00:01
So in this question, we are finding the parametric equations for the line that passes through the point 3, 4 ,5, that's parallel to the plane x plus y plus z equals negative 15, and perpendicular to the line, x equals 15 plus t, y equals 12 minus t, and z equals 3 t.
00:20
So first of all, let's get the direction of the line in question.
00:26
So i have to be perpendicular to the normal vector for this plane.
00:33
So the normal vector for this plane would just be the coefficients of x, y, and z, namely 1 -1.
00:43
I also have to be perpendicular to the line.
00:48
So i am perpendicular to the line 1, negative 1, 3.
00:54
So if i need the direction of the line to be perpendicular to both of these, i'm going to need my cross -product.
01:02
I need n1 cross -n2.
01:07
And so let's see.
01:08
Cross the top row, i've got ijk.
01:11
Across the middle row, 1 -1, and across the bottom row, 1 -1, 3.
01:19
What are we getting? we're getting i times the quantity of 3 minus negative 1...