Write equation of plane with normal vector and point

[jwpaine]

Write an equation of the plane with normal vector n = <-6,2,3> passing through point (-8, -4, -7) in scalar form

The solution is entered, online, in a text box, followed by "= 76"

This is what I have done:

$\displaystyle n\cdot<x,y,z> = n\cdot<x_0, y_0, z_0>$
$\displaystyle <-6,2,3><x,y,z> = <-6,2,3><-8,-4,-7>$
$\displaystyle -6x+2y+3z -19 = 0$
$\displaystyle -6x+2y+3z -19 + 76 = 76$
$\displaystyle -6x+2y+3z +57 = 76$

I tried my answer of $\displaystyle -6x+2y+3z +57$ but the computer says it's wrong.

Any help would be greatly appreciated.

John

 

[galactus]

Hey JW:

You're OK except for your d value. You appear to have it, then get 57. I am not getting that 76 thing. What is that for?.

The vector n=(a,b,c) is the normal which is perpendicular to the plane.

So, as you correctly have, the plane has equation:

$\displaystyle -6x+2y+3z+d=0$

We have to find d:

Plug in the given 'passing through' point in for x, y, z and solve for d. We find d = -19.

$\displaystyle -6x+2y+3z-19$

You could also just write it as:

$\displaystyle -6(x+8)+2(y+4)+3(z+7)=0$ and expand out and you will get the same.

 

[jwpaine]

Thanks - just had to double check I was doing everything correctly.

As for my d value, the computer wants the answer in terms of "entered answer" = 76, that's why I added 76 to both sides.

Can't seem to get this darn web homework to accept an answer. I've emailed my professor.

John


EDIT: So -6x+2y+3z= 19.
the TA wrote back and said "But 76 = 19*4 ==> 4*( -6x+2y+3z= 19) ==> -24x+8y+12z= 76"

And the computer accepts that answer. But the answer I had should have been correct.