JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}\), \(\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}\) and a vector \(\vec{c}\) be such that \(2(\vec{a}\times\vec{b}) + 3(\vec{b}\times\vec{c}) = \vec{0}\). If \(\vec{a}\cdot\vec{c} = 15\), then \(\vec{c}\cdot(\hat{i}+\hat{j}-3\hat{k})\) is equal to:
- A \(-6\)
- B \(-5\)
- C \(-4\)
- D \(-3\)
Answer & Solution
Correct Answer
(B) \(-5\)
Step-by-step Solution
Detailed explanation
Given \(2(\vec{a}\times\vec{b}) + 3(\vec{b}\times\vec{c}) = \vec{0}\) \(\Rightarrow 2(\vec{a}\times\vec{b}) - 3(\vec{c}\times\vec{b}) = \vec{0}\) \(\Rightarrow (2\vec{a} - 3\vec{c}) \times \vec{b} = \vec{0}\) Since the cross product is zero, the vectors are collinear:…
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