Linear Algebra Examples

$y = 3 x + 2$ , $x - 4 y = 9$

Step 1

Subtract $3 x$ from both sides of the equation.

$y - 3 x = 2, x - 4 y = 9$

Step 2

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 3

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

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Step 3.1

$P_{1}$ is $y - 3 x = 2$ . 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 3.2

$P_{2}$ is $x - 4 y = 9$ . Find the normal vector $n_{2} = ⟨ e, f, g ⟩$ from the plane equation of the form $e x + f y + g z = h$ .

$n_{2} = ⟨ 1, - 4, 0 ⟩$

Step 3.3

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.

$- 3 \cdot 1 + 1 \cdot - 4 + 0 \cdot 0$

Step 3.4

Simplify the dot product.

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Step 3.4.1

Remove parentheses.

$- 3 \cdot 1 + 1 \cdot - 4 + 0 \cdot 0$

Step 3.4.2

Simplify each term.

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Step 3.4.2.1

Multiply $- 3$ by $1$ .

$- 3 + 1 \cdot - 4 + 0 \cdot 0$

Step 3.4.2.2

Multiply $- 4$ by $1$ .

$- 3 - 4 + 0 \cdot 0$

Step 3.4.2.3

Multiply $0$ by $0$ .

$- 3 - 4 + 0$

Step 3.4.3

Simplify by adding and subtracting.

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Step 3.4.3.1

Subtract $4$ from $- 3$ .

$- 7 + 0$

Step 3.4.3.2

Add $- 7$ and $0$ .

$- 7$

Step 4

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 $- 7$ for the values of $a$ , $b$ , and $c$ . This set of parametric equations represents the line through the origin that is perpendicular to $P_{1}$ $y - 3 x = 2$ .

$x = 0 + - 3 \cdot t$

$y = 0 + 1 \cdot t$

$z = 0 + 0 \cdot t$

Step 5

Substitute the expression for $x$ , $y$ , and $z$ into the equation for $P_{2}$ $x - 4 y = 9$ .

$(0 - 3 \cdot t) - 4 (0 + 1 \cdot t) = 9$

Step 6

Solve the equation for $t$ .

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Step 6.1

Simplify $(0 - 3 \cdot t) - 4 (0 + 1 \cdot t)$ .

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Step 6.1.1

Combine the opposite terms in $(0 - 3 \cdot t) - 4 (0 + 1 \cdot t)$ .

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Step 6.1.1.1

Subtract $3 \cdot t$ from $0$ .

$- 3 \cdot t - 4 (0 + 1 \cdot t) = 9$

Step 6.1.1.2

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

$- 3 \cdot t - 4 (1 \cdot t) = 9$

Step 6.1.2

Multiply $t$ by $1$ .

$- 3 t - 4 t = 9$

Step 6.1.3

Subtract $4 t$ from $- 3 t$ .

$- 7 t = 9$

Step 6.2

Divide each term in $- 7 t = 9$ by $- 7$ and simplify.

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Step 6.2.1

Divide each term in $- 7 t = 9$ by $- 7$ .

$\frac{- 7 t}{- 7} = \frac{9}{- 7}$

Step 6.2.2

Simplify the left side.

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Step 6.2.2.1

Cancel the common factor of $- 7$ .

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Step 6.2.2.1.1

Cancel the common factor.

$\frac{- 7 t}{- 7} = \frac{9}{- 7}$

Step 6.2.2.1.2

Divide $t$ by $1$ .

$t = \frac{9}{- 7}$

Step 6.2.3

Simplify the right side.

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Step 6.2.3.1

Move the negative in front of the fraction.

$t = - \frac{9}{7}$

Step 7

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

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Step 7.1

Solve the equation for $x$ .

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Step 7.1.1

Remove parentheses.

$x = 0 - 3 \cdot (- 1 (\frac{9}{7}))$

Step 7.1.2

Remove parentheses.

$x = 0 - 3 \cdot (- \frac{9}{7})$

Step 7.1.3

Simplify $0 - 3 \cdot (- \frac{9}{7})$ .

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Step 7.1.3.1

Multiply $- 3 (- \frac{9}{7})$ .

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Step 7.1.3.1.1

Multiply $- 1$ by $- 3$ .

$x = 0 + 3 (\frac{9}{7})$

Step 7.1.3.1.2

Combine $3$ and $\frac{9}{7}$ .

$x = 0 + \frac{3 \cdot 9}{7}$

Step 7.1.3.1.3

Multiply $3$ by $9$ .

$x = 0 + \frac{27}{7}$

Step 7.1.3.2

Add $0$ and $\frac{27}{7}$ .

$x = \frac{27}{7}$

Step 7.2

Solve the equation for $y$ .

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Step 7.2.1

Remove parentheses.

$y = 0 + 1 \cdot (- 1 (\frac{9}{7}))$

Step 7.2.2

Remove parentheses.

$y = 0 + 1 \cdot (- \frac{9}{7})$

Step 7.2.3

Simplify $0 + 1 \cdot (- \frac{9}{7})$ .

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Step 7.2.3.1

Multiply $- \frac{9}{7}$ by $1$ .

$y = 0 - \frac{9}{7}$

Step 7.2.3.2

Subtract $\frac{9}{7}$ from $0$ .

$y = - \frac{9}{7}$

Step 7.3

Solve the equation for $z$ .

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Step 7.3.1

Remove parentheses.

$z = 0 + 0 \cdot (- 1 (\frac{9}{7}))$

Step 7.3.2

Remove parentheses.

$z = 0 + 0 \cdot (- \frac{9}{7})$

Step 7.3.3

Simplify $0 + 0 \cdot (- \frac{9}{7})$ .

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Step 7.3.3.1

Multiply $0 (- \frac{9}{7})$ .

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Step 7.3.3.1.1

Multiply $- 1$ by $0$ .

$z = 0 + 0 (\frac{9}{7})$

Step 7.3.3.1.2

Multiply $0$ by $\frac{9}{7}$ .

$z = 0 + 0$

Step 7.3.3.2

Add $0$ and $0$ .

$z = 0$

Step 7.4

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

$x = \frac{27}{7}$

$y = - \frac{9}{7}$

$z = 0$

$x = \frac{27}{7}$

$y = - \frac{9}{7}$

$z = 0$

Step 8

Using the values calculated for $x$ , $y$ , and $z$ , the intersection point is found to be $(\frac{27}{7}, - \frac{9}{7}, 0)$ .

$(\frac{27}{7}, - \frac{9}{7}, 0)$