Find a nonzero vector orthogonal to the plane through the points P, Q, and R.? Consider the points below.
P(0,-3,0), Q(4,1,-3), R(5,3,1)
To find a nonzero vector orthogonal to the plane through the points P, Q, and R, we can use the cross product of two vectors in the plane.
Let's first find two vectors in the plane: PQ and PR.
1. Vector PQ:
PQ = Q - P
= (4, 1, -3) - (0, -3, 0)
= (4, 4, -3)
2. Vector PR:
PR = R - P
= (5, 3, 1) - (0, -3, 0)
= (5, 6, 1)
Next, we will calculate the cross product of PQ and PR to find a vector orthogonal to the plane.
3. Cross product:
N = PQ x PR
= (4, 4, -3) x (5, 6, 1)
To find the cross product, we can use the determinant method:
| i j k |
N = | 4 4 -3 |
| 5 6 1 |
N = ((4 * 1) - (4 * 6))i - ((4 * 5) - (4 * 1))j + ((4 * 6) - (5 * 4))k
= (-20)i - (16)j + (4)k
Therefore, a nonzero vector orthogonal to the plane is (-20, -16, 4).
To find a nonzero vector orthogonal to the plane through the points P, Q, and R, we can use the cross product of two vectors in the plane.
First, we need to find two vectors in the plane. We can do this by subtracting the coordinates of one point from the coordinates of the other points. Let's take the vectors PQ and PR:
PQ = Q - P = (4, 1, -3) - (0, -3, 0) = (4, 4, -3)
PR = R - P = (5, 3, 1) - (0, -3, 0) = (5, 6, 1)
Next, we can calculate the cross product between PQ and PR:
Cross product = PQ × PR = (4, 4, -3) × (5, 6, 1)
To find the cross product, we can use the determinant of a 3x3 matrix:
|i j k|
|4 4 -3|
|5 6 1|
Expanding the determinant, we have:
4 * (6 * 1 - 1 * 6) - 4 * (5 * 1 - 1 * 5) - 3 * (5 * 6 - 6 * 5)
= 4 * 0 - 4 * 0 - 3 * 0
= 0
Since the cross product is equal to zero, this means that PQ and PR are parallel or linearly dependent. Therefore, there is no nonzero vector orthogonal to the plane through the points P, Q, and R.
vector PQ = <-3,4,4>
vector PR = <1,6,5>
Now take the cross-product between these two vectors, using whichever method you learned.
To test your answer, the dot product of your answer vector with both PQ and PR must be zero.