Linear Algebra Examples

y = 3 x + 2

$y = 3 x + 2$ ,

x - 4 y = 9

$x - 4 y = 9$

Step 1

Subtract

3 x

$3 x$ from both sides of the equation.

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

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

Step 2

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 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}

$P_{1}$ is

y - 3 x = 2

$y - 3 x = 2$ . 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} = ⟨ - 3, 1, 0 ⟩

$n_{1} = ⟨ - 3, 1, 0 ⟩$

Step 3.2

P_{2}

$P_{2}$ is

x - 4 y = 9

$x - 4 y = 9$ . 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} = ⟨ 1, - 4, 0 ⟩

$n_{2} = ⟨ 1, - 4, 0 ⟩$

Step 3.3

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.

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

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

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

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

Step 3.4.2

Simplify each term.

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Step 3.4.2.1

Multiply

- 3

$- 3$ by

1

$1$ .

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

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

Step 3.4.2.2

Multiply

- 4

$- 4$ by

1

$1$ .

- 3 - 4 + 0 \cdot 0

$- 3 - 4 + 0 \cdot 0$

Step 3.4.2.3

Multiply

0

$0$ by

0

$0$ .

- 3 - 4 + 0

$- 3 - 4 + 0$

- 3 - 4 + 0

$- 3 - 4 + 0$

Step 3.4.3

Simplify by adding and subtracting.

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Step 3.4.3.1

Subtract

4

$4$ from

- 3

$- 3$ .

- 7 + 0

$- 7 + 0$

Step 3.4.3.2

Add

- 7

$- 7$ and

0

$0$ .

- 7

$- 7$

- 7

$- 7$

- 7

$- 7$

- 7

$- 7$

Step 4

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

- 7

$- 7$ 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}

$P_{1}$

y - 3 x = 2

$y - 3 x = 2$ .

x = 0 + - 3 \cdot t

$x = 0 + - 3 \cdot t$

y = 0 + 1 \cdot t

$y = 0 + 1 \cdot t$

z = 0 + 0 \cdot t

$z = 0 + 0 \cdot t$

Step 5

Substitute the expression for

x

$x$ ,

y

$y$ , and

z

$z$ into the equation for

P_{2}

$P_{2}$

x - 4 y = 9

$x - 4 y = 9$ .

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

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

Step 6

Solve the equation for

t

$t$ .

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Step 6.1

Simplify

(0 - 3 \cdot t) - 4 (0 + 1 \cdot t)

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

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Step 6.1.1

Combine the opposite terms in

(0 - 3 \cdot t) - 4 (0 + 1 \cdot t)

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

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Step 6.1.1.1

Subtract

3 \cdot t

$3 \cdot t$ from

0

$0$ .

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

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

Step 6.1.1.2

Add

0

$0$ and

1 \cdot t

$1 \cdot t$ .

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

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

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

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

Step 6.1.2

Multiply

t

$t$ by

1

$1$ .

- 3 t - 4 t = 9

$- 3 t - 4 t = 9$

Step 6.1.3

Subtract

4 t

$4 t$ from

- 3 t

$- 3 t$ .

- 7 t = 9

$- 7 t = 9$

- 7 t = 9

$- 7 t = 9$

Step 6.2

Divide each term in

- 7 t = 9

$- 7 t = 9$ by

- 7

$- 7$ and simplify.

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Step 6.2.1

Divide each term in

- 7 t = 9

$- 7 t = 9$ by

- 7

$- 7$ .

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

$- 7$ .

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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

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

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