Question

find the equation of a straight line through the point (-2,4,-5) and paralal to the line whose direction ratio (3,5,6)​

Answer 1

Answer:

Answer

The equation of the plane through (2,2,−1) is a(x−2)+b(y−2)+c(z+1)=0 ……..(1)

This plane passes through (3,4,2)

a(3−2)+b(4−2)+c(2+1)=0

a+2b+3c=0 ………….(2)

Again plane (1) is parallel to the line whose direction ratios are (7,0,6).

It means that the normal of plane (1) is perpendicular to the line whose direction ratios are (7,0,6).

7a+0b+6c=0 ………..(3)

Solving (1),(2) and (3)

∣

∣

∣

∣

∣

∣

∣

∣

x−2

1

7

y−2

2

0

z+1

3

6

∣

∣

∣

∣

∣

∣

∣

∣

=0

12(x−2)+15(y−2)−14(z+1)=0

12x+15y−14z−68=0.