Question

Q. 8 Find the direction ratios of the line :2x=(1-y)/(2)=(3z+4)/(6)

Answer 1

Answer:

The equation of the plane passing through the intersection of the planes x+2y+z=3 and 2x−y−z=5 is given by

(x+2y+z−3)+λ(2x−y−z−5)=0

x(2λ+1)+y(2−λ)+z(1−λ)−3−5λ=0

It passes through (2,1,3). Therefore,

2(2λ+1)+(2−λ)+3(1−λ)−3−5λ=0

4λ+2+2−λ+3−3λ−3−5λ=0

4−5λ=0

λ=

5

4

Substituting this value in equation of plane, we get,

13x+6y+z−35=0 as the equation of required plane.

Direction ratios of the normal to this plane are proportional to 13,6,1.

Step-by-step explanation:

l hope it will help u ☺️