The equations of a line passing through the point (-1,0,3) and perpendicular to the plane 4x+ 3y - 5z = 12are

a. (x-1)/4 = y/3 = (z+3)/-5

b. 5(3x -1) = 20(y -1) = -4(z -4)

c. (x+1)/-5 = y/3 = (z -3)/4

d. none of these

Question

The equations of a line passing through the point (-1,0,3) and
perpendicular to the plane 4x+ 3y - 5z = 12are
a (x-1)/4 = y/3 = (z+3)/-5
b 5(3x -1) = 20(y -1) = -4(z -4)
c (x+1)/-5 = y/3 = (z -3)/4
d none of these

Ajanta Trivedi · Accepted Answer

the line is parallel to the normal of the plane $4 x + 3 y - 5 z = 12$ ..........(1)

and the line passes through the point (-1, 0, 3)

therefore the required equation of the line is :

$\frac{x - (- 1)}{4} = \frac{y - 0}{3} = \frac{z - 3}{- 5} \Rightarrow \frac{x + 1}{4} = \frac{y}{3} = \frac{z - 3}{- 5}$

hope this helps you.

cheers!!

The equations of a line passing through the point (-1,0,3) and perpendicular to the plane 4x+ 3y - 5z = 12are a. (x-1)/4 = y/3 = (z+3)/-5 b. 5(3x -1) = 20(y -1) = -4(z -4) c. (x+1)/-5 = y/3 = (z -3)/4 d. none of these

The equations of a line passing through the point (-1,0,3) and perpendicular to the plane 4x+ 3y - 5z = 12are

a. (x-1)/4 = y/3 = (z+3)/-5

b. 5(3x -1) = 20(y -1) = -4(z -4)

c. (x+1)/-5 = y/3 = (z -3)/4

d. none of these