Solutions To Simple Linear Algebra Problems (4)

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

Solution:  Here, we want:

(x, y, z) = P· N

Using the ordered triples in 3-space, we can write the requirement for the plane perpendicular to N and passing through P in a plane:

x – y + 3z = P ·N

And: P ·N = [(4 ·1) + (-1) · (2) + 2 · (-1)] = -1

Therefore P· N = -1 and:

x – y + 3z = -1

2)    2)Find the equation of the plane perpendicular to the vector N at (-3, -2, 4), and passing through the point:  P= (2, p, -5)

3)    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

3) Find a vector parallel to the line of interception of two planes and the cosine of the angle between them:

2x – y + z = 1 and 3x + y + 2 = 2

Soln.

A = (2, -1, 1) and B = (3,  1,  1)

The vector (line) parallel to the line of interception is:

 (B – A)  =   [(3-2),  (1- (-1),  (1- 1)] =   1, 2, 0

Or: x + 2y = 0

The cosine of the angle between them:

cos(Θ ) = A · B/ [A]{B] 

A · B  =  [(2 · 3 ) + (-1 · (1)) + (1) · (1)]  = 6

[A]{B] =  [Ö {(2) ²  + (- 1) ²  + {(1) ² }] · [Ö {(3) ²  + (1) ²  + {(1) ² }]

[A]{B] =  Ö (6)  · Ö (11) =  Ö (66)

Then:  cos(Θ ) =  6/ Ö (66)

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

 x + y + z = 1 and x – y – z = 5

-- 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