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JEE Mains · Maths · STD 12 - 10. vector algebra
If the vector \(\vec b = 3\hat j + 4\hat k\) is written as the sum of a vector \({\vec {b_1}}\) , parallel to \(\vec a = \hat i + \hat j\) and a vector \({\vec {b_2}}\) , perpendicular to \(\vec a\) , then \({\vec {b_1}} \times {\vec {b_2}}\) is equal to
- A \( - 3\hat i + 3\hat j - 9\hat k\)
- B \( 6\hat i - 6\hat j + \frac{9}{2}\hat k\)
- C \( - 6\hat i + 6\hat j - \frac{9}{2}\hat k\)
- D \(3\hat i - 3\hat j + 9\hat k\)
Answer & Solution
Correct Answer
(B) \( 6\hat i - 6\hat j + \frac{9}{2}\hat k\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{\mathrm{b}_{1}}=\frac{(\overrightarrow{\mathrm{b}_{1}} \cdot \overrightarrow{\mathrm{a}}) \hat a}{1}\) \(=\left\{\frac{(3 \hat{j}+4 \hat{k}) \cdot(\hat{i}+\hat{j})}{\sqrt{2}}\right\}\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)\)…
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