# Thread: Calculus 4: equation of a plane

1. ## Calculus 4: equation of a plane

Hello everyone,

A plane has equation z=5x-2y+7

a. Find the value of W making the vector Wi+j+.5k normal to the plane

b. Find a value of a so that the point (a+1, a, a-1) lies on the plane

My attempt:

z=5x-2y+6
Wi+1j+.5k

a. 1j--> -2y
-.5--> 7

5(x-01)-2(y-0)+5(z-0)+7=0

b.

5(x-a+1)-2(y-a)+0(a-1)+7=0

I now just need to solve for a, righ?

Is this anywhere close?

Thank you very much

2. Originally Posted by Chocolatelover2
Hello everyone,

A plane has equation z=5x-2y+7

a. Find the value of W making the vector Wi+j+.5k normal to the plane

b. Find a value of a so that the point (a+1, a, a-1) lies on the plane

My attempt:

z=5x-2y+6
Wi+1j+.5k

a. 1j--> -2y
-.5--> 7

5(x-01)-2(y-0)+5(z-0)+7=0
why are you working so hard?

the plane is -5x + 2y + z = 7

thus the normal vector is <-5,2,1> we want a vector that is parallel to this with coordinates <w, 1 , 1/2>

note that <w, 1, 1/2> = k<-5, 2, 1> for some constant k

b.

5(x-a+1)-2(y-a)+0(a-1)+7=0

I now just need to solve for a, righ?

Is this anywhere close?

Thank you very much
plug in x = a + 1, y = a, z = a - 1, then we have

-5(a + 1) + 2a + (a - 1) = 7

just solve for a

3. Thank you very much

That's a lot easier!

For a, I just need to find the equation that is parallel to 5x-2y-z=-7 through (-5,2,1) and (w, 1, 1/2), right? Would I use a matrix?

Thank you

4. Originally Posted by Chocolatelover2
Thank you very much

That's a lot easier!

For a, I just need to find the equation that is parallel to 5x-2y-z=-7 through (-5,2,1) and (w, 1, 1/2), right? Would I use a matrix?

Thank you
all you are doing is finding a value for w that makes <w, 1, 1/2> parallel to <-5, 2, 1>

all you have to do is find the constant k that i mentioned. it is not hard, you do not need matrices. just set up a simple equation to relate them