MHT CET · Maths · Vector Algebra
The position vectors of the points \(A, B, C\) are \(\hat{i}+2 \hat{j}-\hat{k}, \hat{i}+\hat{j}+\hat{k}\), \(2 \hat{i}+3 \hat{j}+2 \hat{k}\) respectively. If \(A\) is chosen as the origin, then the cross product of position vectors of \(B\) and \(C\) are
- A \(-5 \hat{i}+2 \hat{j}+\hat{k}\)
- B \(-\hat{i}+0 \hat{j}-\hat{k}\)
- C \(\hat{i}-\hat{\mathrm{k}}\)
- D \(5 \hat{i}-2 \hat{j}-\hat{k}\)
Answer & Solution
Correct Answer
(A) \(-5 \hat{i}+2 \hat{j}+\hat{k}\)
Step-by-step Solution
Detailed explanation
\( \vec{r}_B' = \vec{B} - \vec{A} = (\hat{i}+\hat{j}+\hat{k}) - (\hat{i}+2 \hat{j}-\hat{k}) = - \hat{j} + 2\hat{k} \) \( \vec{r}_C' = \vec{C} - \vec{A} = (2 \hat{i}+3 \hat{j}+2 \hat{k}) - (\hat{i}+2 \hat{j}-\hat{k}) = \hat{i} + \hat{j} + 3\hat{k} \)
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