Algebra Examples

$7 x + 3 = y - 1$ , $y = 3 x$

Get each plane equation in standard form.

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Move $y$ to the left side of the equation because it contains a variable.

$7 x + 3 - y = - 1$

$y = 3 x$

Move $3 x$ to the left side of the equation because it contains a variable.

$7 x + 3 - y = - 1$

$y - 3 x = 0$

Move all terms not containing a variable to the right side of the equation.

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Move $3$ to the right side of the equation because it does not contain a variable.

$7 x - y = - 3 - 1$

$y - 3 x = 0$

Subtract $1$ from $- 3$ to get $- 4$ .

$7 x - y = - 4$

$y - 3 x = 0$

$7 x - y = - 4$

$y - 3 x = 0$

$7 x - y = - 4$

$y - 3 x = 0$

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 $7 x - y = - 4$ . Find the normal vector $n_{1} = 〈 a, b, c 〉$ from the plane equation of the form $a x + b y + c z = d$ .

$n_{1} = 〈 7, - 1, 0 〉$

$P_{2}$ is $y - 3 x = 0$ . Find the normal vector $n_{2} = 〈 e, f, g 〉$ from the plane equation of the form $e x + f y + g z = h$ .

$n_{2} = 〈 - 3, 1, 0 〉$

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.

$7 \cdot - 3 - 1 \cdot 1 + 0 \cdot 0$

Simplify the dot product.

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Remove parentheses.

$7 \cdot - 3 - 1 \cdot 1 + 0 \cdot 0$

Simplify each term.

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Multiply $7$ by $- 3$ to get $- 21$ .

$- 21 - 1 \cdot 1 + 0 \cdot 0$

Multiply $- 1$ by $1$ to get $- 1$ .

$- 21 - 1 + 0 \cdot 0$

Multiply $0$ by $0$ to get $0$ .

$- 21 - 1 + 0$

Simplify by subtracting numbers.

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Subtract $1$ from $- 21$ to get $- 22$ .

$- 22 + 0$

Add $- 22$ and $0$ to get $- 22$ .

$- 22$

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 $- 22$ for the values of $a$ , $b$ , and $c$ . This set of parametric equations represents the line through the origin that is perpendicular to $P_{1}$ $7 x - y = - 4$ .

$x = 0 + 7 \cdot t$

$y = 0 + - 1 \cdot t$

$z = 0 + 0 \cdot t$

Substitute the expression for $x$ , $y$ , and $z$ into the equation for $P_{2}$ $y - 3 x = 0$ .

$(0 - 1 \cdot t) - 3 (0 + 7 \cdot t) = 0$

Solve the equation for $t$ .

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Simplify the left side.

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Simplify each term.

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Multiply $- 1$ by $t$ to get $- 1 t$ .

$0 - 1 t - 3 (7 \cdot t) = 0$

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

$0 - t - 3 (7 \cdot t) = 0$

Multiply $7$ by $t$ to get $7 t$ .

$0 - t - 3 (7 t) = 0$

Multiply $7$ by $- 3$ to get $- 21$ .

$0 - t - 21 t = 0$

Simplify by adding terms.

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Subtract $t$ from $0$ to get $- t$ .

$- t - 21 t = 0$

Subtract $21 t$ from $- t$ to get $- 22 t$ .

$- 22 t = 0$

Divide each term by $- 22$ and simplify.

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Divide each term in $- 22 t = 0$ by $- 22$ .

$- \frac{22 t}{- 22} = \frac{0}{- 22}$

Simplify the left side of the equation by cancelling the common factors.

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Reduce the expression by cancelling the common factors.

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Factor $22$ out of $- 22$ .

$- \frac{22 (t)}{22 (- 1)} = \frac{0}{- 22}$

Cancel the common factor.

$- \frac{22 t}{22 \cdot - 1} = \frac{0}{- 22}$

Rewrite the expression.

$- \frac{t}{- 1} = \frac{0}{- 22}$

Move the negative one from the denominator of $\frac{t}{- 1}$ .

$- (- 1 \cdot t) = \frac{0}{- 22}$

Simplify the expression.

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Multiply $- 1$ by $t$ to get $- 1 t$ .

$- (- 1 t) = \frac{0}{- 22}$

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

$t = \frac{0}{- 22}$

Divide $0$ by $- 22$ to get $0$ .

$t = 0$

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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Simplify each term.

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Multiply $7$ by $0$ to get $7 (0)$ .

$x = 0 + 7 (0)$

Multiply $7$ by $0$ to get $0$ .

$x = 0 + 0$

Add $0$ and $0$ to get $0$ .

$x = 0$

Subtract $0$ from $0$ to get $0$ .

$y = 0$

Solve the equation for $z$ .

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Simplify each term.

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Multiply $0$ by $0$ to get $0 (0)$ .

$z = 0 + 0 (0)$

Multiply $0$ by $0$ to get $0$ .

$z = 0 + 0$

Add $0$ and $0$ to get $0$ .

$z = 0$

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

$x = 0$

$y = 0$

$z = 0$

$x = 0$

$y = 0$

$z = 0$

Using the values calculated for $x$ , $y$ , and $z$ , the intersection point is found to be $(0, 0, 0)$ .

$(0, 0, 0)$

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