JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a}=a_i \hat{i}+a_2 \hat{j}+a_3 \hat{k}\) and \(\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\) be two vectors such that \(|\vec{a}|=1 ; \quad \vec{a} \cdot \vec{b}=2\) and \(|\vec{b}|=4\). If \(\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}\), then the angle between \(\vec{b}\) and \(\vec{c}\) is equal to :
- A \(\cos ^{-1}\left(\frac{2}{\sqrt{3}}\right)\)
- B \(\cos ^{-1}\left(-\frac{1}{\sqrt{3}}\right)\)
- C \(\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)\)
- D \(\cos ^{-1}\left(\frac{2}{3}\right)\)
Answer & Solution
Correct Answer
(C) \(\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)\)
Step-by-step Solution
Detailed explanation
Given \(|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2\) \(\vec{c}=2(\vec{a} \times \vec{b})-3 \vec{b}\) Dot product with \(\overrightarrow{\mathrm{a}}\) on both sides \(\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=-6\) Dot product with \(\vec{b}\) on both sides…
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