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# Vectors and Planes - Problem 3

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

We’re talking about the vector equation of a plane. Here’s a problem; find an equation of the plane tangent to the sphere (x minus y) squared, plus (z plus 5) squared equals 49 at the point <3, 6, -2>.

I’ve drawn a sphere, I’ve located its centre which is <1, 0, -5> right here, point c. And this is the point A of tangency, <3, 6, -2>, the plane touches the sphere only at this point. Recall the vector equation of a plane, it’s n, the normal vector, dot r minus r0 equals zero.

Now r is the position vector for any arbitrary point x, y, z, on the plane and r0 is the position vector for the one point that we know is on the plane. In this case the one point we know is on the plane is <3, 6, -2>. So I’m going to make r0 <3, 6, -2>. Now r is just <x, y, z>. What about n, the normal vector?

Well it turns out that if a plane is tangent to its sphere, it works much like when a line is tangent to a circle. The radius drawn to the point of tangency is perpendicular to the tangent plane. This actually, this vector CA, works as a normal vector. So I’m going to make my normal vector, CA. And that’s going to be, in components, 3 minus 1 which is 2, 6 minus 0, 6 and -2, minus -5, -2 plus 5, 3. Now we’re ready to write the vector equation.

From here it’s going to be <2, 6, 3> dot <x, y, z>. That’s my n.r. Now I need my r0 which is <3, 6, -2>, and that dot product has to equal zero. Remember that’s what says that these two vectors are actually perpendicular. Let me simplify this difference. We’ve got x minus 3, y minus 6 and z minus -2, z plus 2, and I have <2, 6, 3> here. Now this equation by itself is a vector equation for the plane, and so if that’s all you wanted you could stop here.

But let’s go a little further and find the rectangular equation also. The way we do that is we just perform this dot product. So it’s 2 times x minus 3, plus 6 times y minus 6, plus 3, times z plus 2. I just need to simplify this. So 2x minus 6, plus 6 y minus 36, plus 3z plus 6 equals 0. These 6s cancel, and I can move this 36 to the right hand side and I get 2x plus 6y plus 3z, equals 36. And that is a rectangular equation for the plane.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

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