JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+\beta \hat{\mathrm{k}}, \alpha, \beta \in \mathrm{R}\). Let a vector \(\overrightarrow{\mathrm{b}}\) be such that the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{4}\) and \(|\vec{b}|^2=6\), If \(\vec{a} \cdot \vec{b}=3 \sqrt{2}\), then the value of \(\left(\alpha^2+\beta^2\right)|\vec{a} \times \vec{b}|^2\) is equal to
- A \(90\)
- B \(75\)
- C \(95\)
- D \(85\)
Answer & Solution
Correct Answer
(A) \(90\)
Step-by-step Solution
Detailed explanation
\(|\vec{b}|^2=6 ;|\vec{a}||\vec{b}| \cos \theta=3 \sqrt{2}\) \(|\vec{a}|^2|\vec{b}|^2 \cos ^2 \theta=18\) \(|\vec{a}|^2=6\) Also \(1+\alpha^2+\beta^2=6\) \(\left(\alpha^2+\beta^2\right)|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta\) \(=(5)(6)(6)\left(\frac{1}{2}\right)\)…
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