Examples

$3 x - y = - 4$ , $x - 2 y = - 3$

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 $3 x - y = - 4$ . Find the normal vector $n_{1} = ⟨ a, b, c ⟩$ from the plane equation of the form $a x + b y + c z = d$ .

$n_{1} = ⟨ 3, - 1, 0 ⟩$

Step 2.2

$P_{2}$ is $x - 2 y = - 3$ . Find the normal vector $n_{2} = ⟨ e, f, g ⟩$ from the plane equation of the form $e x + f y + g z = h$ .

$n_{2} = ⟨ 1, - 2, 0 ⟩$

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.

$3 \cdot 1 - 1 \cdot - 2 + 0 \cdot 0$

Step 2.4

Simplify the dot product.

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Step 2.4.1

Remove parentheses.

$3 \cdot 1 - 1 \cdot - 2 + 0 \cdot 0$

Step 2.4.2

Simplify each term.

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Step 2.4.2.1

Multiply $3$ by $1$ .

$3 - 1 \cdot - 2 + 0 \cdot 0$

Step 2.4.2.2

Multiply $- 1$ by $- 2$ .

$3 + 2 + 0 \cdot 0$

Step 2.4.2.3

Multiply $0$ by $0$ .

$3 + 2 + 0$

Step 2.4.3

Simplify by adding numbers.

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Step 2.4.3.1

Add $3$ and $2$ .

$5 + 0$

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 $P_{1}$ $3 x - y = - 4$ .

$x = 0 + 3 \cdot t$

$y = 0 + - 1 \cdot t$

$z = 0 + 0 \cdot t$

Step 4

Substitute the expression for $x$ , $y$ , and $z$ into the equation for $P_{2}$ $x - 2 y = - 3$ .

$(0 + 3 \cdot t) - 2 (0 - 1 \cdot t) = - 3$

Step 5

Solve the equation for $t$ .

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Step 5.1

Simplify $(0 + 3 \cdot t) - 2 (0 - 1 \cdot t)$ .

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Step 5.1.1

Combine the opposite terms in $(0 + 3 \cdot t) - 2 (0 - 1 \cdot t)$ .

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Step 5.1.1.1

Add $0$ and $3 \cdot t$ .

$3 \cdot t - 2 (0 - 1 \cdot t) = - 3$

Step 5.1.1.2

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

$3 \cdot t - 2 (- 1 \cdot t) = - 3$

Step 5.1.2

Simplify each term.

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Step 5.1.2.1

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

$3 t - 2 (- t) = - 3$

Step 5.1.2.2

Multiply $- 1$ by $- 2$ .

$3 t + 2 t = - 3$

Step 5.1.3

Add $3 t$ and $2 t$ .

$5 t = - 3$

Step 5.2

Divide each term in $5 t = - 3$ by $5$ and simplify.

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Step 5.2.1

Divide each term in $5 t = - 3$ by $5$ .

$\frac{5 t}{5} = \frac{- 3}{5}$

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.

$\frac{5 t}{5} = \frac{- 3}{5}$

Step 5.2.2.1.2

Divide $t$ by $1$ .

$t = \frac{- 3}{5}$

Step 5.2.3

Simplify the right side.

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Step 5.2.3.1

Move the negative in front of the fraction.

$t = - \frac{3}{5}$

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.

$x = 0 + 3 \cdot (- 1 (\frac{3}{5}))$

Step 6.1.2

Remove parentheses.

$x = 0 + 3 \cdot (- \frac{3}{5})$

Step 6.1.3

Simplify $0 + 3 \cdot (- \frac{3}{5})$ .

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Step 6.1.3.1

Simplify each term.

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Step 6.1.3.1.1

Multiply $3 (- \frac{3}{5})$ .

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Step 6.1.3.1.1.1

Multiply $- 1$ by $3$ .

$x = 0 - 3 (\frac{3}{5})$

Step 6.1.3.1.1.2

Combine $- 3$ and $\frac{3}{5}$ .

$x = 0 + \frac{- 3 \cdot 3}{5}$

Step 6.1.3.1.1.3

Multiply $- 3$ by $3$ .

$x = 0 + \frac{- 9}{5}$

Step 6.1.3.1.2

Move the negative in front of the fraction.

$x = 0 - \frac{9}{5}$

Step 6.1.3.2

Subtract $\frac{9}{5}$ from $0$ .

$x = - \frac{9}{5}$

Step 6.2

Solve the equation for $y$ .

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Step 6.2.1

Remove parentheses.

$y = 0 - 1 \cdot (- 1 (\frac{3}{5}))$

Step 6.2.2

Remove parentheses.

$y = 0 - 1 \cdot (- \frac{3}{5})$

Step 6.2.3

Simplify $0 - 1 \cdot (- \frac{3}{5})$ .

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Step 6.2.3.1

Multiply $- 1 (- \frac{3}{5})$ .

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Step 6.2.3.1.1

Multiply $- 1$ by $- 1$ .

$y = 0 + 1 (\frac{3}{5})$

Step 6.2.3.1.2

Multiply $\frac{3}{5}$ by $1$ .

$y = 0 + \frac{3}{5}$

Step 6.2.3.2

Add $0$ and $\frac{3}{5}$ .

$y = \frac{3}{5}$

Step 6.3

Solve the equation for $z$ .

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Step 6.3.1

Remove parentheses.

$z = 0 + 0 \cdot (- 1 (\frac{3}{5}))$

Step 6.3.2

Remove parentheses.

$z = 0 + 0 \cdot (- \frac{3}{5})$

Step 6.3.3

Simplify $0 + 0 \cdot (- \frac{3}{5})$ .

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Step 6.3.3.1

Multiply $0 (- \frac{3}{5})$ .

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Step 6.3.3.1.1

Multiply $- 1$ by $0$ .

$z = 0 + 0 (\frac{3}{5})$

Step 6.3.3.1.2

Multiply $0$ by $\frac{3}{5}$ .

$z = 0 + 0$

Step 6.3.3.2

Add $0$ and $0$ .

$z = 0$

Step 6.4

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

$x = - \frac{9}{5}$

$y = \frac{3}{5}$

$z = 0$

$x = - \frac{9}{5}$

$y = \frac{3}{5}$

$z = 0$

Step 7

Using the values calculated for $x$ , $y$ , and $z$ , the intersection point is found to be $(- \frac{9}{5}, \frac{3}{5}, 0)$ .

$(- \frac{9}{5}, \frac{3}{5}, 0)$

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