Question

Consider the points below: P(3, 0, 3), Q(−2, 1, 4), R(4, 2, 5).

(a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R.
(b) Find the area of the triangle PQR.
a. i+j+k
b. i-j+2k
c. i+2j+k
d. i+j-k

Answer 1

To find a nonzero vector orthogonal to the plane through the points P, Q, and R, we can find two vectors in the plane and take their cross product. The cross product of these two vectors gives us a nonzero vector orthogonal to the plane. The area of the triangle PQR can be found using the formula: Area = 1/2 * |PQ x PR|.

Step-by-step explanation:

To find a nonzero vector orthogonal to the plane through the given points, we can first find two vectors in the plane and then take their cross product. Let's take vectors PQ and PR. PQ = Q - P = (-2 - 3, 1 - 0, 4 - 3) = (-5, 1, 1) and PR = R - P = (4 - 3, 2 - 0, 5 - 3) = (1, 2, 2). We can now find the cross product of these two vectors: PQ x PR = (-5, 1, 1) x (1, 2, 2).

Using the cross product formula, we get: PQ x PR = ((1 * 2) - (2 * 1))i - ((-5 * 2) - (1 * 1))j + ((-5 * 1) - (1 * 1))k = 0i - (-11)j - (-6)k = 11j - 6k.

Therefore, a nonzero vector orthogonal to the plane through the points P, Q, and R is 11j - 6k.

The area of a triangle can be found using the formula: Area = 1/2 * |PQ x PR|. Substituting the values we found for PQ and PR, we get: Area = 1/2 * |(-5, 1, 1) x (1, 2, 2)| = 1/2 * |11j - 6k| = 1/2 * sqrt((0^2) + (11^2) + (-6^2)) = 1/2 * sqrt(121 + 36) = 1/2 * sqrt(157) = sqrt(157)/2.