Algebra Examples

$4 x - y = 2$ , $6 x - 2 y = - 1$

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 $(x, y, z)$ .

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

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$P_{1}$ is $4 x - y = 2$ . Find the normal vector $n_{1} = 〈 a, b, c 〉$ from the plane equation of the form $a x + b y + c z = d$ .

$n_{1} = 〈 4, - 1, 0 〉$

$P_{2}$ is $6 x - 2 y = - 1$ . Find the normal vector $n_{2} = 〈 e, f, g 〉$ from the plane equation of the form $e x + f y + g z = h$ .

$n_{2} = 〈 6, - 2, 0 〉$

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.

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

Simplify the dot product.

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Remove parentheses.

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

Simplify each term.

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Multiply $4$ by $6$ .

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

Multiply $- 1$ by $- 2$ .

$24 + 2 + 0 \cdot 0$

Multiply $0$ by $0$ .

$24 + 2 + 0$

Simplify by adding numbers.

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Add $24$ and $2$ .

$26 + 0$

Add $26$ and $0$ .

$26$

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 $26$ for the values of $a$ , $b$ , and $c$ . This set of parametric equations represents the line through the origin that is perpendicular to $P_{1}$ $4 x - y = 2$ .

$x = 0 + 4 \cdot t$

$y = 0 + - 1 \cdot t$

$z = 0 + 0 \cdot t$

Substitute the expression for $x$ , $y$ , and $z$ into the equation for $P_{2}$ $6 x - 2 y = - 1$ .

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

Solve the equation for $t$ .

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

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Combine the opposite terms in $6 (0 + 4 \cdot t) - 2 (0 - 1 \cdot t)$ .

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Add $0$ and $4 \cdot t$ .

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

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

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

Simplify each term.

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Multiply $4$ by $6$ .

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

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

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

Multiply $- 1$ by $- 2$ .

$24 t + 2 t = - 1$

Add $24 t$ and $2 t$ .

$26 t = - 1$

Divide each term by $26$ and simplify.

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Divide each term in $26 t = - 1$ by $26$ .

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

Cancel the common factor of $26$ .

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Cancel the common factor.

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

Divide $t$ by $1$ .

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

Move the negative in front of the fraction.

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

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

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Solve the equation for $x$ .

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Remove parentheses.

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

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

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Simplify each term.

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Cancel the common factor of $2$ .

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Move the leading negative in $- \frac{1}{26}$ into the numerator.

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

Factor $2$ out of $4$ .

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

Factor $2$ out of $26$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Combine $2$ and $\frac{- 1}{13}$ .

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

Multiply $2$ by $- 1$ .

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

Move the negative in front of the fraction.

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

Subtract $\frac{2}{13}$ from $0$ .

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

Solve the equation for $y$ .

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Remove parentheses.

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

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

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

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Multiply $- 1$ by $- 1$ .

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

Multiply $\frac{1}{26}$ by $1$ .

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

Add $0$ and $\frac{1}{26}$ .

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

Solve the equation for $z$ .

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Remove parentheses.

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

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

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

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Multiply $- 1$ by $0$ .

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

Multiply $0$ by $\frac{1}{26}$ .

$z = 0 + 0$

Add $0$ and $0$ .

$z = 0$

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

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

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

$z = 0$

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

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

$z = 0$

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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