Question

If A=4+33-2 and B = 8 +67-4k, the angle between and B is:

Answer 1

$\theta=\cos^{-1}\left(\frac{2251}{\sqrt{1109\,4569}}\right)\approx 0^\circ.$

Answer 2

We interpret the given as follows (note that “k” here denotes the unit vector):

$$\vec{A}=4\,\hat{i}+33\,\hat{j}-2\,\hat{k},\quad \vec{B}=8\,\hat{i}+67\,\hat{j}-4\,\hat{k}.$$

The angle $\theta$ between $\vec{A}$ and $\vec{B}$ is given by

$$\cos\theta=\frac{\vec{A}\cdot\vec{B}}{\|\vec{A}\|\|\vec{B}\|}.$$

Step 1. Compute the dot product:

$$\begin{aligned} \vec{A}\cdot\vec{B} &= (4)(8) + (33)(67) + (-2)(-4)\\[1mm] &= 32 + 2211 + 8\\[1mm] &= 2251. \end{aligned}$$

Step 2. Compute the magnitudes:

$$\|\vec{A}\| = \sqrt{4^2+33^2+(-2)^2} = \sqrt{16+1089+4} = \sqrt{1109}.$$ $$\|\vec{B}\| = \sqrt{8^2+67^2+(-4)^2} = \sqrt{64+4489+16} = \sqrt{4569}.$$

Step 3. Write the expression for the angle:

$$\cos\theta=\frac{2251}{\sqrt{1109}\,\sqrt{4569}}.$$

Thus,

$$\theta=\cos^{-1}\left(\frac{2251}{\sqrt{1109\,4569}}\right).$$

A quick numerical check shows that

$$\sqrt{1109}\approx33.3,\quad \sqrt{4569}\approx67.6,\quad \text{and}\quad 33.3\times67.6\approx2251.$$

So, $\cos\theta\approx \frac{2251}{2251}\approx 1$ and consequently

$$\theta\approx \cos^{-1}(1) \approx 0^\circ.$$