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Calculus/equation of plane

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Expert: - 10/26/2009


Question
find the equation of plane which passes through the point (3,4,5),
has an x intercept equal to -5 and is perpendicular to the plane 2x+3y-z=8

Answer
If we solve both for z, that gives z = (-2x+y)/3 and z = (x+2y+3)/2.
Setting these equal gives (-2x+y)/3 = (x+2y+3)/2.
Multiply by 6 gives -4x + 2y = 3x + 6y + 9.
Moving the variables to the left by adding -3x - 6y gives -7x - 4y = 9.

The equation perpendiucular to the yz plane would be


Plane have the form ax + by + cz = 1.
If the value on the right is not 1, divide the whole equation by that value.

It passes through (3,4,5), so 3a + 4b + 5c = 1.
The x intercept is where y and z are both 0, so this says that a(-5) = 1, or a = -0.2.

The equation is now -0.2x + by + cz = 1.
Using x=3, y=4, and z=5, we have -0.6 + 4b + 5c = 1.
This can be changed to 4b + 5c = 1.6

We know that the cross product of a vector in our plane with 2x+3y-z=8 is 0.
That cross product of (-0.2 b c) with (2 3 -1) is 3c + b + 0.2 - 2c + 2b - .6.
This reduces to 3b + c = 0.4.

The equations to solve for b and c are 4b + 5c = 1.6 and 3b + c = 0.4.

Take c = 0.4 - 3b from the 2nd equation and put it into the 1st to find b.
Then use either (or both, to check) to find c.
Since a was known already, the equation for the plane is known.

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