Solutions To Linear Algebra Applications (To Lines, Planes)

1) Find the equation of the plane perpendicular to the vector N at (-3, -2, 4), and passing through the point:

  P= (2, p, -5)

Solution:

 Using the given triples, we obtain the equation as:

 -3x – 2y + 4z = P · N

Whence: P · N = [(-3)(2) + (2)( p) + 4(-5)] = -26 + 2 p  = -2(13 + p)

So: 

-3x -2y +4z = -2(13 + p)

Or reduced to:  -3x/2 – y + 2z = 13 + p

 

2) Find the cosine of the angle between the planes:

 x + y + z = 1 and x – y – z = 5

Solution:

The respective vectors we need, from the coefficients of the equations are:

A = (1, 1, 1) and B = (1, -1, -1)

Then: cos(Θ  ) = A · B/ [A]{B] =

{(1 ·1) + (1 · (-1)) + (1) · (-1)} / {(1) ² + (1) ²

 + (1) ²] [(1) ²  + (- 1) ²  +( 1) ²]}

Or cos (Θ ) = -1 / {Ö (3) Ö (3)} = -1/ 3