JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a}, \vec{b}, \vec{c}\) be three non-coplanar vectors such that \(\overrightarrow{ a } \times \overrightarrow{ b }=4 \overrightarrow{ c }, \overrightarrow{ b } \times \overrightarrow{ c }=9 \overrightarrow{ a }\) and \(\overrightarrow{ c } \times \overrightarrow{ a }=\alpha \overrightarrow{ b }, \alpha>0\) . If \(|\vec{a}|+|\vec{b}|+|\vec{c}|=36\), then \(\alpha\) is equal to\(....\)
- A \(33\)
- B \(34\)
- C \(35\)
- D \(36\)
Answer & Solution
Correct Answer
(D) \(36\)
Step-by-step Solution
Detailed explanation
\(|\vec{a}|+|\vec{b}|+|\vec{c}|=36\) \(\frac{5}{2} \lambda+6=36\) \(\lambda=12\) \(\alpha=\frac{|\vec{c}||\vec{a}|}{|\vec{b}|}=\frac{3 \times 12}{2} \times \frac{12}{6}\) \(\alpha=36\)
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