JEE Mains · Maths · STD 12 - 10. vector algebra
If \(\hat x,\,\hat y\) and \(\hat z\) are three unit vectors in three dimensional space , then the minimum value of \({\left| {\hat x + \hat y} \right|^2}\, + \,{\left| {\hat y + \hat z} \right|^2}\, + \,{\left| {\hat z + \hat x} \right|^2}\)
- A \(\frac {3}{2}\)
- B \(3\)
- C \(3\sqrt 3\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\((\hat{x}+\hat{y}+\hat{z})^{2} \geq 0\) \(\Rightarrow 3+2 \Sigma \hat{x} . \hat{y} \geq 0\) \(\Rightarrow 2 \Sigma \hat{x} \hat{y} \geq-3\) Now, \(|\hat{x}+\hat{y}|^{2}+|\hat{y}+\hat{z}|^{2}+|\hat{z}+\hat{x}|^{2}\) \(=6+2 \Sigma \hat{x} \cdot \hat{y} \geq 6+(-3)\)…
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