Examples

$x - y = 4$ , $4 x - y = - 5$

Step 1

To find the intersection of the line through a point $(p, q, r)$ perpendicular to plane $P_{1}$ $a x + b y + c z = d$ and plane $P_{2}$ $e x + f y + g z = h$ :

1. Find the normal vectors of plane $P_{1}$ and plane $P_{2}$ where the normal vectors are $n_{1} = ⟨ a, b, c ⟩$ and $n_{2} = ⟨ e, f, g ⟩$ . Check to see if the dot product is 0.

2. Create a set of parametric equations such that $x = p + a t$ , $y = q + b t$ , and $z = r + c t$ .

3. Substitute these equations into the equation for plane $P_{2}$ such that $e (p + a t) + f (q + b t) + g (r + c t) = h$ and solve for $t$ .

4. Using the value of $t$ , solve the parametric equations $x = p + a t$ , $y = q + b t$ , and $z = r + c t$ for $t$ to find the intersection $(x, y, z)$ .

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

$P_{1}$ is $x - y = 4$ . Find the normal vector $n_{1} = ⟨ a, b, c ⟩$ from the plane equation of the form $a x + b y + c z = d$ .

$n_{1} = ⟨ 1, - 1, 0 ⟩$

Step 2.2

$P_{2}$ is $4 x - y = - 5$ . Find the normal vector $n_{2} = ⟨ e, f, g ⟩$ from the plane equation of the form $e x + f y + g z = h$ .

$n_{2} = ⟨ 4, - 1, 0 ⟩$

Step 2.3

Calculate the dot product of $n_{1}$ and $n_{2}$ by summing the products of the corresponding $x$ , $y$ , and $z$ values in the normal vectors.

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

Step 2.4

Simplify the dot product.

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Step 2.4.1

Remove parentheses.

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

Step 2.4.2

Simplify each term.

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Step 2.4.2.1

Multiply $4$ by $1$ .

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

Step 2.4.2.2

Multiply $- 1$ by $- 1$ .

$4 + 1 + 0 \cdot 0$

Step 2.4.2.3

Multiply $0$ by $0$ .

$4 + 1 + 0$

Step 2.4.3

Simplify by adding numbers.

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Step 2.4.3.1

Add $4$ and $1$ .

$5 + 0$

Step 2.4.3.2

Add $5$ and $0$ .

$5$

Step 3

Next, build a set of parametric equations $x = p + a t$ , $y = q + b t$ , and $z = r + c t$ using the origin $(0, 0, 0)$ for the point $(p, q, r)$ and the values from the normal vector $5$ for the values of $a$ , $b$ , and $c$ . This set of parametric equations represents the line through the origin that is perpendicular to $P_{1}$ $x - y = 4$ .

$x = 0 + 1 \cdot t$

$y = 0 + - 1 \cdot t$

$z = 0 + 0 \cdot t$

Step 4

Substitute the expression for $x$ , $y$ , and $z$ into the equation for $P_{2}$ $4 x - y = - 5$ .

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

Step 5

Solve the equation for $t$ .

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Step 5.1

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

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Step 5.1.1

Combine the opposite terms in $4 (0 + 1 \cdot t) - (0 - 1 \cdot t)$ .

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Step 5.1.1.1

Add $0$ and $1 \cdot t$ .

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

Step 5.1.1.2

Subtract $1 \cdot t$ from $0$ .

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

Step 5.1.2

Simplify each term.

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Step 5.1.2.1

Multiply $t$ by $1$ .

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

Step 5.1.2.2

Rewrite $- 1 t$ as $- t$ .

$4 t - - t = - 5$

Step 5.1.2.3

Multiply $- - t$ .

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Step 5.1.2.3.1

Multiply $- 1$ by $- 1$ .

$4 t + 1 t = - 5$

Step 5.1.2.3.2

Multiply $t$ by $1$ .

$4 t + t = - 5$

Step 5.1.3

Add $4 t$ and $t$ .

$5 t = - 5$

Step 5.2

Divide each term in $5 t = - 5$ by $5$ and simplify.

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Step 5.2.1

Divide each term in $5 t = - 5$ by $5$ .

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

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $5$ .

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Step 5.2.2.1.1

Cancel the common factor.

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

Step 5.2.2.1.2

Divide $t$ by $1$ .

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

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Divide $- 5$ by $5$ .

$t = - 1$

Step 6

Solve the parametric equations for $x$ , $y$ , and $z$ using the value of $t$ .

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Step 6.1

Solve the equation for $x$ .

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Step 6.1.1

Remove parentheses.

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

Step 6.1.2

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

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Step 6.1.2.1

Multiply $- 1$ by $1$ .

$x = 0 - 1$

Step 6.1.2.2

Subtract $1$ from $0$ .

$x = - 1$

Step 6.2

Solve the equation for $y$ .

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Step 6.2.1

Remove parentheses.

$y = 0 - 1 \cdot - 1$

Step 6.2.2

Simplify $0 - 1 \cdot - 1$ .

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Step 6.2.2.1

Multiply $- 1$ by $- 1$ .

$y = 0 + 1$

Step 6.2.2.2

Add $0$ and $1$ .

$y = 1$

Step 6.3

Solve the equation for $z$ .

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Step 6.3.1

Remove parentheses.

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

Step 6.3.2

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

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Step 6.3.2.1

Multiply $0$ by $- 1$ .

$z = 0 + 0$

Step 6.3.2.2

Add $0$ and $0$ .

$z = 0$

Step 6.4

The solved parametric equations for $x$ , $y$ , and $z$ .

$x = - 1$

$y = 1$

$z = 0$

$x = - 1$

$y = 1$

$z = 0$

Step 7

Using the values calculated for $x$ , $y$ , and $z$ , the intersection point is found to be $(- 1, 1, 0)$ .

$(- 1, 1, 0)$

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