If ${\bf{\vec a}},{\bf{\vec b}},{\bf{\vec c}}$ are unit coplanar vectors, then $\left[ {2{\bf{\vec a}} - {\bf{\vec b}}\;2\;{\bf{\vec b}} - {\bf{\vec c}}\;2\;{\bf{\vec c}} - {\bf{\vec a}}} \right]$ is equal to

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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\mathbf{\vec a},\mathbf{\vec b},\mathbf{\vec c}

are unit coplanar vectors, then

\left[ {2\mathbf{\vec a} - \mathbf{\vec b}\;2\;\mathbf{\vec b} - \mathbf{\vec c}\;2\;\mathbf{\vec c} - \mathbf{\vec a}} \right]

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because if three vectors are coplanar, any linear combination of them will also be coplanar, resulting in a scalar triple product of zero.

More Questions from SCALAR TRIPLE PRODUCT

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\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0

for some non-zero vector

\vec r,

then the value of

\left[ {\vec a\vec b\vec c} \right],

is The shortest distance between the lines

\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}}

and

\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}

and

\mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}.

Then

\left[ {\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}} \right]

depends on If

\vec{a}, \vec{b}

are non-zero vectors, then

((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b})

equals

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

\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}