81.9k views
2 votes
Find an equation of the plane through the point (1, -5, 3) and perpendicular to the vector (2, -3, -3).

User Dazed
by
9.4k points

1 Answer

4 votes
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
User Est
by
8.4k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories