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

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