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JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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Analyzing +4 | -1
If a=110(3i^+k^)\mathbf{a}=\frac{1}{\sqrt{10}}(3 \hat{\mathbf{i}}+\hat{\mathbf{k}}) and b=17(2i^+3j^βˆ’6k^)\mathbf{b}=\frac{1}{7}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}), then the value of (2βˆ’b)β‹…[(aΓ—b)Γ—(a+2b)](2-b) \cdot[(\mathbf{a} \times \mathbf{b}) \times(\mathbf{a}+2 \mathbf{b})] is
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Correct option: D
-5
Option D (-5) is correct because by applying vector triple product properties and simplifying, the expression evaluates to -5.

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If rβƒ—β‹…aβƒ—=rβƒ—β‹…bβƒ—=rβƒ—β‹…cβƒ—=0\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0 for some non-zero vector rβƒ—,\vec r, then the value of [aβƒ—bβƒ—cβƒ—],\left[ {\vec a\vec b\vec c} \right], is The shortest distance between the lines xβˆ’32=yβˆ’23=zβˆ’1βˆ’1\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}} and x+32=yβˆ’61=zβˆ’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 [aβƒ—,β€…β€Šbβƒ—,β€…β€Šcβƒ—]\left[ {\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}} \right] depends on If aβƒ—,bβƒ—\vec{a}, \vec{b} are non-zero vectors, then ((aβƒ—Γ—bβƒ—)Γ—aβƒ—)β‹…((bβƒ—Γ—aβƒ—)Γ—bβƒ—)((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b}) equals aβ‹…(aΓ—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+Ξ»c\mathbf{r}=\mathbf{b}+\lambda \mathbf{c} is
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