Question

If A=4i+3j-2k and B = 8i+6j-4k, the angle between A and B is:

• A. 90°
• B. 45°
• C. 0°
• D. 180°

Answer 1

Answer: (C)

0°

Answer 2

Solution:

Given:

$$\vec{A}=4\hat{i}+3\hat{j}-2\hat{k},\quad \vec{B}=8\hat{i}+6\hat{j}-4\hat{k}.$$

Step 1: Compute the dot product

$$\vec{A}\cdot\vec{B} = (4)(8) + (3)(6) + (-2)(-4) = 32 + 18 + 8 = 58.$$

Step 2: Compute the magnitudes

$$\|\vec{A}\| = \sqrt{4^2+3^2+(-2)^2} = \sqrt{16+9+4} = \sqrt{29},$$ $$\|\vec{B}\| = \sqrt{8^2+6^2+(-4)^2} = \sqrt{64+36+16} = \sqrt{116} = 2\sqrt{29}.$$

Step 3: Calculate the cosine of the angle $\theta$

$$\cos\theta = \frac{\vec{A}\cdot\vec{B}}{\|\vec{A}\|\|\vec{B}\|} = \frac{58}{\sqrt{29}\cdot 2\sqrt{29}} = \frac{58}{2\cdot29} = \frac{58}{58} = 1.$$

Thus,

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