Question

If $|a|=1, |b|=4, a\cdot b = 2$ and $c = 2a \times b - 3b$, then the angle between $b$ and $c$ is

• A. $\frac{\pi}{6}$
• B. $\frac{5\pi}{6}$
• C. $\frac{\pi}{3}$
• D. $\frac{2\pi}{3}$

Hint:

Use $b \cdot (a \times b) = 0$ directly to simplify vector problems.

Answer 1

The Correct Option is B

To find the angle between vectors \( \mathbf{b} \) and \( \mathbf{c} \), we first need to determine the expression for \( \mathbf{c} \) given by \( \mathbf{c} = 2\mathbf{a} \times \mathbf{b} - 3\mathbf{b} \).

1. We know that \( |\mathbf{a}| = 1 \) and \( |\mathbf{b}| = 4 \), with the dot product \( \mathbf{a} \cdot \mathbf{b} = 2 \).
2. Recall that the cross product \( \mathbf{a} \times \mathbf{b} \) is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). Therefore, \( 2\mathbf{a} \times \mathbf{b} \) is also perpendicular to both.
3. The vector \( \mathbf{c} = 2\mathbf{a} \times \mathbf{b} - 3\mathbf{b} \) is a linear combination of \( 2\mathbf{a} \times \mathbf{b} \) and \( -3\mathbf{b} \).
4. Next, compute the magnitude of \( \mathbf{c} \):

\(|\mathbf{c}| = \sqrt{|2\mathbf{a} \times \mathbf{b}|^2 + (-3|\mathbf{b}|)^2}\)

1. The magnitude of \( \mathbf{a} \times \mathbf{b} \) is given by:

\(|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}| |\mathbf{b}| \sin \theta\)

1. where \(\theta\) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \).
2. Calculate \( \theta \) using the dot product formula:

\(\cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} = \frac{2}{4} = \frac{1}{2}\)

1. This implies \(\theta = \frac{\pi}{3}\).
2. Substitute back to find \( |\mathbf{a} \times \mathbf{b}| \):

\(|\mathbf{a} \times \mathbf{b}| = 4 \sin \frac{\pi}{3} = 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3}\)

1. Therefore,

\(|2\mathbf{a} \times \mathbf{b}| = 2 \cdot 2\sqrt{3} = 4\sqrt{3}\)

1. Now calculate \( |\mathbf{c}| \):

\(|\mathbf{c}| = \sqrt{(4\sqrt{3})^2 + (-3 \cdot 4)^2} = \sqrt{48 + 144} = \sqrt{192} = 8\sqrt{3}\)

1. To find the cosine of the angle \(\alpha\) between \( \mathbf{b} \) and \( \mathbf{c} \):

\(\mathbf{b} \cdot \mathbf{c} = \mathbf{b} \cdot (2\mathbf{a} \times \mathbf{b} - 3\mathbf{b}) = -3 |\mathbf{b}|^2 = -3 \cdot 16 = -48\)

1. Substitute into the angle formula:

\(\cos \alpha = \frac{\mathbf{b} \cdot \mathbf{c}}{|\mathbf{b}| |\mathbf{c}|} = \frac{-48}{4 \times 8\sqrt{3}} = -\frac{1}{2}\)

1. Thus, \(\alpha = \frac{5\pi}{6}\).

This confirms that the angle between \( \mathbf{b} \) and \( \mathbf{c} \) is indeed \(\frac{5\pi}{6}\).