Find the Intersection of the Line Perpendicular to Plane 1 Through the Origin and Plane 2 -(y-4)=x+9 , x-8/3y=0

$- (y - 4) = x + 9$ , $x - \frac{8}{3} y = 0$

Step 1

Get each plane equation in standard form.

Simplify.

Apply the distributive property.

$- y - - 4 = x + 9, x - \frac{8}{3} y = 0$

Multiply $- 1$ by $- 4$ .

$- y + 4 = x + 9, x - \frac{8}{3} y = 0$

Subtract $4$ from both sides of the equation.

$- y = x + 9 - 4, x - \frac{8}{3} y = 0$

Subtract $x$ from both sides of the equation.

$- y - x = 9 - 4, x - \frac{8}{3} y = 0$

Simplify each term.

Combine $y$ and $\frac{8}{3}$ .

$- y - x = 9 - 4, x - \frac{y \cdot 8}{3} = 0$

Move $8$ to the left of $y$ .

$- y - x = 9 - 4, x - \frac{8 y}{3} = 0$

Subtract $4$ from $9$ .

$- y - x = 5, x - \frac{8 y}{3} = 0$

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 $- y - x = 5$ . Find the normal vector $n_{1} = 〈 a, b, c 〉$ from the plane equation of the form $a x + b y + c z = d$ .

$n_{1} = 〈 - 1, - 1, 0 〉$

$P_{2}$ is $x - \frac{8 y}{3} = 0$ . Find the normal vector $n_{2} = 〈 e, f, g 〉$ from the plane equation of the form $e x + f y + g z = h$ .

$n_{2} = 〈 1, - \frac{8}{3}, 0 〉$

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.

$- 1 \cdot 1 - 1 \cdot (- \frac{8}{3}) + 0 \cdot 0$

Simplify the dot product.

Remove parentheses.

$- 1 \cdot 1 - 1 \cdot (- \frac{8}{3}) + 0 \cdot 0$

Simplify each term.

Multiply $- 1$ by $1$ .

$- 1 - 1 \cdot (- \frac{8}{3}) + 0 \cdot 0$

Multiply $- 1 (- \frac{8}{3})$ .

Multiply $- 1$ by $- 1$ .

$- 1 + 1 (\frac{8}{3}) + 0 \cdot 0$

Multiply $\frac{8}{3}$ by $1$ .

$- 1 + \frac{8}{3} + 0 \cdot 0$

Multiply $0$ by $0$ .

$- 1 + \frac{8}{3} + 0$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- 1 \cdot \frac{3}{3} + \frac{8}{3} + 0$

Combine $- 1$ and $\frac{3}{3}$ .

$\frac{- 1 \cdot 3}{3} + \frac{8}{3} + 0$

Combine the numerators over the common denominator.

$\frac{- 1 \cdot 3 + 8}{3} + 0$

Simplify the numerator.

Multiply $- 1$ by $3$ .

$\frac{- 3 + 8}{3} + 0$

Add $- 3$ and $8$ .

$\frac{5}{3} + 0$

Add $\frac{5}{3}$ and $0$ .

$\frac{5}{3}$

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 $\frac{5}{3}$ for the values of $a$ , $b$ , and $c$ . This set of parametric equations represents the line through the origin that is perpendicular to $P_{1}$ $- y - x = 5$ .

$x = 0 + - 1 \cdot t$

$y = 0 + - 1 \cdot t$

$z = 0 + 0 \cdot t$

Step 5

Substitute the expression for $x$ , $y$ , and $z$ into the equation for $P_{2}$ $x - \frac{8 y}{3} = 0$ .

$(0 - 1 \cdot t) - \frac{8 (0 - 1 \cdot t)}{3} = 0$

Step 6

Solve the equation for $t$ .

Simplify $(0 - 1 \cdot t) - \frac{8 (0 - 1 \cdot t)}{3}$ .

Combine the opposite terms in $(0 - 1 \cdot t) - \frac{8 (0 - 1 \cdot t)}{3}$ .

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

$- 1 \cdot t - \frac{8 (0 - 1 \cdot t)}{3} = 0$

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

$- 1 \cdot t - \frac{8 (- 1 \cdot t)}{3} = 0$

Simplify each term.

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

$- t - \frac{8 (- 1 \cdot t)}{3} = 0$

Multiply $- 1$ by $8$ .

$- t - \frac{- 8 t}{3} = 0$

Move the negative in front of the fraction.

$- t - - \frac{(8) t}{3} = 0$

Multiply $- - \frac{8 t}{3}$ .

Multiply $- 1$ by $- 1$ .

$- t + 1 \frac{8 t}{3} = 0$

Multiply $\frac{8 t}{3}$ by $1$ .

$- t + \frac{8 t}{3} = 0$

To write $- t$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$- t \cdot \frac{3}{3} + \frac{8 t}{3} = 0$

Simplify terms.

Combine $- t$ and $\frac{3}{3}$ .

$\frac{- t \cdot 3}{3} + \frac{8 t}{3} = 0$

Combine the numerators over the common denominator.

$\frac{- t \cdot 3 + 8 t}{3} = 0$

Simplify the numerator.

Factor $t$ out of $- t \cdot 3 + 8 t$ .

Factor $t$ out of $- t \cdot 3$ .

$\frac{t (- 1 \cdot 3) + 8 t}{3} = 0$

Factor $t$ out of $8 t$ .

$\frac{t (- 1 \cdot 3) + t \cdot 8}{3} = 0$

Factor $t$ out of $t (- 1 \cdot 3) + t \cdot 8$ .

$\frac{t (- 1 \cdot 3 + 8)}{3} = 0$

Multiply $- 1$ by $3$ .

$\frac{t (- 3 + 8)}{3} = 0$

Add $- 3$ and $8$ .

$\frac{t \cdot 5}{3} = 0$

Move $5$ to the left of $t$ .

$\frac{5 t}{3} = 0$

Set the numerator equal to zero.

$5 t = 0$

Divide each term in $5 t = 0$ by $5$ and simplify.

Divide each term in $5 t = 0$ by $5$ .

$\frac{5 t}{5} = \frac{0}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 t}{5} = \frac{0}{5}$

Divide $t$ by $1$ .

$t = \frac{0}{5}$

Simplify the right side.

Divide $0$ by $5$ .

$t = 0$

Step 7

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

Solve the equation for $x$ .

Remove parentheses.

$x = 0 - 1 \cdot 0$

Subtract $0$ from $0$ .

$x = 0$

Solve the equation for $y$ .

Remove parentheses.

$y = 0 - 1 \cdot 0$

Subtract $0$ from $0$ .

$y = 0$

Solve the equation for $z$ .

Remove parentheses.

$z = 0 + 0 \cdot (0)$

Simplify $0 + 0 \cdot (0)$ .

Multiply $0$ by $0$ .

$z = 0 + 0$

Add $0$ and $0$ .

$z = 0$

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

$x = 0$

$y = 0$

$z = 0$

$x = 0$

$y = 0$

$z = 0$

Step 8

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

$(0, 0, 0)$

Find the Intersection of the Line Perpendicular to Plane 1 Through the Origin and Plane 2 -(y-4)=x+9 , x-8/3y=0

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