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JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy
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Analyzing +4 | -1
If a⃗=110(3i^+k^)\vec{a}=\frac{1}{\sqrt{10}}(3 \hat{i}+\hat{k}) and b⃗=17(2i^+3j^−6k^)\vec{b}=\frac{1}{7}(2 \hat{i}+3 \hat{j}-6 \hat{k}), then the value of (2a⃗−b⃗)[(a⃗×b⃗)×(a⃗+2b⃗)](2 \vec{a}-\vec{b})[(\vec{a} \times \vec{b}) \times(\vec{a}+2 \vec{b})] is
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Correct option: D
-5
Option D (-5) is correct because by simplifying the given expression using vector identities and the provided values for a⃗\vec{a} and b⃗\vec{b}, the result is -5.

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If r⃗⋅a⃗=r⃗⋅b⃗=r⃗⋅c⃗=0\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0 for some non-zero vector r⃗,\vec r, then the value of [a⃗b⃗c⃗],\left[ {\vec a\vec b\vec c} \right], is The shortest distance between the lines x−32=y−23=z−1−1\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}} and x+32=y−61=z−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 [a⃗,  b⃗,  c⃗]\left[ {\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}} \right] depends on If a⃗,b⃗\vec{a}, \vec{b} are non-zero vectors, then ((a⃗×b⃗)×a⃗)⋅((b⃗×a⃗)×b⃗)((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b}) equals a⋅(a×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+λc\mathbf{r}=\mathbf{b}+\lambda \mathbf{c} is
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