JEE Mains · Maths · STD 12 - 10. vector algebra
Let the vectors \(\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}\) and \(\vec{b} = \hat{i} + 3\hat{j} + \hat{k}\). For some \(\lambda, \mu \in \mathbb{R}\), let \(\vec{c} = \lambda \vec{a} + \mu \vec{b}\). If \(\vec{c} \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) = 10\) and \(\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = -2\), then \(|\vec{c}|^2\) is equal to:
- A \(8\)
- B \(12\)
- C \(14\)
- D \(15\)
Answer & Solution
Correct Answer
(B) \(12\)
Step-by-step Solution
Detailed explanation
Given \(\vec{a} = -\hat{i} + \hat{j} + 3\hat{k}\) and \(\vec{b} = \hat{i} + 3\hat{j} + \hat{k}\). The vector \(\vec{c}\) is given by: \(\vec{c} = \lambda \vec{a} + \mu \vec{b} = (-\lambda + \mu)\hat{i} + (\lambda + 3\mu)\hat{j} + (3\lambda + \mu)\hat{k}\) Using the first…
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