Find an equation of the plane the plane through the point 1...

Transcript

00:01 In this video, we want to write the equation of a plane, and we know a point that it passes through and a line that it's perpendicular to.

00:08 So in general, if you can use the standard form of the equation, where we'll have a times x minus x not plus b times y minus y not plus c times z minus z not is equal to zero, where the point that it passes through, is x -not, y -not, z -not, and then it's perpendicular to the vector containing a, b, c.

00:46 So this point that the plane passes through is already given as 1 ,0, comma, 7.

00:53 So then in order to plug in, all we need to do is find the normal vector.

00:57 And we know our particular plane is perpendicular to a line that was given by a parametric equation.

01:05 X is equal to 5 t, y is equal to 5 minus t, and z is equal to 3 plus 4 t.

01:19 So if we wanted to write the equation in terms of x, y, and z, let's see.

01:24 So x is equal to 5t, so t is equal to x over 5.

01:32 So then we have y is equal to 5 minus x over 5, or just to make the fractions go and multiply everything 3 by 5.

01:41 We get 5y is equal to 25 minus x.

01:47 And then z is equal to 3 plus 4 times x over 5, subbing in x for t.

01:56 Multiplying through by 5, we get 5 z is equal to 15 plus 4x.

02:14 We can add these two together.

02:16 So we get 5y plus 5 z is equal to 40 minus 4x.

02:24 X plus 3x, move the x over to the other side.

02:29 So 3x plus 5y plus 5z is equal to 40.

02:39 All right.

02:39 So from this standard equation, we see that the normal vector to this line is negative 3, 5, and 5.

02:51 Just taking the coefficients...

Question

Find an equation of the plane. The plane through the point (1, 0, 7) and perpendicular to the line x = 5t, y = 5 − t, z = 3 + 4t

Question

Find an equation of the plane. The plane through the point (1, 0, 7) and perpendicular to the line x = 5t, y = 5 − t, z = 3 + 4t

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