Answer to Question #149068 in Physics for Hauwa Argando

Question

Two vectors A=5i-6j+4k and B=-3i+8j-2k
Find :
a- (A . B)
b- (AxB) and its magnitude
c- (A - 2B) and its magnitude
d- The angle between vectors A and B

Answer 1

Let $A = 5\mathbf i - 6\mathbf j +4\mathbf k$  and $B= -3 \mathbf i +8\mathbf j -2\mathbf k$

a) The dot product is:

$$(A\cdot B) = 5\cdot (-3) + (-6)\cdot 8 + 4\cdot (-2) = -71$$

b) The cross product is:

$$(A\times B) = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ 5 & -6 & 4 \\ -3 & 8 & -2 \end{vmatrix} =\\ =\mathbf i ((-6)\cdot(-2) - 8\cdot 4) - \mathbf j (5\cdot (-2)-4\cdot (-3)) + \mathbf k (5\cdot 8-(-6)\cdot (-3)) =\\ = -20 \mathbf i - 2\mathbf j + 22 \mathbf k$$

It's magnitude is:

$$|(A\times B)| = \sqrt{(-20)^2 +(-2)^2 + 22^2} \approx 29.8$$

c) Combination (A - 2B) is given as

$$A-2B = 5\mathbf i - 6\mathbf j +4\mathbf k - 2(-3 \mathbf i +8\mathbf j -2\mathbf k ) =11 \mathbf i -22\mathbf j +8\mathbf k$$

Its mabnitude:

$$|A-2B| = \sqrt{(11)^2 +(-22)^2 + 8^2} \approx 25.9$$

d) The angle between vectors A and B is given as

$$\theta = \arccos\left(\dfrac{(A\cdot B)}{|A|\cdot |B|} \right)$$

The magnitudes of the vectors are:

$$|A| = \sqrt{5^2 +(-6)^2 + 4^2} \approx 8.1\\ |B| = \sqrt{(-3)^2 +8^2 + (-2)^2} \approx 8.8$$

Thus, the angle is:

$$\theta = \arccos\left(\dfrac{-71}{8.1\cdot 8.8} \right) \approx 174.9\degree$$

Answer. a) -71, b) $-20 \mathbf i - 2\mathbf j + 22 \mathbf k$ and 29.8, c) $11 \mathbf i -22\mathbf j +8\mathbf k$ and 25.9, d) $174.9\degree$.