What is the line of intersection between the planes #3x+y-4z=2# and #x+y=18#?

Question

Caroline Aucoin · Accepted Answer

#y+2z-26=0# Equation of the line of intersection between the planes #3x+y-4z=2# and #x+y=18# can be obtained by putting the value of #x# from second equation into first. As #x+y=18#, #x=18-y# and putting this in #3x+y-4z=2#, we get #3(18-y)+y-4z=2# or #54-3y+y-4z=2# or #-2y-4z+52=0# or dividing each term by #-2#, we get #y+2z-26=0#

Autumn Armstrong · Answer

undefined To find the line of intersection between the planes $3x+y-4z=2$ and $x+y=18$, follow these steps:

1. Solve one of the equations for one of the variables.
2. Substitute the expression obtained in step 1 into the other equation to find the value of another variable.
3. Use the values found in steps 1 and 2 to find the third variable.
4. Write the solution as a parametric equation representing a line.

Solving $x + y = 18$ for $x$, we get $x = 18 - y$.

Substituting $x = 18 - y$ into $3x + y - 4z = 2$, we get $3(18 - y) + y - 4z = 2$, which simplifies to $54 - 2y - 4z = 2$.

Rearranging terms, we get $4z = 52 - 2y$, and dividing by 4, we have $z = \frac{52 - 2y}{4}$.

Now, we have expressions for $x$, $y$, and $z$ in terms of $y$:

$x = 18 - y$, $y = y$, $z = \frac{52 - 2y}{4}$.

So, the line of intersection between the planes is parametrically represented as:

$$ x = 18 - t $$
$$ y = t $$
$$ z = \frac{52 - 2t}{4} $$

where $t$ is any real number.