JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\overrightarrow{ a }\) be a non-zero vector parallel to the line of intersection of the two planes described by \(\hat{i}+\hat{j}, \hat{i}+\hat{k}\) and \(\hat{i}-\hat{j}, \hat{j}-\hat{k}\). If \(\theta\) is the angle between the vector \(\vec{a}\) and the vector \(\vec{b}=2 \hat{i}-2 \hat{j}+\hat{k}\) and \(\vec{a} \cdot \vec{b}=6\) then the ordered pair \((\theta,|\vec{a} \times \vec{b}|)\) is equal to
- A \(\left(\frac{\pi}{4}, 3 \sqrt{6}\right)\)
- B \(\left(\frac{\pi}{3}, 3 \sqrt{6}\right)\)
- C \(\left(\frac{\pi}{3}, 6\right)\)
- D \(\left(\frac{\pi}{4}, 6\right)\)
Answer & Solution
Correct Answer
(D) \(\left(\frac{\pi}{4}, 6\right)\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ n }_1\) and \(\overrightarrow{ n }_2\) are normal vector to the plane \(\hat{i}+\hat{j}, \hat{i}+\hat{k}\) and \(\hat{i}-\hat{j} ; \hat{j}-\hat{k}\) respectively…
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