Algebra Examples

4 x - y = 2

$4 x - y = 2$ ,

6 x - 2 y = - 1

$6 x - 2 y = - 1$

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

4 x - y = 2

$4 x - y = 2$ . 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} = ⟨ 4, - 1, 0 ⟩

$n_{1} = ⟨ 4, - 1, 0 ⟩$

Step 2.2

P_{2}

$P_{2}$ is

6 x - 2 y = - 1

$6 x - 2 y = - 1$ . 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} = ⟨ 6, - 2, 0 ⟩

$n_{2} = ⟨ 6, - 2, 0 ⟩$

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.

4 \cdot 6 - 1 \cdot - 2 + 0 \cdot 0

$4 \cdot 6 - 1 \cdot - 2 + 0 \cdot 0$

Step 2.4

Simplify the dot product.

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Step 2.4.1

Remove parentheses.

4 \cdot 6 - 1 \cdot - 2 + 0 \cdot 0

$4 \cdot 6 - 1 \cdot - 2 + 0 \cdot 0$

Step 2.4.2

Simplify each term.

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Step 2.4.2.1

Multiply

4

$4$ by

6

$6$ .

24 - 1 \cdot - 2 + 0 \cdot 0

$24 - 1 \cdot - 2 + 0 \cdot 0$

Step 2.4.2.2

Multiply

- 1

$- 1$ by

- 2

$- 2$ .

24 + 2 + 0 \cdot 0

$24 + 2 + 0 \cdot 0$

Step 2.4.2.3

Multiply

0

$0$ by

0

$0$ .

24 + 2 + 0

$24 + 2 + 0$

24 + 2 + 0

$24 + 2 + 0$

Step 2.4.3

Simplify by adding numbers.

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Step 2.4.3.1

Add

24

$24$ and

2

$2$ .

26 + 0

$26 + 0$

Step 2.4.3.2

Add

26

$26$ and

0

$0$ .

26

$26$

26

$26$

26

$26$

26

$26$

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

26

$26$ 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}

$P_{1}$

4 x - y = 2

$4 x - y = 2$ .

x = 0 + 4 \cdot t

$x = 0 + 4 \cdot t$

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}

$P_{2}$

6 x - 2 y = - 1

$6 x - 2 y = - 1$ .

6 (0 + 4 \cdot t) - 2 (0 - 1 \cdot t) = - 1

$6 (0 + 4 \cdot t) - 2 (0 - 1 \cdot t) = - 1$

Step 5

Solve the equation for

t

$t$ .

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Step 5.1

Simplify

6 (0 + 4 \cdot t) - 2 (0 - 1 \cdot t)

$6 (0 + 4 \cdot t) - 2 (0 - 1 \cdot t)$ .

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Step 5.1.1

Combine the opposite terms in

6 (0 + 4 \cdot t) - 2 (0 - 1 \cdot t)

$6 (0 + 4 \cdot t) - 2 (0 - 1 \cdot t)$ .

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Step 5.1.1.1

Add

0

$0$ and

4 \cdot t

$4 \cdot t$ .

6 (4 \cdot t) - 2 (0 - 1 \cdot t) = - 1

$6 (4 \cdot t) - 2 (0 - 1 \cdot t) = - 1$

Step 5.1.1.2

Subtract

1 \cdot t

$1 \cdot t$ from

0

$0$ .

6 (4 \cdot t) - 2 (- 1 \cdot t) = - 1

$6 (4 \cdot t) - 2 (- 1 \cdot t) = - 1$

6 (4 \cdot t) - 2 (- 1 \cdot t) = - 1

$6 (4 \cdot t) - 2 (- 1 \cdot t) = - 1$

Step 5.1.2

Simplify each term.

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Step 5.1.2.1

Multiply

4

$4$ by

6

$6$ .

24 t - 2 (- 1 \cdot t) = - 1

$24 t - 2 (- 1 \cdot t) = - 1$

Step 5.1.2.2

Rewrite

- 1 t

$- 1 t$ as

- t

$- t$ .

24 t - 2 (- t) = - 1

$24 t - 2 (- t) = - 1$

Step 5.1.2.3

Multiply

- 1

$- 1$ by

- 2

$- 2$ .

24 t + 2 t = - 1

$24 t + 2 t = - 1$

24 t + 2 t = - 1

$24 t + 2 t = - 1$

Step 5.1.3

Add

24 t

$24 t$ and

2 t

$2 t$ .

26 t = - 1

$26 t = - 1$

26 t = - 1

$26 t = - 1$

Step 5.2

Divide each term in

26 t = - 1

$26 t = - 1$ by

26

$26$ and simplify.

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Step 5.2.1

Divide each term in

26 t = - 1

$26 t = - 1$ by

26

$26$ .

\frac{26 t}{26} = \frac{- 1}{26}

$\frac{26 t}{26} = \frac{- 1}{26}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of

26

$26$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{26 t}{26} = \frac{- 1}{26}$

Step 5.2.2.1.2

Divide

$t$ by

$1$ .

$t = \frac{- 1}{26}$

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.

$t = - \frac{1}{26}$

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.

$x = 0 + 4 \cdot (- 1 (\frac{1}{26}))$

Step 6.1.2

Remove parentheses.

$x = 0 + 4 \cdot (- \frac{1}{26})$

Step 6.1.3

Simplify

$0 + 4 \cdot (- \frac{1}{26})$ .

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Step 6.1.3.1

Simplify each term.

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Step 6.1.3.1.1

Cancel the common factor of

$2$ .

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Step 6.1.3.1.1.1

Move the leading negative in

$- \frac{1}{26}$ into the numerator.

$x = 0 + 4 \cdot \frac{- 1}{26}$

Step 6.1.3.1.1.2

Factor

$2$ out of

$4$ .

$x = 0 + 2 (2) \cdot \frac{- 1}{26}$

Step 6.1.3.1.1.3

Factor

$2$ out of

$26$ .

$x = 0 + 2 \cdot 2 \cdot \frac{- 1}{2 \cdot 13}$

Step 6.1.3.1.1.4

Cancel the common factor.

$x = 0 + 2 \cdot 2 \cdot \frac{- 1}{2 \cdot 13}$

Step 6.1.3.1.1.5

Rewrite the expression.

$x = 0 + 2 \cdot \frac{- 1}{13}$

Step 6.1.3.1.2

Combine

$2$ and

$\frac{- 1}{13}$ .

$x = 0 + \frac{2 \cdot - 1}{13}$

Step 6.1.3.1.3

Multiply

$2$ by

$- 1$ .

$x = 0 + \frac{- 2}{13}$

Step 6.1.3.1.4

Move the negative in front of the fraction.

$x = 0 - \frac{2}{13}$

Step 6.1.3.2

Subtract

$\frac{2}{13}$ from

$0$ .

$x = - \frac{2}{13}$

Step 6.2

Solve the equation for

$y$ .

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Step 6.2.1

Remove parentheses.

$y = 0 - 1 \cdot (- 1 (\frac{1}{26}))$

Step 6.2.2

Remove parentheses.

$y = 0 - 1 \cdot (- \frac{1}{26})$

Step 6.2.3

Simplify

$0 - 1 \cdot (- \frac{1}{26})$ .

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Step 6.2.3.1

Multiply

$- 1 (- \frac{1}{26})$ .

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Step 6.2.3.1.1

Multiply

$- 1$ by

$- 1$ .

$y = 0 + 1 (\frac{1}{26})$

Step 6.2.3.1.2

Multiply

$\frac{1}{26}$ by

$1$ .

$y = 0 + \frac{1}{26}$

Step 6.2.3.2

Add

$0$ and

$\frac{1}{26}$ .

$y = \frac{1}{26}$

Step 6.3

Solve the equation for

$z$ .

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Step 6.3.1

Remove parentheses.

$z = 0 + 0 \cdot (- 1 (\frac{1}{26}))$

Step 6.3.2

Remove parentheses.

$z = 0 + 0 \cdot (- \frac{1}{26})$

Step 6.3.3

Simplify

$0 + 0 \cdot (- \frac{1}{26})$ .

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Step 6.3.3.1

Multiply

$0 (- \frac{1}{26})$ .

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Step 6.3.3.1.1

Multiply

$- 1$ by

$0$ .

$z = 0 + 0 (\frac{1}{26})$

Step 6.3.3.1.2

Multiply

$0$ by

$\frac{1}{26}$ .

$z = 0 + 0$

Step 6.3.3.2

Add

$0$ and

$0$ .

$z = 0$

Step 6.4

The solved parametric equations for

$x$ ,

$y$ , and

$z$ .

$x = - \frac{2}{13}$

$y = \frac{1}{26}$

$z = 0$

$x = - \frac{2}{13}$

$y = \frac{1}{26}$

$z = 0$

Step 7

Using the values calculated for

$x$ ,

$y$ , and

$z$ , the intersection point is found to be

$(- \frac{2}{13}, \frac{1}{26}, 0)$ .

$(- \frac{2}{13}, \frac{1}{26}, 0)$

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