Question

(1.1) Let U and V be the planes given by:
U : λx + 5y − 2λz − 3 = 0,
V : −λx + y + 2z + 1 = 0.
Determine for which value(s) of λ the planes U and V are:
(a) orthogonal,
(b) Parallel.
(1.2) Find an equation for the plane that passes through the origin (0, 0, 0) and is parallel to the
plane −x + 3y − 2z = 6.
(1.3) Find the distance between the point (−1, −2, 0) and the plane 3x − y + 4z = −2.

Answer 1

Solution.

1.1

a)

The planes are orthogonal when

$-\lambda^2+5-4\lambda=0$

$\lambda^2+4\lambda-5=0$

$\lambda_1=1, \lambda_2=-5$

b)

The planes are parallel when

$\lambda/-\lambda=5=-2\lambda/2$

Such $\lambda$ does not exist, the planes can not be parallel.

b)

1.2

$N(3,-1,4)$ normal vector.

An equation for the plane that passes through the origin (x0, y0, z0) and normal vector (A,B,C) is$A(x-x_0)+B(y-y_0)+C(z-z_0)=0.$

Thus we will have

−x + 3y − 2z =0

1.3

$d=\frac{|-3+2+2|}{\sqrt{3^2+1+4^2}} ​ =\frac{1}{ 26}. ​$