00:01
Let's determine the angle between the two planes.
00:03
So we are given the two planes 2x plus 3y minus z plus 9 equals 0.
00:08
This is plane 1 and plane 2 is x plus 2y plus 4 equals 0.
00:13
Observe that in plane 2 we don't have the z term.
00:18
So i'm going to write this as 0 .z.
00:23
So this can be written as 0 .c plus 4 equals 0.
00:28
And to determine the angle between the two planes, we utilize this formula.
00:33
That is cosine theta equals the numerator, we take the absolute value of the dot product of the two normals.
00:41
And in the denominator, we have the product of the magnitude of the normals.
00:47
So we have to first to find the normals from the given planes.
00:51
The normals is basically the coefficients of x, y, and z.
00:58
So for the plane 1, we write the normal vector.
01:01
For the components of the normal vector, we take the coefficients of the variables.
01:06
So it is 2, 3 and negative 1.
01:09
This is the normal 1 vector.
01:12
Now let's determine the normal 2 vector.
01:15
This we obtained from the plane 2.
01:17
So here we take the components of the normal 2 from the coefficients of the normal 2 from the coefficients of the variables.
01:23
So it is 1, 2 and 0.
01:27
So we have obtained the two normal vectors, which means we can find the dot product as well as their magnitudes.
01:34
So first let's find the dot product of these two normals.
01:38
That is n1 .n2 vector and this equals, we have to find the product of the corresponding terms and then add them together.
01:48
So two times one is 2 plus.
01:51
Then 3 times 2 is 6 then negative 1 times 0 is 0 we have to take the absolute value of this dot product so let's take the absolute value and this equals 2 plus 6 equals 8 in the absolute value of 8 equals 8 so we have applied the numerator term to determine the angles now in the denominator we have the product of the normals so let's first define the i'm sorry in the denominator we're we have the product of the magnitudes of the normals.
02:24
So let's find the magnitude of normal one.
02:28
Magnitude of normal one.
02:32
This equals the square root of the sum of the squares of the components.
02:37
For normal one, the components are 2, 3, negative 1.
02:40
So it will be 2 square plus 3 squared plus negative 1 square.
02:48
And so this equals 2 square is 4, 3 square is 9.
02:52
To 1 square is 1 and so this equals 9 plus 1 is 10 10 plus 4 is 14 so the magnitude of normal 1 is root 14 now let's find the magnitude of normal 2 by using the components of normal 2 the components of normal 2 are 1 to 0 so therefore the magnitude equals 1 square plus 2 square plus 0 squared and so this equals 1 squared is 1, 2 square is 4, 0 is 0.
03:28
So this equals 4 plus 1 is root 5.
03:31
And so this is the magnitude of normal 2.
03:34
Therefore, we can now write down the cosine theta, where theta is the angle between the two planes...