Plane Equation

**Del** · April 21st 2008, 12:48 PM

Write an equation of the plane with normal vector

passing through the point

in scalar form:
-6x+(-7)y+(-3)z+180

This answer is incorrect. I found d to be 60, but I can't get it right - please help!

**Spec** · April 21st 2008, 02:01 PM

D is indeed 60. If you subtract 180 from 240 you get 60, and

$(-6)x + (-7)y + (-3)z = D$
$(-6)\cdot 2 + (-7)\cdot (-9) + (-3)\cdot (-3) = 60 = D$

**Plato** · April 21st 2008, 02:02 PM

If $n = \left\langle { - 6, - 7, - 3} \right\rangle$ is the normal the we can use any multiple, so let $n = \left\langle { 6, 7, 3} \right\rangle$ .
The general plane is $6x + 7y + 3z + d = 0$ .
Now substitute the point to find d: $6(2) + 7( - 9) + 3( - 3) + d = 0 \Rightarrow \quad d = 60$ .

Dee, May I ask: "Do you have an instructor and/or a textbook?

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