Algebra Examples

$7 x - y = - 4$ , $3 x - y = 0$

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 $3 x - y = 0$ . 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 $3$ .

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

Multiply $- 1$ by $- 1$ .

$21 + 1 + 0 \cdot 0$

Multiply $0$ by $0$ .

$21 + 1 + 0$

Simplify by adding numbers.

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Add $21$ and $1$ .

$22 + 0$

Add $22$ and $0$ .

$22$

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 $P_{2}$ $3 x - y = 0$ .

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

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 $7$ by $t$ .

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

Add $0$ and $7 t$ .

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

Multiply $7$ by $3$ .

$21 t - (0 - 1 \cdot t) = 0$

Simplify each term.

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Multiply $- 1$ by $t$ .

$21 t - (0 - 1 t) = 0$

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

$21 t - (0 - t) = 0$

Subtract $t$ from $0$ .

$21 t + t = 0$

Simplify $- - t$ .

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Multiply $- 1$ by $- 1$ .

$21 t + 1 t = 0$

Multiply $t$ by $1$ .

$21 t + t = 0$

Add $21 t$ and $t$ .

$22 t = 0$

Divide each term by $22$ and simplify.

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

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

Reduce the expression by cancelling the common factors.

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Cancel the common factor.

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

Divide $t$ by $1$ .

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

Divide $0$ by $22$ .

$t = 0$

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

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Simplify the right side.

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Simplify each term.

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Multiply $7$ by $0$ .

$x = 0 + 7 (0)$

Multiply $7$ by $0$ .

$x = 0 + 0$

Add $0$ and $0$ .

$x = 0$

Subtract $0$ from $0$ .

$y = 0$

Simplify the right side.

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Simplify each term.

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Multiply $0$ by $0$ .

$z = 0 + 0 (0)$

Multiply $0$ by $0$ .

$z = 0 + 0$

Add $0$ and $0$ .

$z = 0$

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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