Vectors: Question 8

Worked solution

Step 1: Find the scalar product

$\mathbf{a}\cdot\mathbf{b} = (1)(2) + (2)(-2) + (2)(1) = 2 - 4 + 2 = 0$

Step 2: Find the magnitudes

$|\mathbf{a}| = \sqrt{1^2+2^2+2^2} = \sqrt{9} = 3, \qquad |\mathbf{b}| = \sqrt{2^2+(-2)^2+1^2} = \sqrt{9} = 3$

Step 3: Solve for the angle

$\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{0}{3\times3} = 0 \implies \theta = \cos^{-1}(0) = 90^\circ$

Step 4: Recompute independently as a check

Re-adding the scalar product term-by-term in reverse order: $(2)(1) + (-2)(2) + (1)(2) = 2 - 4 + 2 = 0$, matches. A scalar product of exactly zero for two nonzero vectors always means the vectors are perpendicular, i.e. $\theta = 90^\circ$; it does not mean the angle is undefined.

Why the other options are wrong

• B: equal magnitudes do not imply the vectors point the same way; the scalar product (not the magnitudes) determines the angle, and here it is $0$, not the maximum possible value $|\mathbf{a}||\mathbf{b}|=9$ that $0^\circ$ would require.
• C: comes from a sign slip in the scalar product ($1(2)+2(2)+2(1)=8$, using $+2$ instead of $-2$ for the middle term of $\mathbf{b}$), giving $\cos\theta = \tfrac{8}{9}$, i.e. $\theta \approx 27.3^\circ$, not the correct $90^\circ$.
• D: a zero scalar product is a perfectly meaningful, well-defined result. It is precisely the condition for perpendicularity, so the angle $90^\circ$ can and must be stated.

Final answer

$\boxed{\theta = 90^\circ} \quad \text{(Option A)}$