JEE Mains · Maths · STD 12 - 10. vector algebra
Consider three vectors \(\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}\). Let \(|\overrightarrow{\mathrm{a}}|=2,|\overrightarrow{\mathrm{b}}|=3\) and \(\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}\). If \(\alpha \in\left[0, \frac{\pi}{3}\right]\) is the angle between the vectors \(\vec{b}\) and \(\vec{c}\), then the minimum value of \(27|\overrightarrow{c}-\overrightarrow{a}|^2\) is equal to :
- A \(110\)
- B \(105\)
- C \(124\)
- D \(121\)
Answer & Solution
Correct Answer
(C) \(124\)
Step-by-step Solution
Detailed explanation
\( |\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{c}}|^2+|\overrightarrow{\mathrm{a}}|^2-2 \overline{\mathrm{a}} \cdot \overline{\mathrm{c}} \) \( =|\overrightarrow{\mathrm{c}}|^2+4-0 \)…
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