JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec \alpha \, = \,3\hat i\, + \hat j\) and \(\vec \beta \, = \,2\hat i\, - \hat j + 3\hat k.\) If \(\vec \beta \, = \,{\vec \beta _1} - {\vec \beta _2},\) where \({\vec \beta _1}\) is parallel to \(\vec \alpha \) and \(\vec \beta_2 \) is perpendicular to \(\vec \alpha ,\) then \({\vec \beta _1} \times {\vec \beta _2}\) is equal to
- A \(\frac{1}{2}( - 3\hat i + 9\hat j + 5\hat k)\)
- B \(\frac{1}{2}( 3\hat i - 9\hat j + 5\hat k)\)
- C \(- 3\hat i + 9\hat j + 5\hat k\)
- D \(3\hat i - 9\hat j - 5\hat k\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}( - 3\hat i + 9\hat j + 5\hat k)\)
Step-by-step Solution
Detailed explanation
\(\vec{\alpha}=3 \hat{i}+\hat{j}\) \(\vec{\beta}=2 \hat{i}-\hat{j}+3 \hat{k}\) \(\vec{\beta}=\vec{\beta}_{1}-\vec{\beta}_{2}\) \(\overrightarrow {{\beta _1}} = \lambda (3\hat i + \hat j),\overrightarrow {{\beta _2}} = \lambda (3\hat i + \hat j) - 2\hat i + \hat j - 3\hat k\)…
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