Question:
Let \(\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}\) and \(\vec{b} = 7\hat{i} + \hat{j} - 6\hat{k}\).
If \(\vec{r}\times\vec{a} = \vec{r}\times\vec{b}\), \(\vec{r}\cdot(\hat{i} + 2\hat{j} + \hat{k}) = -3\), then \(\vec{r}\cdot(2\hat{i} - 3\hat{j} + \hat{k})\) is equal to :
Let \(\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}\) and \(\vec{b} = 7\hat{i} + \hat{j} - 6\hat{k}\).
If \(\vec{r}\times\vec{a} = \vec{r}\times\vec{b}\), \(\vec{r}\cdot(\hat{i} + 2\hat{j} + \hat{k}) = -3\), then \(\vec{r}\cdot(2\hat{i} - 3\hat{j} + \hat{k})\) is equal to :
If \(\vec{r}\times\vec{a} = \vec{r}\times\vec{b}\), \(\vec{r}\cdot(\hat{i} + 2\hat{j} + \hat{k}) = -3\), then \(\vec{r}\cdot(2\hat{i} - 3\hat{j} + \hat{k})\) is equal to :
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\(\vec{r}\times(\vec{a}-\vec{b}) = \vec{0}\) means \(\vec{r}\) is parallel to \(\vec{a}-\vec{b}\); fix the scalar using the given dot product.
Updated On: Nov 4, 2026
- \(10\)
- \(8\)
- \(13\)
- \(12\)
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The Correct Option is D
Solution and Explanation
\(\vec{r}\times\vec{a} - \vec{r}\times\vec{b} = \vec{0}\Rightarrow \vec{r}\times(\vec{a}-\vec{b}) = \vec{0}\), so \(\vec{r} = \lambda(\vec{a}-\vec{b})\).
\(\vec{a}-\vec{b} = -5\hat{i} - 4\hat{j} + 10\hat{k}\), so \(\vec{r} = \lambda(-5\hat{i} - 4\hat{j} + 10\hat{k})\).
\(\vec{r}\cdot(\hat{i}+2\hat{j}+\hat{k}) = \lambda(-5 - 8 + 10) = -3\lambda = -3\Rightarrow \lambda = 1\).
\(\vec{r}\cdot(2\hat{i}-3\hat{j}+\hat{k}) = -10 + 12 + 10 = 12\).
Answer: (4) 12
\(\vec{a}-\vec{b} = -5\hat{i} - 4\hat{j} + 10\hat{k}\), so \(\vec{r} = \lambda(-5\hat{i} - 4\hat{j} + 10\hat{k})\).
\(\vec{r}\cdot(\hat{i}+2\hat{j}+\hat{k}) = \lambda(-5 - 8 + 10) = -3\lambda = -3\Rightarrow \lambda = 1\).
\(\vec{r}\cdot(2\hat{i}-3\hat{j}+\hat{k}) = -10 + 12 + 10 = 12\).
Answer: (4) 12
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