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JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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If \mathbf{\vec a}\cdot\widehat \mathbf{i} = 4\mathbf{\vec a}\cdot\widehat \mathbf{i} = 4 then \left( {\mathbf{\vec a} \times \widehat \mathbf{j}} \right)\cdot\left( {2\widehat \mathbf{j} - 3\mathbf{\hat k}} \right)\left( {\mathbf{\vec a} \times \widehat \mathbf{j}} \right)\cdot\left( {2\widehat \mathbf{j} - 3\mathbf{\hat k}} \right) is equal to
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Correct option: D
โˆ’12- 12
Option D (โˆ’12- 12) is correct because it is the result of the scalar triple product calculation using the given dot product.

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If rโƒ—โ‹…aโƒ—=rโƒ—โ‹…bโƒ—=rโƒ—โ‹…cโƒ—=0\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0 for some non-zero vector rโƒ—,\vec r, then the value of [aโƒ—bโƒ—cโƒ—],\left[ {\vec a\vec b\vec c} \right], is The shortest distance between the lines xโˆ’32=yโˆ’23=zโˆ’1โˆ’1\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}} and x+32=yโˆ’61=zโˆ’53,\frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3}, is Let \mathbf{\vec a} = \widehat \mathbf{i} - \mathbf{\hat k},\;\mathbf{\vec b} = x\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - x} \right)\mathbf{\hat k}\mathbf{\vec a} = \widehat \mathbf{i} - \mathbf{\hat k},\;\mathbf{\vec b} = x\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - x} \right)\mathbf{\hat k} and \mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}.\mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}. Then [aโƒ—,โ€…โ€Šbโƒ—,โ€…โ€Šcโƒ—]\left[ {\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}} \right] depends on If aโƒ—,bโƒ—\vec{a}, \vec{b} are non-zero vectors, then ((aโƒ—ร—bโƒ—)ร—aโƒ—)โ‹…((bโƒ—ร—aโƒ—)ร—bโƒ—)((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b}) equals aโ‹…(aร—b)=a \cdot(a \times b)= The length of the perpendicular from the origin to the plane passing through the point a and containing the line r=b+ฮปc\mathbf{r}=\mathbf{b}+\lambda \mathbf{c} is
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