Suppose L is a line in space passing through a point P₀ (x ₀, y ₀, z ₀) parallel to the vector v = Ai + Bj + Ck. Then L is the set of all points P (x, y, z) for which P₀P is parallel to v.

FACT:

Vector equation for the line through P₀ (x ₀, y ₀, z ₀) parallel to v is

P₀P = tv

-¥ < t < ¥

or

(x - x ₀)i + (y - y ₀)j + (z - z ₀)k = (Ai + Bj + Ck)t

or

x - x ₀ = At ® x = x ₀ + At

y - y ₀ = Bt ® y = y ₀ + Bt

z - z ₀ = Ct ® z = z ₀ + Ct

EXAMPLE 1:

Find the parametric equations for the line through P (2, 1, -1) parallel to the vector -2i + j - k.

SOLUTION:

First I will determine the values for A, B, and C from the given vector.

A = -2, B = 1, and C = -1

The parametric equations for the lines are the following.

x = 2 - 2t

y = 1 + t

z = -1 -t

EXAMPLE 2:

Find the parametric equations for the line through the point (3, -2, 1) parallel to the line x = 1 + 2t, y = 2 - t, z = 3t.

SOLUTION:

Using the A, B, and C from the given line, then A = 2, B = -1, and C = 3.

Here are the equations for the new line.

x = 3 + 2t

y = -2 - t

z = 1 + 3t

EXAMPLE 3:

Find the parametric equations for the line through the point P (2, 4, -1) and Q (-1, 2, 5).

SOLUTION:

First, I have to determine the vector PQ.

PQ = (-1 - 2)i + (2 - 4)j + (5 - (-1))k = -3i - 2j + 6k

So A = -3, B = -2, and C = 6.

Now using either P or Q, I will write the parametric equations for the line. I will use the point P.

x = 2 - 3t

y = 4 - 2t

z = -1 + 6t

EXAMPLE 4:

Find the parameterization for the line segment joining the points (1, 0, -1) and (2, -4, 1).

SOLUTION:

Solving this problem is nothing different than the previous example; the only difference is that we have to state the range the t-values will take on.

Let P be (1, 0, -1) and Q be (2, -4, 1).

PQ = (2 - 1)i + (-4 - 0)j + (1- (-1))k = i - 4j + 2k

Therefore, A = 1, B = -4, and C = 2.

x = 1 + t

y = -4t

z = -1 + 2t

Now to determine the range for the t-values.

Let 1 + t = 1 ® t = 0.

When t = 0, then x = 1, y = 0, and z = -1. This is point P.

Let 1 + t = 2 ® t = 1.

When t = 1, then x = 2, y = -4, and z = 1. This is point Q.

x = 1 + t

y = -4t

z = -1 + 2t

0 £ t £ 1

To find the distance from a point S to a line that passes through the point P parallel to a vector v, we will find the length of the component of PS normal to the line. (see figure 1)

figure 1

EXAMPLE 5:

Find the distance from (-1, 4, 3) to the line x = 10 + 4t, y = -3, z = 4t.

SOLUTION:

The vector parallel to the line is of the form Ai + Bj + Ck.

Therefore v = 4i + 4k.

To find a point on the line, let P be of the form (x ₀, y ₀, z ₀) or (10, -3, 0).

Now find the vector PS where S is (-1, 4, 3).

PS = (-1 - 10)i + (4 - (-3))j + (3 - 0)k = -11i + 7j + 3k

Suppose plane M passes through a point P₀ (x ₀, y ₀, z ₀) and is normal to the nonzero vector n = Ai + Bj + Ck. (see figure 2) Then M is the set of all points P (x, y, z) for which P₀P is orthogonal to n. (i.e. P lies on M if and only if n · P₀P = 0) We will use the equation

n · P₀P = 0.

figure 2

FACT:

Ai + Bj + Ck is normal to the plane Ax + By + Cz = 0.

EXAMPLE 6:

Find the equation of the plane through (2, 4, -1) parallel to the plane 2x - y + 3z = 5.

SOLUTION:

First of all, we must determine the vector n that is normal to the plane 2x - y + 3z = 5.

n = 2i - j + 3k

Since the new plane is to be parallel to the plane 2x - y + 3z = 5, then n will also be normal to the new plane. Now to find the vector P₀P where P is (x, y, z).

P₀P = (x - 2)i + (y - 4)j + (z + 1)k

Now to use n · P₀P = 0.

(x - 2)(2) + (y - 4)(-1) + (z + 1)(3) = 0

2x - 4 - y + 4 + 3z + 3 = 0

2x - y + 3z = -3

EXAMPLE 7:

Find the equation of the plane through (1, 1, 1), (2, -1, 0), and (3, 1, 5).

SOLUTION:

All three of these points are lying in this new plane, so let P₀ be (1, 1, 1), Q be (2, -1, 0) and R be (3, 1, 5). Since I need a vector n perpendicular to this new plane, I will have to use the three points to find two vectors so I can cross them to find n.

P₀Q = (2 - 1)i + (-1 - 1)j + (0 -1)k = i - 2j - k

P₀R = (3 - 1)i + (1 - 1)j + (5 - 1)k = 2i + 4k

Now to find the cross product of these two vectors.

Let P be (x, y, z), then P₀P = (x - 1)i + (y - 1)j + (z - 1)k.

n · P₀P = 0 ® -8(x - 1) - 6(y - 1) + 4(z - 1) = 0

-8x + 8 - 6y + 6 + 4z - 4 = 0

-8x - 6y + 4z = -10

8x + 6y - 4z = 10

EXAMPLE 8:

Find the point of intersection of the lines x = t, y = -t + 2, z = t + 1, and x = 2s + 2, y = s + 3, z = 5s + 6, and then find the plane determined by these lines.

SOLUTION:

Line 1: x = t, y = -t + 2, z = t + 1

Line 2: x = 2s + 2, y = s + 3, z = 5s + 6

I need to find the point of intersection of these two lines. To do it, I will solve the equations simultaneously.

When t = 0 and s = -1, then z = 1 and z = 5(-1) + 6 = 1 true!

The point of intersection is (0, 2, 1).

Now to determine the equation of the plane.

First of all n₁ is parallel to line 1 and n₁ = i + j + k.

Next, n₂ is parallel to line 2 and n₂ = 2i + j + 5k.

Since n₁ is parallel to line 1, and n₂ is parallel to line 2, then n₁ x n₂ is perpendicular to the plane.

Using the point P₀ (0, 2, 1) and P (x, y, z) I am going to find the vector P₀P.

P₀P = (x - 0)i + (y - 2)j + (z - 1)k = xi + (y - 2)j + (z - 1)k

n · P₀P = 0 ® 4x - 3(y - 2) - 1(z - 1) = 0

4x - 3y + 6 - z + 1 = 0

4x - 3y - z = -7

EXAMPLE 9:

Find the equation of the line through (2, -1, 1) perpendicular to the plane 3x - 2y + 6z = 5.

SOLUTION:

First of all, we need to find n which is the vector perpendicular to the plane.

n = 3i - 2j + 6k

(Remember that Ai + Bj + Ck is normal to the plane Ax + By + Cz = 0.)

The vector n is parallel to the line we must find, so here is the equation for the line.

x = 2 + 3t

y = -1 - 2t

z = 1 + 6t

EXAMPLE 10:

Find the distance from the point S (1, 0, -1) to the plane -4x + y + z = 4.

SOLUTION:

Let use find a point on the plane. A point that is easy to find is one of the three intercepts. I will find the y-intercept and that is when x = z = 0. Therefore, y = 4.

Let P (0, 4, 0). (The y-intercept.)

So PS = (1 - 0)i + (0 - 4)j + (-1 - 0)k = i - 4j - k

n = -4i + j + k

I need the direction of n because I am going to use the following formula.

The angle between two intersecting planes is defined to be the (acute) angle determined by their normal vectors. (see figure 3)

figure 3

EXAMPLE 11:

Find the angles between the planes 5x + y - z = 10 and x - 2y + 3z = -1.

SOLUTION:

For plane 1: (5x + y - z = 10)

n₁ = 5i + j - k

For plane 2: (x - 2y + 3z = -1)

n₂ = i - 2j + 3k

n₁ · n₂ = (5)(1) + (1)(-2) + (-1)(3) = 0

EXAMPLE 12:

Find the point in which the line x = 1 + 2t, y = 1 + 5t, z = 3t and the plane x + y + z = 2.

SOLUTION:

The point where the line meets the plane has the form (1 + 2t, 1 + 5t, 3t). Plug this point into the equation for the plane.

1 + 2t + 1 + 5t + 3t = 2

10t = 0

t = 0

Plug t = 0 into (1 + 2t, 1 + 5t, 3t), so the point of intersection is (1, 1, 0).

EXAMPLE 13:

Find the parameterization for the line in which the planes 5x - 2y = 11 and 4y - 5z = -17 intersect.

SOLUTION:

Plane 1: (5x - 2y = 11)

n₁ = 5i - 2j

Plane 2: (4y - 5z = -17)

n₂ = 4j - 5z

n₁ x n₂ is the vector in the direction of the desired line.

Let x = 0, so -5z = 5 or z = -1.

Now solve for y.

4y + 5 = -17 ® 4y = -22 ® y = -11/2

Therefore, the point is (0, -11/2. -1) and the line of intersection is the following.

 x = 10t

y = -11/2 + 25t

z = -1 + 20t