Question

The angle $\theta$ between the vectors $\overrightarrow{a}=5\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{i}+\hat{j}-\hat{k}$ is:

• (A) ${\cos }^{-1}\left(\frac{1}{3}\right)$
• (B) ${\cos }^{-1}\left(\frac{2}{3}\right)$
• (C) ${\cos }^{-1}\left(\frac{4}{7}\right)$
• (D) None of these

Answer 1

The Correct Option is A

To find the angle \( \theta \) between the vectors \( \vec{a} \) and \( \vec{b} \), we use the dot product formula:

\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta

Given vectors:

• \( \vec{a} = 5\hat{i} - \hat{j} + \hat{k} \)
• \( \vec{b} = \hat{i} + \hat{j} - \hat{k} \)

Step 1: Calculate the dot product \( \vec{a} \cdot \vec{b} \).

The dot product is calculated as:

(5\hat{i} - \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) = 5 \cdot 1 + (-1) \cdot 1 + 1 \cdot (-1)

This simplifies to:

5 - 1 - 1 = 3

Step 2: Calculate the magnitudes of \( \vec{a} \) and \( \vec{b} \).

|\vec{a}| = \sqrt{5^2 + (-1)^2 + 1^2} = \sqrt{27}

|\vec{b}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}

Step 3: Use the dot product formula to find \( \cos \theta \).

\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta

3 = \sqrt{27} \cdot \sqrt{3} \cdot \cos \theta

Simplify to find \( \cos \theta \):

3 = \sqrt{81} \cdot \cos \theta \Rightarrow 3 = 9 \cdot \cos \theta \Rightarrow \cos \theta = \frac{1}{3}

Step 4: Determine \( \theta \) using the inverse cosine function:

The angle \( \theta \) can thus be written as:

\theta = \cos^{-1} \left(\frac{1}{3}\right)

The correct answer is option (A): \cos^{-1}\left(\frac{1}{3}\right).