Examples

x - y = 4

$x - y = 4$ ,

4 x - y = - 5

$4 x - y = - 5$

Step 1

To find the intersection of the line through a point

(p, q, r)

$(p, q, r)$ perpendicular to plane

P_{1}

$P_{1}$

a x + b y + c z = d

$a x + b y + c z = d$ and plane

P_{2}

$P_{2}$

e x + f y + g z = h

$e x + f y + g z = h$ :

1. Find the normal vectors of plane

P_{1}

$P_{1}$ and plane

P_{2}

$P_{2}$ where the normal vectors are

n_{1} = ⟨ a, b, c ⟩

$n_{1} = ⟨ a, b, c ⟩$ and

n_{2} = ⟨ e, f, g ⟩

$n_{2} = ⟨ e, f, g ⟩$ . Check to see if the dot product is 0.

2. Create a set of parametric equations such that

x = p + a t

$x = p + a t$ ,

y = q + b t

$y = q + b t$ , and

z = r + c t

$z = r + c t$ .

3. Substitute these equations into the equation for plane

P_{2}

$P_{2}$ such that

e (p + a t) + f (q + b t) + g (r + c t) = h

$e (p + a t) + f (q + b t) + g (r + c t) = h$ and solve for

t

$t$ .

4. Using the value of

t

$t$ , solve the parametric equations

x = p + a t

$x = p + a t$ ,

y = q + b t

$y = q + b t$ , and

z = r + c t

$z = r + c t$ for

t

$t$ to find the intersection

(x, y, z)

$(x, y, z)$ .

Step 2

Find the normal vectors for each plane and determine if they are perpendicular by calculating the dot product.

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Step 2.1

P_{1}

$P_{1}$ is

x - y = 4

$x - y = 4$ . Find the normal vector

n_{1} = ⟨ a, b, c ⟩

$n_{1} = ⟨ a, b, c ⟩$ from the plane equation of the form

a x + b y + c z = d

$a x + b y + c z = d$ .

n_{1} = ⟨ 1, - 1, 0 ⟩

$n_{1} = ⟨ 1, - 1, 0 ⟩$

Step 2.2

P_{2}

$P_{2}$ is

4 x - y = - 5

$4 x - y = - 5$ . Find the normal vector

n_{2} = ⟨ e, f, g ⟩

$n_{2} = ⟨ e, f, g ⟩$ from the plane equation of the form

e x + f y + g z = h

$e x + f y + g z = h$ .

n_{2} = ⟨ 4, - 1, 0 ⟩

$n_{2} = ⟨ 4, - 1, 0 ⟩$

Step 2.3

Calculate the dot product of

n_{1}

$n_{1}$ and

n_{2}

$n_{2}$ by summing the products of the corresponding

x

$x$ ,

y

$y$ , and

z

$z$ values in the normal vectors.

1 \cdot 4 - 1 \cdot - 1 + 0 \cdot 0

$1 \cdot 4 - 1 \cdot - 1 + 0 \cdot 0$

Step 2.4

Simplify the dot product.

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Step 2.4.1

Remove parentheses.

1 \cdot 4 - 1 \cdot - 1 + 0 \cdot 0

$1 \cdot 4 - 1 \cdot - 1 + 0 \cdot 0$

Step 2.4.2

Simplify each term.

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Step 2.4.2.1

Multiply

4

$4$ by

1

$1$ .

4 - 1 \cdot - 1 + 0 \cdot 0

$4 - 1 \cdot - 1 + 0 \cdot 0$

Step 2.4.2.2

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

4 + 1 + 0 \cdot 0

$4 + 1 + 0 \cdot 0$

Step 2.4.2.3

Multiply

0

$0$ by

0

$0$ .

4 + 1 + 0

$4 + 1 + 0$

4 + 1 + 0

$4 + 1 + 0$

Step 2.4.3

Simplify by adding numbers.

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Step 2.4.3.1

Add

4

$4$ and

1

$1$ .

5 + 0

$5 + 0$

Step 2.4.3.2

Add

5

$5$ and

0

$0$ .

5

$5$

5

$5$

5

$5$

5

$5$

Step 3

Next, build a set of parametric equations

x = p + a t

$x = p + a t$ ,

y = q + b t

$y = q + b t$ , and

z = r + c t

$z = r + c t$ using the origin

(0, 0, 0)

$(0, 0, 0)$ for the point

(p, q, r)

$(p, q, r)$ and the values from the normal vector

5

$5$ for the values of

a

$a$ ,

b

$b$ , and

c

$c$ . This set of parametric equations represents the line through the origin that is perpendicular to

P_{1}

$P_{1}$

x - y = 4

$x - y = 4$ .

x = 0 + 1 \cdot t

$x = 0 + 1 \cdot t$

y = 0 + - 1 \cdot t

$y = 0 + - 1 \cdot t$

z = 0 + 0 \cdot t

$z = 0 + 0 \cdot t$

Step 4

Substitute the expression for

x

$x$ ,

y

$y$ , and

z

$z$ into the equation for

P_{2}

$P_{2}$

4 x - y = - 5

$4 x - y = - 5$ .

4 (0 + 1 \cdot t) - (0 - 1 \cdot t) = - 5

$4 (0 + 1 \cdot t) - (0 - 1 \cdot t) = - 5$

Step 5

Solve the equation for

t

$t$ .

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Step 5.1

Simplify

4 (0 + 1 \cdot t) - (0 - 1 \cdot t)

$4 (0 + 1 \cdot t) - (0 - 1 \cdot t)$ .

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Step 5.1.1

Combine the opposite terms in

4 (0 + 1 \cdot t) - (0 - 1 \cdot t)

$4 (0 + 1 \cdot t) - (0 - 1 \cdot t)$ .

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Step 5.1.1.1

Add

0

$0$ and

1 \cdot t

$1 \cdot t$ .

4 (1 \cdot t) - (0 - 1 \cdot t) = - 5

$4 (1 \cdot t) - (0 - 1 \cdot t) = - 5$

Step 5.1.1.2

Subtract

1 \cdot t

$1 \cdot t$ from

0

$0$ .

4 (1 \cdot t) - (- 1 \cdot t) = - 5

$4 (1 \cdot t) - (- 1 \cdot t) = - 5$

4 (1 \cdot t) - (- 1 \cdot t) = - 5

$4 (1 \cdot t) - (- 1 \cdot t) = - 5$

Step 5.1.2

Simplify each term.

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Step 5.1.2.1

Multiply

t

$t$ by

1

$1$ .

4 t - (- 1 \cdot t) = - 5

$4 t - (- 1 \cdot t) = - 5$

Step 5.1.2.2

Rewrite

- 1 t

$- 1 t$ as

- t

$- t$ .

4 t - - t = - 5

$4 t - - t = - 5$

Step 5.1.2.3

Multiply

- - t

$- - t$ .

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Step 5.1.2.3.1

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

4 t + 1 t = - 5

$4 t + 1 t = - 5$

Step 5.1.2.3.2

Multiply

t

$t$ by

1

$1$ .

4 t + t = - 5

$4 t + t = - 5$

4 t + t = - 5

$4 t + t = - 5$

4 t + t = - 5

$4 t + t = - 5$

Step 5.1.3

Add

4 t

$4 t$ and

t

$t$ .

5 t = - 5

$5 t = - 5$

5 t = - 5

$5 t = - 5$

Step 5.2

Divide each term in

5 t = - 5

$5 t = - 5$ by

5

$5$ and simplify.

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Step 5.2.1

Divide each term in

5 t = - 5

$5 t = - 5$ by

5

$5$ .

\frac{5 t}{5} = \frac{- 5}{5}

$\frac{5 t}{5} = \frac{- 5}{5}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of

5

$5$ .

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Step 5.2.2.1.1

Cancel the common factor.

\frac{5 t}{5} = \frac{- 5}{5}

$\frac{5 t}{5} = \frac{- 5}{5}$

Step 5.2.2.1.2

Divide

t

$t$ by

1

$1$ .

t = \frac{- 5}{5}

$t = \frac{- 5}{5}$

t = \frac{- 5}{5}

$t = \frac{- 5}{5}$

t = \frac{- 5}{5}

$t = \frac{- 5}{5}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Divide

- 5

$- 5$ by

5

$5$ .

t = - 1

$t = - 1$

t = - 1

$t = - 1$

t = - 1

$t = - 1$

t = - 1

$t = - 1$

Step 6

Solve the parametric equations for

x

$x$ ,

y

$y$ , and

z

$z$ using the value of

t

$t$ .

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Step 6.1

Solve the equation for

x

$x$ .

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Step 6.1.1

Remove parentheses.

x = 0 + 1 \cdot (- 1)

$x = 0 + 1 \cdot (- 1)$

Step 6.1.2

Simplify

0 + 1 \cdot (- 1)

$0 + 1 \cdot (- 1)$ .

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Step 6.1.2.1

Multiply

- 1

$- 1$ by

1

$1$ .

x = 0 - 1

$x = 0 - 1$

Step 6.1.2.2

Subtract

1

$1$ from

0

$0$ .

x = - 1

$x = - 1$

x = - 1

$x = - 1$

x = - 1

$x = - 1$

Step 6.2

Solve the equation for

y

$y$ .

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Step 6.2.1

Remove parentheses.

y = 0 - 1 \cdot - 1

$y = 0 - 1 \cdot - 1$

Step 6.2.2

Simplify

0 - 1 \cdot - 1

$0 - 1 \cdot - 1$ .

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Step 6.2.2.1

Multiply

- 1

$- 1$ by

- 1

$- 1$ .

y = 0 + 1

$y = 0 + 1$

Step 6.2.2.2

Add

0

$0$ and

1

$1$ .

y = 1

$y = 1$

y = 1

$y = 1$

y = 1

$y = 1$

Step 6.3

Solve the equation for

z

$z$ .

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Step 6.3.1

Remove parentheses.

z = 0 + 0 \cdot (- 1)

$z = 0 + 0 \cdot (- 1)$

Step 6.3.2

Simplify

0 + 0 \cdot (- 1)

$0 + 0 \cdot (- 1)$ .

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Step 6.3.2.1

Multiply

0

$0$ by

- 1

$- 1$ .

z = 0 + 0

$z = 0 + 0$

Step 6.3.2.2

Add

0

$0$ and

0

$0$ .

z = 0

$z = 0$

z = 0

$z = 0$

z = 0

$z = 0$

Step 6.4

The solved parametric equations for

x

$x$ ,

y

$y$ , and

z

$z$ .

x = - 1

$x = - 1$

y = 1

$y = 1$

z = 0

$z = 0$

x = - 1

$x = - 1$

y = 1

$y = 1$

z = 0

$z = 0$

Step 7

Using the values calculated for

x

$x$ ,

y

$y$ , and

z

$z$ , the intersection point is found to be

(- 1, 1, 0)

$(- 1, 1, 0)$ .

(- 1, 1, 0)

$(- 1, 1, 0)$

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