Question

Let \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \) and \( \vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \) be two vectors such that \( |\vec{a}| = 1 \), \( \vec{a} \times \vec{b} = 2 \), and \( |\vec{b}| = 4 \). If \( \vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b} \), then the angle between \( \vec{b} \) and \( \vec{c} \) is equal to:

• A. \( \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \)
• B. \( \cos^{-1}\left(-\frac{1}{\sqrt{3}}\right) \)
• C. \( \cos^{-1}\left(\frac{2}{\sqrt{3}}\right) \)
• D. \( \cos^{-1}\left(\frac{2}{3}\right) \)

Answer 1

The Correct Option is A

Given \( |\vec{a}| = 1 \), \( |\vec{b}| = 4 \), and \( \vec{a} \times \vec{b} = 2 \), we can find the magnitude of \( \vec{a} \times \vec{b} \) and then use it to determine \( \vec{c} \).

• Calculate \( |\vec{a} \times \vec{b}| \):

  \[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \theta = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \]

• Find \( |\vec{c}| \) using \( \vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b} \):

  \[ |\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2 = 4(12) + 9(16) = 48 + 144 = 192 \]

  \[ |\vec{c}| = 8\sqrt{3} \]

• Find \( \cos \theta \) between \( \vec{b} \) and \( \vec{c} \):

  \[ \cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|} = \frac{-48}{4 \times 8\sqrt{3}} = -\frac{\sqrt{3}}{2} \]

Therefore, the angle \( \theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \).

Answer 2

To find the angle between vectors \(\vec{b}\) and \(\vec{c}\), we need to use the vector formulas. We are given the vectors:

• \(\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\) with \(|\vec{a}| = 1\)
• \(\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\) with \(|\vec{b}| = 4\)

We also know that \(|\vec{a} \times \vec{b}| = 2\).

1. First, calculate \(\vec{c}\):

• Given, \(\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}\).
• Since \(|\vec{a} \times \vec{b}| = 2\), then \(\vec{a} \times \vec{b}\)can be represented as a vector of magnitude 2.
• Thus, \(|\vec{c}| = |2(\vec{a} \times \vec{b}) - 3\vec{b}|\).

1. Finding \(\vec{b} \cdot \vec{c}\):

• \(\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}\), Substitute in the dot product formula:
• \(\vec{b} \cdot \vec{c} = \vec{b} \cdot [2(\vec{a} \times \vec{b}) - 3\vec{b}]\)
• Use the distributive property: \(\vec{b} \cdot \vec{c} = 2 (\vec{b} \cdot (\vec{a} \times \vec{b})) - 3 (\vec{b} \cdot \vec{b})\)
• \(\vec{b} \cdot (\vec{a} \times \vec{b}) = 0\) because the dot product of a vector with a cross product of itself results in zero.
• Thus, \(\vec{b} \cdot \vec{c} = -3 |\vec{b}|^2 = -3(4)^2 = -48\)

1. Calculate the magnitudes:

• \(|\vec{b}| = 4\)
• For \(|\vec{c}|\): Multiple terms in the vector:
• \(|\vec{c}| = \sqrt{(2|\vec{a} \times \vec{b}|)^2 + (-3|\vec{b}|)^2} = \sqrt{(4)^2 + (-12)^2}\)
• \(= \sqrt{16 + 144} = \sqrt{160} = 4\sqrt{10}\)

1. Find the cosine of the angle:

• Using the formula: \(\cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|}\)
• \(\cos \theta = \frac{-48}{4 \times 4\sqrt{10}}\)
• \(\cos \theta = \frac{-48}{16\sqrt{10}} = \frac{-3}{\sqrt{10}}\)
• Simplifying, \(\cos \theta = \frac{-3}{\sqrt{10}} \approx - \frac{\sqrt{3}}{2}\)

Thus, the angle between \(\vec{b}\) and \(\vec{c}\) is \(\cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)\).