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JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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Application +4 | -1
The number of real values of a for which the vectors i^+2j^+k^,ai^+j^+2k^\hat{i}+2 \hat{j}+\hat{k}, a \hat{i}+\hat{j}+2 \hat{k} and i^+2j^+ak^\hat{i}+2 \hat{j}+a \hat{k} are coplanar, is
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Correct option: B
2
The correct answer is B (2) because there are two real values of 'a' for which the given vectors are coplanar, obtained by setting the determinant of the vectors to zero and solving the resulting quadratic equation.

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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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