac{1}{8})$

if the line is parallel to the plane $2x-y+mz-2=0$

$\therefore \frac{5}{4}-1+\frac{m}{8}=0$

$\therefore \frac{1}{4}+\frac{m}{8}=0$

$2+m=0$

$m=-2$