Show that the point (−1,−1,1) lies on both planes. Then find a vector parametric equation for the line of intersection.

Question

(a) Find a vector parallel to the line of intersection of the planes −4x+2y−z=1 and 3x−2y+2z=1.

v→ = 2i + 5j + 2k

(b) Show that the point (−1,−1,1) lies on both planes. Then find a vector parametric equation for the line of intersection.

r(t)→ = ?

PS. I found the a part but I couldn't find the b part.Thank you for everyone who helps.

Answer 1

A fast way to find the parametric equations of the line of intersection between two planes is to solve the system

−4x +2y =1+z

3x−2y =1.- 2z assuming z is known.

Adding the equations we get -x = 2 -z ⇒ x = -2 + z ( Equation I )

Solving for y we get y = -7/2 + 5/2z

Setting z= t

we obtain

x = -2 + t

y = -7/2 + 5/2t

z = t

The directing vector of this line is u = < 1, 5/2, 1 > or any scalar multiple say <2, 5 ,2 >

The other method is to take the cross product of the normal vectors of the given planes

u = < -4 , 2 , -1 > x < 3, -2, 2 > = <2, 5 ,2 >

Now since the point  (−1,−1,1) lies in both planes

then lies also on the line of their intersection.

Hence the parametric equations of the line of intersection are as follws

x = -1 +2 t

y = -1 +5 t

z = 1 + 2t

Notice that the newly found set of equations describe the same line.