JEE Mains · Maths · STD 12 - 10. vector algebra
If \(\lambda>0\), let \(\theta\) be the angle between the vectors \(\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}\) and \(\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}\). If the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) are mutually perpendicular, then the value of \((14 \cos \theta)^2\) is equal to
- A \(25\)
- B \(20\)
- C \(50\)
- D \(40\)
Answer & Solution
Correct Answer
(A) \(25\)
Step-by-step Solution
Detailed explanation
\( (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0, \lambda>0 \) \( |\vec{a}|^2-|\vec{b}|^2=0 \rightarrow 1+\lambda^2+9=9+1+4 \) \( \therefore \lambda=2, \cos \theta=\frac{\vec{a}-\vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\frac{3-\lambda-6}{\sqrt{14} \cdot \sqrt{14}} \)…
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