JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\overrightarrow{ a }, \overrightarrow{ b }, \overrightarrow{ c }\) be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is \(14\) and \((\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=168\) then \(|\vec{a}|+|\vec{b}|+|\vec{c}|\) is equal to.
- A \(10\)
- B \(14\)
- C \(16\)
- D \(18\)
Answer & Solution
Correct Answer
(C) \(16\)
Step-by-step Solution
Detailed explanation
\(|\vec{a}\|\vec{b}\| \vec{c}|=14\) \(\overrightarrow{ a }^{\wedge} \overrightarrow{ b }=\overrightarrow{ b } \wedge \overrightarrow{ c }=\overrightarrow{ c }^{\wedge} \overrightarrow{ a }=\theta=\frac{2 \pi}{3}\) So,…
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