Q77 / 180 · COMEDK 2022 · Mathematics
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
- Correct answerA$\cos \theta=\dfrac{4}{21}$
- B$\cos \theta=\dfrac{3}{19}$
- C$\cos \theta=\dfrac{2}{19}$
- D$\cos \theta=\dfrac{5}{21}$
Solution
The angle $\theta$ between two vectors $\vec{a}$ and $\vec{b}$ is given by the formula $\cos \theta = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
Given $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$.
Calculate the dot product $\vec{a} \cdot \vec{b} = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4$.
Calculate the magnitude $|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Calculate the magnitude $|\vec{b}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.
Substitute these values into the formula: $\cos \theta = \dfrac{4}{3 \times 7} = \dfrac{4}{21}$.
Answer: $\cos \theta=\dfrac{4}{21}$
Given $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$.
Calculate the dot product $\vec{a} \cdot \vec{b} = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4$.
Calculate the magnitude $|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Calculate the magnitude $|\vec{b}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.
Substitute these values into the formula: $\cos \theta = \dfrac{4}{3 \times 7} = \dfrac{4}{21}$.
Answer: $\cos \theta=\dfrac{4}{21}$
Hence, the correct answer is option: 'A'.