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3 years ago

15

Find an equation of the plane through the point (1, -5, 3) and perpendicular to the vector (2, -3, -3).

1 answer:

AnnyKZ [126]3 years ago

6 0

<span>2x - 3y - 3z = 8
We've been given the normal vector to the plane <2, -3, -3> and a point within the plane (1, -5, 3). In general if you've been given both the normal vector <a,b,c> and a point (e,f,g) within the plane, the expression for the plane will be: ax + by + cz = d
and you can compute d by:
d = ae + bf + cg
So let's calculate d:
d = ae + bf + cg
d = 2*1 + -3*-5 + -3*3
d = 2 + 15 + -9
d = 8
And the equation for the plane is 2x - 3y - 3z = 8</span>

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Substituting we have:

$88 = \frac {(6 + 2h) * h} {2}\\176 = 6h + 2h ^ 2\\2h ^ 2 + 6h-176 = 0$

We divide between 2 on both sides:

$h ^ 2 + 3h-88 = 0$

We factor by looking for two numbers that, when multiplied, are obtained -88 and when added together, +3 is obtained.

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We have two roots:

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We choose the positive value.

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Answer:

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Work Shown:

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