Let \(\vec{a}=2\hat{\imath}+3\hat{\jmath}-4\hat{k}\) and \(\vec{b}=3\hat{\imath}-5\hat{\jmath}+6\hat{k}\). Let \(\vec{c}\) be a vector such that \(\vec{c}\times\vec{a}=\vec{b}\times\vec{c}\) and \(\vec{c}\cdot(2\vec{a}-3\vec{b})=238\sqrt{2}\) then \(|\vec{c}|^2\) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 264
Since \(\vec{b}\times\vec{c} = -\,\vec{c}\times\vec{b}\), the first condition gives
\( \vec{c}\times\vec{a} + \vec{c}\times\vec{b} = \vec{0} \)
\( \vec{c}\times(\vec{a} + \vec{b}) = \vec{0} \)
A zero cross product means \(\vec{c}\) is parallel to \(\vec{a} + \vec{b}\) (\(\vec{c} \ne \vec{0}\), since its dot product with \(2\vec{a} - 3\vec{b}\) is not 0). So \(\vec{c} = \lambda(\vec{a} + \vec{b})\), where
\( \vec{a} + \vec{b} = 5\hat{\imath} - 2\hat{\jmath} + 2\hat{k} \)
\( 2\vec{a} - 3\vec{b} = (4 - 9)\hat{\imath} + (6 + 15)\hat{\jmath} + (-8 - 18)\hat{k} = -5\hat{\imath} + 21\hat{\jmath} - 26\hat{k} \)
Using the second condition:
\( \lambda\,[5(-5) + (-2)(21) + 2(-26)] = 238\sqrt{2} \)
\( \lambda\,(-25 - 42 - 52) = 238\sqrt{2} \)
\( -119\lambda = 238\sqrt{2} \)
\( \lambda = -2\sqrt{2} \)
Hence
\( |\vec{c}|^2 = \lambda^2\,|\vec{a} + \vec{b}|^2 \)
\( = 8\,(25 + 4 + 4) \)
\( = 8 \times 33 = 264 \)