Question

If \( \theta \) be the angle between the vectors \( a = 2\hat{i} + 2\hat{j} - \hat{k} \) and \( b = 6\hat{i} - 3\hat{j} + 2\hat{k} \), then

• A. \( \cos \theta = \frac{4}{21} \)
• B. \( \cos \theta = \frac{3}{19} \)
• C. \( \cos \theta = \frac{2}{19} \)
• D. \( \cos \theta = \frac{5}{21} \)

Hint:

When finding the angle between two vectors, remember to use the formula \( \cos \theta = \frac{a \cdot b}{|a| |b|} \). This requires calculating both the dot product and the magnitudes of the vectors.

Answer 1

The Correct Option is A

We are given two vectors \( a = 2\hat{i} + 2\hat{j} - \hat{k} \) and \( b = 6\hat{i} - 3\hat{j} + 2\hat{k} \). The formula for the cosine of the angle between two vectors is: \[ \cos \theta = \frac{a \cdot b}{|a| |b|} \] First, calculate the dot product \( a \cdot b \): \[ a \cdot b = (2 \times 6) + (2 \times -3) + (-1 \times 2) = 12 - 6 - 2 = 4 \] Now, calculate the magnitudes of \( a \) and \( b \): \[ |a| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 \] \[ |b| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7 \] Now, substitute into the formula for \( \cos \theta \): \[ \cos \theta = \frac{4}{3 \times 7} = \frac{4}{21} \] Thus, the correct answer is \( \cos \theta = \frac{4}{21} \).