Find the Intersection of the Line Perpendicular to Plane 1 Through the Origin and Plane 2 (x+3)/4+(y-1)/3=1 , 2x-y=12

$\frac{x + 3}{4} + \frac{y - 1}{3} = 1$ , $2 x - y = 12$

Get each plane equation in standard form.

Simplify the left side.

Simplify $\frac{x + 3}{4} + \frac{y - 1}{3}$ .

To write $\frac{x + 3}{4}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{x + 3}{4} \cdot \frac{3}{3} + \frac{y - 1}{3} = 1$

$2 x - y = 12$

To write $\frac{y - 1}{3}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{x + 3}{4} \cdot \frac{3}{3} + \frac{y - 1}{3} \cdot \frac{4}{4} = 1$

$2 x - y = 12$

Write each expression with a common denominator of $12$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{x + 3}{4}$ by $\frac{3}{3}$ .

$\frac{(x + 3) \cdot 3}{4 \cdot 3} + \frac{y - 1}{3} \cdot \frac{4}{4} = 1$

$2 x - y = 12$

Multiply $4$ by $3$ .

$\frac{(x + 3) \cdot 3}{12} + \frac{y - 1}{3} \cdot \frac{4}{4} = 1$

$2 x - y = 12$

Multiply $\frac{y - 1}{3}$ by $\frac{4}{4}$ .

$\frac{(x + 3) \cdot 3}{12} + \frac{(y - 1) \cdot 4}{3 \cdot 4} = 1$

$2 x - y = 12$

Multiply $3$ by $4$ .

$\frac{(x + 3) \cdot 3}{12} + \frac{(y - 1) \cdot 4}{12} = 1$

$2 x - y = 12$

$\frac{(x + 3) \cdot 3}{12} + \frac{(y - 1) \cdot 4}{12} = 1$

$2 x - y = 12$

Combine the numerators over the common denominator.

$\frac{(x + 3) \cdot 3 + (y - 1) \cdot 4}{12} = 1$

$2 x - y = 12$

Simplify the numerator.

Apply the distributive property.

$\frac{x \cdot 3 + 3 \cdot 3 + (y - 1) \cdot 4}{12} = 1$

$2 x - y = 12$

Move $3$ to the left of $x$ .

$\frac{3 \cdot x + 3 \cdot 3 + (y - 1) \cdot 4}{12} = 1$

$2 x - y = 12$

Multiply $3$ by $3$ .

$\frac{3 \cdot x + 9 + (y - 1) \cdot 4}{12} = 1$

$2 x - y = 12$

Apply the distributive property.

$\frac{3 x + 9 + y \cdot 4 - 1 \cdot 4}{12} = 1$

$2 x - y = 12$

Move $4$ to the left of $y$ .

$\frac{3 x + 9 + 4 \cdot y - 1 \cdot 4}{12} = 1$

$2 x - y = 12$

Multiply $- 1$ by $4$ .

$\frac{3 x + 9 + 4 \cdot y - 4}{12} = 1$

$2 x - y = 12$

Subtract $4$ from $9$ .

$\frac{3 x + 4 y + 5}{12} = 1$

$2 x - y = 12$

$\frac{3 x + 4 y + 5}{12} = 1$

$2 x - y = 12$

$\frac{3 x + 4 y + 5}{12} = 1$

$2 x - y = 12$

$\frac{3 x + 4 y + 5}{12} = 1$

$2 x - y = 12$

Split the fraction $\frac{3 x + 4 y + 5}{12}$ into two fractions.

$\frac{3 x + 4 y}{12} + \frac{5}{12} = 1$

$2 x - y = 12$

Split the fraction $\frac{3 x + 4 y}{12}$ into two fractions.

$\frac{3 x}{12} + \frac{4 y}{12} + \frac{5}{12} = 1$

$2 x - y = 12$

Cancel the common factor of $3$ and $12$ .

Factor $3$ out of $3 x$ .

$\frac{3 (x)}{12} + \frac{4 y}{12} + \frac{5}{12} = 1$

$2 x - y = 12$

Cancel the common factors.

Factor $3$ out of $12$ .

$\frac{3 x}{3 \cdot 4} + \frac{4 y}{12} + \frac{5}{12} = 1$

$2 x - y = 12$

Cancel the common factor.

$\frac{3 x}{3 \cdot 4} + \frac{4 y}{12} + \frac{5}{12} = 1$

$2 x - y = 12$

Rewrite the expression.

$\frac{x}{4} + \frac{4 y}{12} + \frac{5}{12} = 1$

$2 x - y = 12$

$\frac{x}{4} + \frac{4 y}{12} + \frac{5}{12} = 1$

$2 x - y = 12$

$\frac{x}{4} + \frac{4 y}{12} + \frac{5}{12} = 1$

$2 x - y = 12$

Cancel the common factor of $4$ and $12$ .

Factor $4$ out of $4 y$ .

$\frac{x}{4} + \frac{4 (y)}{12} + \frac{5}{12} = 1$

$2 x - y = 12$

Cancel the common factors.

Factor $4$ out of $12$ .

$\frac{x}{4} + \frac{4 y}{4 \cdot 3} + \frac{5}{12} = 1$

$2 x - y = 12$

Cancel the common factor.

$\frac{x}{4} + \frac{4 y}{4 \cdot 3} + \frac{5}{12} = 1$

$2 x - y = 12$

Rewrite the expression.

$\frac{x}{4} + \frac{y}{3} + \frac{5}{12} = 1$

$2 x - y = 12$

$\frac{x}{4} + \frac{y}{3} + \frac{5}{12} = 1$

$2 x - y = 12$

$\frac{x}{4} + \frac{y}{3} + \frac{5}{12} = 1$

$2 x - y = 12$

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

Subtract $\frac{5}{12}$ from both sides of the equation.

$\frac{x}{4} + \frac{y}{3} = 1 - \frac{5}{12}$

$2 x - y = 12$

Write $1$ as a fraction with a common denominator.

$\frac{x}{4} + \frac{y}{3} = \frac{12}{12} - \frac{5}{12}$

$2 x - y = 12$

Combine the numerators over the common denominator.

$\frac{x}{4} + \frac{y}{3} = \frac{12 - 5}{12}$

$2 x - y = 12$

Subtract $5$ from $12$ .

$\frac{x}{4} + \frac{y}{3} = \frac{7}{12}$

$2 x - y = 12$

$\frac{x}{4} + \frac{y}{3} = \frac{7}{12}$

$2 x - y = 12$

$\frac{x}{4} + \frac{y}{3} = \frac{7}{12}$

$2 x - y = 12$

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.

$P_{1}$ is $\frac{x}{4} + \frac{y}{3} = \frac{7}{12}$ . Find the normal vector $n_{1} = 〈 a, b, c 〉$ from the plane equation of the form $a x + b y + c z = d$ .

$n_{1} = 〈 \frac{1}{4}, \frac{1}{3}, 0 〉$

$P_{2}$ is $2 x - y = 12$ . Find the normal vector $n_{2} = 〈 e, f, g 〉$ from the plane equation of the form $e x + f y + g z = h$ .

$n_{2} = 〈 2, - 1, 0 〉$

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.

$\frac{1}{4} \cdot 2 + \frac{1}{3} \cdot - 1 + 0 \cdot 0$

Simplify the dot product.

Remove parentheses.

$\frac{1}{4} \cdot 2 + \frac{1}{3} \cdot - 1 + 0 \cdot 0$

Simplify each term.

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$\frac{1}{2 (2)} \cdot 2 + \frac{1}{3} \cdot - 1 + 0 \cdot 0$

Cancel the common factor.

$\frac{1}{2 \cdot 2} \cdot 2 + \frac{1}{3} \cdot - 1 + 0 \cdot 0$

Rewrite the expression.

$\frac{1}{2} + \frac{1}{3} \cdot - 1 + 0 \cdot 0$

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

$\frac{1}{2} + \frac{- 1}{3} + 0 \cdot 0$

Move the negative in front of the fraction.

$\frac{1}{2} - \frac{1}{3} + 0 \cdot 0$

Multiply $0$ by $0$ .

$\frac{1}{2} - \frac{1}{3} + 0$

To write $\frac{1}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{1}{2} \cdot \frac{3}{3} - \frac{1}{3} + 0$

To write $- \frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{2} \cdot \frac{3}{3} - \frac{1}{3} \cdot \frac{2}{2} + 0$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{2}$ by $\frac{3}{3}$ .

$\frac{3}{2 \cdot 3} - \frac{1}{3} \cdot \frac{2}{2} + 0$

Multiply $2$ by $3$ .

$\frac{3}{6} - \frac{1}{3} \cdot \frac{2}{2} + 0$

Multiply $\frac{1}{3}$ by $\frac{2}{2}$ .

$\frac{3}{6} - \frac{2}{3 \cdot 2} + 0$

Multiply $3$ by $2$ .

$\frac{3}{6} - \frac{2}{6} + 0$

Combine the numerators over the common denominator.

$\frac{3 - 2}{6} + 0$

Subtract $2$ from $3$ .

$\frac{1}{6} + 0$

Add $\frac{1}{6}$ and $0$ .

$\frac{1}{6}$

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 $\frac{1}{6}$ for the values of $a$ , $b$ , and $c$ . This set of parametric equations represents the line through the origin that is perpendicular to $P_{1}$ $\frac{x}{4} + \frac{y}{3} = \frac{7}{12}$ .

$x = 0 + \frac{1}{4} t$

$y = 0 + \frac{1}{3} t$

$z = 0 + 0 \cdot t$

Substitute the expression for $x$ , $y$ , and $z$ into the equation for $P_{2}$ $2 x - y = 12$ .

$2 (0 + \frac{1}{4} t) - (0 + \frac{1}{3} t) = 12$

Solve the equation for $t$ .

Simplify $2 (0 + \frac{1}{4} t) - (0 + \frac{1}{3} t)$ .

Combine the opposite terms in $2 (0 + \frac{1}{4} t) - (0 + \frac{1}{3} t)$ .

Add $0$ and $\frac{1}{4} t$ .

$2 (\frac{1}{4} t) - (0 + \frac{1}{3} t) = 12$

Add $0$ and $\frac{1}{3} t$ .

$2 (\frac{1}{4} t) - (\frac{1}{3} t) = 12$

Simplify each term.

Combine $\frac{1}{4}$ and $t$ .

$2 \frac{t}{4} - (\frac{1}{3} t) = 12$

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

$2 \frac{t}{2 (2)} - (\frac{1}{3} t) = 12$

Cancel the common factor.

$2 \frac{t}{2 \cdot 2} - (\frac{1}{3} t) = 12$

Rewrite the expression.

$\frac{t}{2} - (\frac{1}{3} t) = 12$

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

$\frac{t}{2} - \frac{t}{3} = 12$

To write $\frac{t}{2}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$\frac{t}{2} \cdot \frac{3}{3} - \frac{t}{3} = 12$

To write $- \frac{t}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{t}{2} \cdot \frac{3}{3} - \frac{t}{3} \cdot \frac{2}{2} = 12$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{t}{2}$ by $\frac{3}{3}$ .

$\frac{t \cdot 3}{2 \cdot 3} - \frac{t}{3} \cdot \frac{2}{2} = 12$

Multiply $2$ by $3$ .

$\frac{t \cdot 3}{6} - \frac{t}{3} \cdot \frac{2}{2} = 12$

Multiply $\frac{t}{3}$ by $\frac{2}{2}$ .

$\frac{t \cdot 3}{6} - \frac{t \cdot 2}{3 \cdot 2} = 12$

Multiply $3$ by $2$ .

$\frac{t \cdot 3}{6} - \frac{t \cdot 2}{6} = 12$

Combine the numerators over the common denominator.

$\frac{t \cdot 3 - t \cdot 2}{6} = 12$

Simplify the numerator.

Factor $t$ out of $t \cdot 3 - t \cdot 2$ .

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

$\frac{t (3) - t \cdot 2}{6} = 12$

Factor $t$ out of $- t \cdot 2$ .

$\frac{t (3) + t (- 1 \cdot 2)}{6} = 12$

Factor $t$ out of $t (3) + t (- 1 \cdot 2)$ .

$\frac{t (3 - 1 \cdot 2)}{6} = 12$

Multiply $- 1$ by $2$ .

$\frac{t (3 - 2)}{6} = 12$

Subtract $2$ from $3$ .

$\frac{t \cdot 1}{6} = 12$

Multiply $t$ by $1$ .

$\frac{t}{6} = 12$

Multiply both sides of the equation by $6$ .

$6 \frac{t}{6} = 6 \cdot 12$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$6 \frac{t}{6} = 6 \cdot 12$

Rewrite the expression.

$t = 6 \cdot 12$

Simplify the right side.

Multiply $6$ by $12$ .

$t = 72$

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

Solve the equation for $x$ .

Multiply $\frac{1}{4}$ by $72$ .

$x = 0 + \frac{1}{4} \cdot 72$

Remove parentheses.

$x = 0 + \frac{1}{4} \cdot (72)$

Simplify $0 + \frac{1}{4} \cdot (72)$ .

Cancel the common factor of $4$ .

Factor $4$ out of $72$ .

$x = 0 + \frac{1}{4} \cdot (4 (18))$

Cancel the common factor.

$x = 0 + \frac{1}{4} \cdot (4 \cdot 18)$

Rewrite the expression.

$x = 0 + 18$

Add $0$ and $18$ .

$x = 18$

Solve the equation for $y$ .

Multiply $\frac{1}{3}$ by $72$ .

$y = 0 + \frac{1}{3} \cdot 72$

Remove parentheses.

$y = 0 + \frac{1}{3} \cdot (72)$

Simplify $0 + \frac{1}{3} \cdot (72)$ .

Cancel the common factor of $3$ .

Factor $3$ out of $72$ .

$y = 0 + \frac{1}{3} \cdot (3 (24))$

Cancel the common factor.

$y = 0 + \frac{1}{3} \cdot (3 \cdot 24)$

Rewrite the expression.

$y = 0 + 24$

Add $0$ and $24$ .

$y = 24$

Solve the equation for $z$ .

Remove parentheses.

$z = 0 + 0 \cdot (72)$

Simplify $0 + 0 \cdot (72)$ .

Multiply $0$ by $72$ .

$z = 0 + 0$

Add $0$ and $0$ .

$z = 0$

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

$x = 18$

$y = 24$

$z = 0$

$x = 18$

$y = 24$

$z = 0$

Using the values calculated for $x$ , $y$ , and $z$ , the intersection point is found to be $(18, 24, 0)$ .

$(18, 24, 0)$

Find the Intersection of the Line Perpendicular to Plane 1 Through the Origin and Plane 2 (x+3)/4+(y-1)/3=1 , 2x-y=12

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