Let \(\vec{a}=2\hat{\imath}+\hat{k}\), \(\vec{b}=\hat{\imath}+\hat{\jmath}+\hat{k}\) and \(\vec{c}=4\hat{\imath}-3\hat{\jmath}+7\hat{k}\). If a vector \(\vec{r}\) satisfies \(\vec{r}\times\vec{b}=\vec{c}\times\vec{b}\) and \(\vec{r}\cdot\vec{a}=0\), then \(\vec{r}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-\hat{\imath}-8\hat{\jmath}+2\hat{k}\)
Moving everything to one side of the first condition:
\((\vec{r}-\vec{c})\times\vec{b} = \vec{0}\)
The cross product of two non-zero vectors is zero only when they are parallel, so \(\vec{r}-\vec{c}\) is a multiple of \(\vec{b}\):
\(\vec{r} = \vec{c}+t\,\vec{b}\)
\( = (4+t)\hat{\imath}+(-3+t)\hat{\jmath}+(7+t)\hat{k}\)
Now use \(\vec{r}\cdot\vec{a}=0\) with \(\vec{a}=2\hat{\imath}+0\hat{\jmath}+\hat{k}\):
\(2(4+t)+(7+t) = 0\)
\(15+3t = 0\)
\(t = -5\)
Hence
\(\vec{r} = (4-5)\hat{\imath}+(-3-5)\hat{\jmath}+(7-5)\hat{k}\)
\( = -\hat{\imath}-8\hat{\jmath}+2\hat{k}\)
Shortcut: Test the options. \(\vec{r}\cdot\vec{a} = 2x + z\) must be 0:
Option (1): \(-2+2 = 0\)
Option (2): \(-2+2 = 0\)
Option (3): \(18+12 = 30\)
Option (4): \(4+2 = 6\)
Only Options (1) and (2) remain. Then \(\vec{r}-\vec{c}\) must be a multiple of \(\vec{b}\):
Option (1): \(\vec{r}-\vec{c} = -5\hat{\imath}+7\hat{\jmath}-5\hat{k}\), not a multiple of \(\vec{b}\)
Option (2): \(\vec{r}-\vec{c} = -5\hat{\imath}-5\hat{\jmath}-5\hat{k} = -5\vec{b}\)