Question

Consider the points P(-4,1,0) and Q(-4, -5,2).

a. Find PQ and state your answer in two forms: (a, b, c) and ai + bj + ck.
b. Find the magnitude of PQ.
c. Find two unit vectors parallel to PQ.

Answer 1

To find the distance between points P(-4,1,0) and Q(-4, -5,2), use the distance formula. The magnitude of PQ is sqrt(40). Two-unit vectors parallel to PQ can be obtained by dividing PQ by its magnitude.

Step-by-step explanation:

To find the distance between points P(-4,1,0) and Q(-4, -5,2), we can use the distance formula. The formula for the distance between two points in 3D space is:

d = sqrt((x2-x1) ^2 + (y2-y1) ^2 + (z2-z1) ^2)

Plugging in the coordinates of P and Q, we get:

d = sqrt ((-4--4) ^2 + (-5-1) ^2 + (2-0) ^2)

d = sqrt (0 + 36 + 4)

d = sqrt (40)

So, the distance PQ is sqrt (40) units. In the specified forms, PQ = (0, 6, 2) and PQ = 6i + 2j + ck.

The magnitude of PQ is sqrt (40).

Two-unit vectors parallel to PQ can be found by dividing PQ by its magnitude. So, one unit vector would be (1/sqrt(40))(6i + 2j + ck).