JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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

are non-coplanar non-null vetors stch that

\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=2

, then

\left\{\left[\begin{array}{llll}\vec{a} \times \vec{b} & \vec{b} \times \vec{c} & \vec{c} \times \vec{a}\end{array}\right]\right\}^2=

View answer and explanation

Correct option: B

Option B (16) is correct because

\left[\begin{array}{lll}\vec{a} \times \vec{b} & \vec{b} \times \vec{c} & \vec{c} \times \vec{a}\end{array}\right] = \left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]^2

, and thus the square of that is

\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]^4 = 2^4 = 16

More Questions from SCALAR TRIPLE PRODUCT

Practice All Chapter MCQs →

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

\vec r,

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

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

\frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3},

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

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

\vec{a}, \vec{b}

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

a \cdot(a \times b)=

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