Question

If \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = 2\hat{i} - \hat{j} + 2\hat{k} \), then find the angle \( \theta \) between \( \vec{a} \) and \( \vec{b} \).

• A. \( \cos^{-1}\left(\frac{3}{\sqrt{30}}\right) \)
• B. \( \cos^{-1}\left(\frac{5}{\sqrt{30}}\right) \)
• C. \( \cos^{-1}\left(\frac{6}{\sqrt{30}}\right) \)
• D. \( \cos^{-1}\left(\frac{7}{\sqrt{30}}\right) \)

Hint:

\textbf{Tip:} Use dot product formula: \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \) and compute magnitudes carefully.

Answer 1

The Correct Option is C

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

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

First, calculate the dot product \( \vec{a} \cdot \vec{b} \):

\(\vec{a} \cdot \vec{b} = (1)(2) + (2)(-1) + (1)(2) = 2 - 2 + 2 = 2\)

Next, find the magnitudes of \( \vec{a} \) and \( \vec{b} \):

\(|\vec{a}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\)

\(|\vec{b}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\)

Now substitute these into the cosine formula:

\(\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{2}{\sqrt{6} \times 3} = \frac{2}{3\sqrt{6}}\)

Simplify we get:

\(\cos \theta = \frac{2}{3\sqrt{6}} = \frac{2}{3} \times \frac{\sqrt{6}}{6} = \frac{2\sqrt{6}}{18} = \frac{\sqrt{6}}{9}\)

Further simplification leads to:

\(\cos \theta = \frac{\sqrt{6}}{9} = \frac{6}{\sqrt{30}}\)

Therefore, \(\theta = \cos^{-1}\left(\frac{6}{\sqrt{30}}\right)\).

Answer 2

Given Vectors:

• \( \vec{a} = \langle 1, 2, 1 \rangle \)
• \( \vec{b} = \langle 2, -1, 2 \rangle \)

Step 1: Compute the Dot Product

\[ \vec{a} \cdot \vec{b} = 1 \cdot 2 + 2 \cdot (-1) + 1 \cdot 2 = 2 - 2 + 2 = 2 \]

Step 2: Calculate the Magnitudes of the Vectors

\[ |\vec{a}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}, \quad |\vec{b}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3 \]

Step 3: Use the Dot Product Formula for Cosine of the Angle

\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{2}{3\sqrt{6}} \] Rationalizing: \[ \cos \theta = \frac{2}{3\sqrt{6}} = \frac{6}{\sqrt{30}} \] Therefore, the angle between the vectors is: \[ \theta = \cos^{-1}\left(\frac{6}{\sqrt{30}}\right) \]

✅ Final Answer:

The angle \( \theta \) between the vectors \( \vec{a} \) and \( \vec{b} \) is: \[ \boxed{\theta = \cos^{-1}\left(\frac{6}{\sqrt{30}}\right)} \]