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JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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Application +4 | -1
If ∣a∣=1,ā€‰ā€‰āˆ£b∣=3   andā€‰ā€‰Ā āˆ£c∣=5|a| = 1,\,\,|b| = 3{\text{ }}\,\,{\text{and}}\,\,{\text{ }}|c| = 5 then the value of [aāˆ’b  bāˆ’c  cāˆ’a][a - b\,\,b - c\,\,c - a] is
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Correct option: A
0
Option A (0) is correct because the scalar triple product of vectors in the form [a-b b-c c-a] is always zero.

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If rāƒ—ā‹…aāƒ—=rāƒ—ā‹…bāƒ—=rāƒ—ā‹…cāƒ—=0\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0 for some non-zero vector rāƒ—,\vec r, then the value of [aāƒ—bāƒ—cāƒ—],\left[ {\vec a\vec b\vec c} \right], is The shortest distance between the lines xāˆ’32=yāˆ’23=zāˆ’1āˆ’1\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}} and x+32=yāˆ’61=zāˆ’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 [aāƒ—,ā€…ā€Šbāƒ—,ā€…ā€Šcāƒ—]\left[ {\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}} \right] depends on If aāƒ—,bāƒ—\vec{a}, \vec{b} are non-zero vectors, then ((aāƒ—Ć—bāƒ—)Ɨaāƒ—)ā‹…((bāƒ—Ć—aāƒ—)Ɨbāƒ—)((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b}) equals aā‹…(aƗb)=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+Ī»c\mathbf{r}=\mathbf{b}+\lambda \mathbf{c} is
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