Hey guys, in this video I want to do this calculus 3 problem. It says,
determine the equation of the plane that is perpendicular to both planes--and then it gives
two planar equations--and passes through a point and then it gives the coordinates for a point 1,
0, 0. Let's go ahead and get started with the video on how to solve this problem.
The solution to this problem is an equation of a plane. Basically what we need to do is we
need to know what the equation of a plane is and how to fill in all the missing variables for this
equation. The equation of a plane has two parts. It's going to have a point that the plane passes
through, which I'm going to call P0, and that has coordinates x0, y0, and z0. And it's going to have
what's called a normal vector, which I'm going to denote as the vector n, and this is by definition
a vector that is perpendicular to the plane. And I'm going to write that as coordinates a, b, c.
A normal vector is a vector that is perpendicular to a plane. If this is the plane equation that
we want to find, the normal vector is going to be a vector that comes out perpendicularly from the
plane. And with these two pieces of information, we can write out the equation of the plane in scalar
form as follows. The equation is going to be a times x minus x0 plus b times y minus y0
plus c times z minus z0 equals zero. And this is the equation of the plane and what we want
to find are the values x0, y0, z0, and a, b, c. Once we plug in those two aspects of the
equation, we will be able to find the solution to this problem. In the problem we're given that x0, y0,
and z0 are 1, 0, 0. We know that the point coordinate P0 is 1, 0, 0. So all we have left to
find are the coordinates a, b, c, which come from the normal vector and we're given that we need to find
a plane that is perpendicular to two planes, which we are given the equations for.
To find the normal vector, we're just going to find a vector that is perpendicular to these
two planes. To find the normal vector n, I'm going to take the cross product of
the coordinates of these two planes and the cross product will give me a perpendicular vector.
The coordinates for the first planar equation are 1, 2, 3 and the coordinates for the second
planar equation are 3, -1, 5. I'm going to write that as the normal vector is equal to the
cross product of n1 by n2 and that is equal to finding the determinant of a three by three matrix
where we write out the coordinates from the two planar equations. The first row is going to be i,
j, k. The second is going to be 1, 2, 3, which are the coordinates from the first planar equation.
And then the third row is going to be 3, -1, 5. Then I'm just going to find the determinant
of this 3 by 3 matrix to give me the coordinates for the normal vector. This is going to be equal
to i times 2 times 5, which is 10 minus 3 times -3, which is +3
minus j times 1 times 5, which is 5 minus 3 times 3, which is 9 plus k
times 1 times -1, which is -1 minus 3 times 2, which is 6. Then if I simplify that
and write it in coordinate form, that's going to give me 13, 4, -7. The normal vector
that we want to plug in is going to be 13, 4, -7 and now we know the point and the
normal vector to plug into the planar equation and we have enough information to solve the problem.
Our final solution is going to be 13 times x minus 1 plus 4 times y minus 0
minus 7 times z minus 0 is equal to zero. If we distribute the coefficients and simplify,
our final answer is going to be 13x plus 4y minus 7z minus 13 is equal to 0
and this is our final answer for the planar equation and we're done with this problem.
*chiptune music*