JEE Mains · Maths · STD 12 - 10. vector algebra
Let a vector \(\vec{a}\) has a magnitude \(9\) . Let a vector \(\vec{b}\) be such that for every \((x, y) \in R \times R-\{(0,0)\}\), the vector \((x \vec{a}+y \vec{b})\) is perpendicular to the vector (6y \(\vec{a}-18 \times \vec{b}\) ). Then the value of \(|\vec{a} \times \vec{b}|\) is equal to.
- A \(9 \sqrt{3}\)
- B \(27 \sqrt{3}\)
- C \(9\)
- D \(81\)
Answer & Solution
Correct Answer
(B) \(27 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(|\vec{a}|=9 \&(x \vec{a}+y \vec{b}) \cdot(6 y \vec{a}-18 x \vec{b})=0\) \(\Rightarrow 6 x y|\bar{a}|^{2}-18 x^{2}(\bar{a} \cdot \bar{b})+6 y^{2}(\bar{a} \cdot \bar{b})-18 x y|\bar{b}|^{2}=0\)…
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