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JEE Mains · Maths · STD 12 - 10. vector algebra

ધારો કે \(\vec{a}=\hat{i}+\alpha \hat{j}+\beta \hat{k}, \alpha, \beta \in \mathbb{R}\). ધારો કે એક સદિશ \(\vec{b}\) એવો છે કે જેથી \(\vec{a}\) અને \(\vec{b}\) વચ્ચેનો ખૂણો \(\frac{\pi}{4}\) હોય અને \(|\vec{b}|^2=6\) હોય જો \(\vec{a} \cdot \vec{b}=3 \sqrt{2}\) હોય, તો \(\left(\alpha^2+\beta^2\right)|\vec{a} \times \vec{b}|^2\) નું મૂલ્ય ........... છે.

  1. A \(90\)
  2. B \(75\)
  3. C \(95\)
  4. D \(85\)
Verified Solution

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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