Question:
If $|a|=1, |b|=4, a\cdot b = 2$ and $c = 2a \times b - 3b$, then the angle between $b$ and $c$ is
If $|a|=1, |b|=4, a\cdot b = 2$ and $c = 2a \times b - 3b$, then the angle between $b$ and $c$ is
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Use $b \cdot (a \times b) = 0$ directly to simplify vector problems.
Updated On: Apr 23, 2026
- $\frac{\pi}{6}$
- $\frac{5\pi}{6}$
- $\frac{\pi}{3}$
- $\frac{2\pi}{3}$
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The Correct Option is B
Solution and Explanation
Concept:
Angle between vectors:
\[
\cos \theta = \frac{b \cdot c}{|b||c|}
\]
Step 1: Find $c$.
\[ c = 2(a \times b) - 3b \]
Step 2: Compute $b \cdot c$.
\[ b \cdot c = 2[b \cdot (a \times b)] - 3(b \cdot b) \] \[ b \cdot (a \times b) = 0 \] \[ \Rightarrow b \cdot c = -3|b|^2 = -3(16) = -48 \]
Step 3: Find $|c|$.
\[ |c|^2 = 4|a \times b|^2 + 9|b|^2 \] \[ |a \times b|^2 = |a|^2|b|^2 - (a \cdot b)^2 = 1 \cdot 16 - 4 = 12 \] \[ |c|^2 = 4(12) + 9(16) = 48 + 144 = 192 \] \[ |c| = \sqrt{192} = 8\sqrt{3} \]
Step 4: Find angle.
\[ \cos \theta = \frac{-48}{4 \cdot 8\sqrt{3}} = -\frac{1}{\sqrt{3}} \] \[ \theta = \frac{5\pi}{6} \] Conclusion:
Angle = $\frac{5\pi}{6}$
Step 1: Find $c$.
\[ c = 2(a \times b) - 3b \]
Step 2: Compute $b \cdot c$.
\[ b \cdot c = 2[b \cdot (a \times b)] - 3(b \cdot b) \] \[ b \cdot (a \times b) = 0 \] \[ \Rightarrow b \cdot c = -3|b|^2 = -3(16) = -48 \]
Step 3: Find $|c|$.
\[ |c|^2 = 4|a \times b|^2 + 9|b|^2 \] \[ |a \times b|^2 = |a|^2|b|^2 - (a \cdot b)^2 = 1 \cdot 16 - 4 = 12 \] \[ |c|^2 = 4(12) + 9(16) = 48 + 144 = 192 \] \[ |c| = \sqrt{192} = 8\sqrt{3} \]
Step 4: Find angle.
\[ \cos \theta = \frac{-48}{4 \cdot 8\sqrt{3}} = -\frac{1}{\sqrt{3}} \] \[ \theta = \frac{5\pi}{6} \] Conclusion:
Angle = $\frac{5\pi}{6}$
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