Find the Intersection of the Line Perpendicular to Plane 1 Through the Origin and Plane 2 2y=x^2+2x , y=4x-23

$2 y = x^{2} + 2 x$ , $y = 4 x - 23$

Step 1

Get each plane equation in standard form.

Move all terms containing variables to the left side of the equation.

Subtract $x^{2}$ from both sides of the equation.

$2 y - x^{2} = 2 x, y = 4 x - 23$

Subtract $2 x$ from both sides of the equation.

$2 y - x^{2} - 2 x = 0, y = 4 x - 23$

Subtract $4 x$ from both sides of the equation.

$2 y - x^{2} - 2 x = 0, y - 4 x = - 23$

Step 2

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 $(x, y, z)$ .

Step 3

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

$P_{1}$ is $2 y - x^{2} - 2 x = 0$ . Find the normal vector $n_{1} = 〈 a, b, c 〉$ from the plane equation of the form $a x + b y + c z = d$ .

$n_{1} = 〈 - 2, 2, 0 〉$

$P_{2}$ is $y - 4 x = - 23$ . Find the normal vector $n_{2} = 〈 e, f, g 〉$ from the plane equation of the form $e x + f y + g z = h$ .

$n_{2} = 〈 - 4, 1, 0 〉$

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.

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

Simplify the dot product.

Remove parentheses.

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

Simplify each term.

Multiply $- 2$ by $- 4$ .

$8 + 2 \cdot 1 + 0 \cdot 0$

Multiply $2$ by $1$ .

$8 + 2 + 0 \cdot 0$

Multiply $0$ by $0$ .

$8 + 2 + 0$

Simplify by adding numbers.

Add $8$ and $2$ .

$10 + 0$

Add $10$ and $0$ .

$10$

Step 4

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 $10$ for the values of $a$ , $b$ , and $c$ . This set of parametric equations represents the line through the origin that is perpendicular to $P_{1}$ $2 y - x^{2} - 2 x = 0$ .

$x = 0 + - 2 \cdot t$

$y = 0 + 2 \cdot t$

$z = 0 + 0 \cdot t$

Step 5

Substitute the expression for $x$ , $y$ , and $z$ into the equation for $P_{2}$ $y - 4 x = - 23$ .

$(0 + 2 \cdot t) - 4 (0 - 2 \cdot t) = - 23$

Step 6

Solve the equation for $t$ .

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

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

Add $0$ and $2 \cdot t$ .

$2 \cdot t - 4 (0 - 2 \cdot t) = - 23$

Subtract $2 \cdot t$ from $0$ .

$2 \cdot t - 4 (- 2 \cdot t) = - 23$

Multiply $- 2$ by $- 4$ .

$2 t + 8 t = - 23$

Add $2 t$ and $8 t$ .

$10 t = - 23$

Divide each term in $10 t = - 23$ by $10$ and simplify.

Divide each term in $10 t = - 23$ by $10$ .

$\frac{10 t}{10} = \frac{- 23}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 t}{10} = \frac{- 23}{10}$

Divide $t$ by $1$ .

$t = \frac{- 23}{10}$

Simplify the right side.

Move the negative in front of the fraction.

$t = - \frac{23}{10}$

Step 7

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

Solve the equation for $x$ .

Remove parentheses.

$x = 0 - 2 \cdot (- 1 (\frac{23}{10}))$

Remove parentheses.

$x = 0 - 2 \cdot (- \frac{23}{10})$

Simplify $0 - 2 \cdot (- \frac{23}{10})$ .

Simplify each term.

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{23}{10}$ into the numerator.

$x = 0 - 2 \cdot \frac{- 23}{10}$

Factor $2$ out of $- 2$ .

$x = 0 + 2 (- 1) \cdot \frac{- 23}{10}$

Factor $2$ out of $10$ .

$x = 0 + 2 \cdot - 1 \cdot \frac{- 23}{2 \cdot 5}$

Cancel the common factor.

$x = 0 + 2 \cdot - 1 \cdot \frac{- 23}{2 \cdot 5}$

Rewrite the expression.

$x = 0 - 1 \cdot \frac{- 23}{5}$

Move the negative in front of the fraction.

$x = 0 - 1 \cdot (- \frac{23}{5})$

Multiply $- 1 (- \frac{23}{5})$ .

Multiply $- 1$ by $- 1$ .

$x = 0 + 1 (\frac{23}{5})$

Multiply $\frac{23}{5}$ by $1$ .

$x = 0 + \frac{23}{5}$

Add $0$ and $\frac{23}{5}$ .

$x = \frac{23}{5}$

Solve the equation for $y$ .

Remove parentheses.

$y = 0 + 2 \cdot (- 1 (\frac{23}{10}))$

Remove parentheses.

$y = 0 + 2 \cdot (- \frac{23}{10})$

Simplify $0 + 2 \cdot (- \frac{23}{10})$ .

Simplify each term.

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{23}{10}$ into the numerator.

$y = 0 + 2 \cdot \frac{- 23}{10}$

Factor $2$ out of $10$ .

$y = 0 + 2 \cdot \frac{- 23}{2 (5)}$

Cancel the common factor.

$y = 0 + 2 \cdot \frac{- 23}{2 \cdot 5}$

Rewrite the expression.

$y = 0 + \frac{- 23}{5}$

Move the negative in front of the fraction.

$y = 0 - \frac{23}{5}$

Subtract $\frac{23}{5}$ from $0$ .

$y = - \frac{23}{5}$

Solve the equation for $z$ .

Remove parentheses.

$z = 0 + 0 \cdot (- 1 (\frac{23}{10}))$

Remove parentheses.

$z = 0 + 0 \cdot (- \frac{23}{10})$

Simplify $0 + 0 \cdot (- \frac{23}{10})$ .

Multiply $0 (- \frac{23}{10})$ .

Multiply $- 1$ by $0$ .

$z = 0 + 0 (\frac{23}{10})$

Multiply $0$ by $\frac{23}{10}$ .

$z = 0 + 0$

Add $0$ and $0$ .

$z = 0$

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

$x = \frac{23}{5}$

$y = - \frac{23}{5}$

$z = 0$

$x = \frac{23}{5}$

$y = - \frac{23}{5}$

$z = 0$

Step 8

Using the values calculated for $x$ , $y$ , and $z$ , the intersection point is found to be $(\frac{23}{5}, - \frac{23}{5}, 0)$ .

$(\frac{23}{5}, - \frac{23}{5}, 0)$

Find the Intersection of the Line Perpendicular to Plane 1 Through the Origin and Plane 2 2y=x^2+2x , y=4x-23

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