Question

The angle θ between the vectors \(\rm \vec{a} = 5\hat{i} - \hat{j} +  \hat{k}\) and \(\rm \vec{b} = \hat{i} + \hat{j} -  \hat{k}\) equals to

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

Answer 1

The Correct Option is A

To determine the angle \( \theta \) between the vectors \(\vec{a} = 5\hat{i} - \hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} + \hat{j} - \hat{k}\), we can use the formula for the cosine of the angle between two vectors:

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

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

   The dot product of two vectors \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) is:

   \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3

   Substitute the given values:

   \vec{a} \cdot \vec{b} = (5)(1) + (-1)(1) + (1)(-1) = 5 - 1 - 1 = 3

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

   The magnitude of a vector \(\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}\) is:

   |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}

   Calculate \(|\vec{a}|\):

   |\vec{a}| = \sqrt{5^2 + (-1)^2 + 1^2} = \sqrt{25 + 1 + 1} = \sqrt{27} = 3\sqrt{3}

   Calculate \(|\vec{b}|\):

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

3. Substitute the values into the cosine formula:

   cos\, \theta = \frac{3}{3\sqrt{3} \cdot \sqrt{3}} = \frac{3}{3 \cdot 3} = \frac{1}{3}

4. Find \(\theta\) using the inverse cosine:

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

Therefore, the correct answer is \cos^{-1}\left(\frac{1}{3}\right).