Find the Intersection of the Line Perpendicular to Plane 1 Through the Origin and Plane 2 3x+4=y , 2y-3=x

$3 x + 4 = y$ , $2 y - 3 = x$

Step 1

Get each plane equation in standard form.

Subtract $4$ from both sides of the equation.

$3 x = y - 4, 2 y - 3 = x$

Subtract $y$ from both sides of the equation.

$3 x - y = - 4, 2 y - 3 = x$

Add $3$ to both sides of the equation.

$3 x - y = - 4, 2 y = x + 3$

Subtract $x$ from both sides of the equation.

$3 x - y = - 4, 2 y - x = 3$

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 $3 x - y = - 4$ . Find the normal vector $n_{1} = 〈 a, b, c 〉$ from the plane equation of the form $a x + b y + c z = d$ .

$n_{1} = 〈 3, - 1, 0 〉$

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

$n_{2} = 〈 - 1, 2, 0 〉$

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.

$3 \cdot - 1 - 1 \cdot 2 + 0 \cdot 0$

Simplify the dot product.

Remove parentheses.

$3 \cdot - 1 - 1 \cdot 2 + 0 \cdot 0$

Simplify each term.

Multiply $3$ by $- 1$ .

$- 3 - 1 \cdot 2 + 0 \cdot 0$

Multiply $- 1$ by $2$ .

$- 3 - 2 + 0 \cdot 0$

Multiply $0$ by $0$ .

$- 3 - 2 + 0$

Simplify by adding and subtracting.

Subtract $2$ from $- 3$ .

$- 5 + 0$

Add $- 5$ and $0$ .

$- 5$

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 $- 5$ for the values of $a$ , $b$ , and $c$ . This set of parametric equations represents the line through the origin that is perpendicular to $P_{1}$ $3 x - y = - 4$ .

$x = 0 + 3 \cdot t$

$y = 0 + - 1 \cdot t$

$z = 0 + 0 \cdot t$

Step 5

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

$2 (0 - 1 \cdot t) - (0 + 3 \cdot t) = 3$

Step 6

Solve the equation for $t$ .

Simplify $2 (0 - 1 \cdot t) - (0 + 3 \cdot t)$ .

Combine the opposite terms in $2 (0 - 1 \cdot t) - (0 + 3 \cdot t)$ .

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

$2 (- 1 \cdot t) - (0 + 3 \cdot t) = 3$

Add $0$ and $3 \cdot t$ .

$2 (- 1 \cdot t) - (3 \cdot t) = 3$

Simplify each term.

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

$2 (- t) - (3 \cdot t) = 3$

Multiply $- 1$ by $2$ .

$- 2 t - (3 \cdot t) = 3$

Multiply $3$ by $- 1$ .

$- 2 t - 3 t = 3$

Subtract $3 t$ from $- 2 t$ .

$- 5 t = 3$

Divide each term in $- 5 t = 3$ by $- 5$ and simplify.

Divide each term in $- 5 t = 3$ by $- 5$ .

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

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

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

Divide $t$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 7

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

Solve the equation for $x$ .

Remove parentheses.

$x = 0 + 3 \cdot (- 1 (\frac{3}{5}))$

Remove parentheses.

$x = 0 + 3 \cdot (- \frac{3}{5})$

Simplify $0 + 3 \cdot (- \frac{3}{5})$ .

Simplify each term.

Multiply $3 (- \frac{3}{5})$ .

Multiply $- 1$ by $3$ .

$x = 0 - 3 (\frac{3}{5})$

Combine $- 3$ and $\frac{3}{5}$ .

$x = 0 + \frac{- 3 \cdot 3}{5}$

Multiply $- 3$ by $3$ .

$x = 0 + \frac{- 9}{5}$

Move the negative in front of the fraction.

$x = 0 - \frac{9}{5}$

Subtract $\frac{9}{5}$ from $0$ .

$x = - \frac{9}{5}$

Solve the equation for $y$ .

Remove parentheses.

$y = 0 - 1 \cdot (- 1 (\frac{3}{5}))$

Remove parentheses.

$y = 0 - 1 \cdot (- \frac{3}{5})$

Simplify $0 - 1 \cdot (- \frac{3}{5})$ .

Multiply $- 1 (- \frac{3}{5})$ .

Multiply $- 1$ by $- 1$ .

$y = 0 + 1 (\frac{3}{5})$

Multiply $\frac{3}{5}$ by $1$ .

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

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

$y = \frac{3}{5}$

Solve the equation for $z$ .

Remove parentheses.

$z = 0 + 0 \cdot (- 1 (\frac{3}{5}))$

Remove parentheses.

$z = 0 + 0 \cdot (- \frac{3}{5})$

Simplify $0 + 0 \cdot (- \frac{3}{5})$ .

Multiply $0 (- \frac{3}{5})$ .

Multiply $- 1$ by $0$ .

$z = 0 + 0 (\frac{3}{5})$

Multiply $0$ by $\frac{3}{5}$ .

$z = 0 + 0$

Add $0$ and $0$ .

$z = 0$

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

$x = - \frac{9}{5}$

$y = \frac{3}{5}$

$z = 0$

$x = - \frac{9}{5}$

$y = \frac{3}{5}$

$z = 0$

Step 8

Using the values calculated for $x$ , $y$ , and $z$ , the intersection point is found to be $(- \frac{9}{5}, \frac{3}{5}, 0)$ .

$(- \frac{9}{5}, \frac{3}{5}, 0)$

Find the Intersection of the Line Perpendicular to Plane 1 Through the Origin and Plane 2 3x+4=y , 2y-3=x

Related Questions

Support LCMGCF.com by disabling your adblocker.