Question

T find the equation of the plane through rhe lines of intersection of planes x+y+z=1and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0

Answer 1

Answer:

The equation of the plane is 7y+4z−5=0

Explanation:

A vector in the plane is

→u=⎛⎜⎝1−13⎞⎟⎠−⎛⎜⎝23−4⎞⎟⎠=⎛⎜⎝−1−47⎞⎟⎠

A vector direction for the x-axis is

→v=⎛⎜⎝100⎞⎟⎠

So,

A normal vector to the plane is

→n=→u×→v

=∣∣

∣

∣∣ˆiˆjˆk−1−47100∣∣

∣

∣∣

=ˆi(0−0)−ˆj(0−7)+ˆk(0+4)

=<0,7,4>

The equation of the plane is

0(x−2)+7(y−3)+4(z+4)=0

7y−21+4z+16=0

7y+4z−5=0

Step-by-step explanation: