JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

\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

View answer and explanation

Correct option: A

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

Option D ((

\vec{a} \cdot \vec{b})|\vec{a} \times \vec{b}|^2

) is correct because expanding both triple vector products and simplifying using dot product properties results in this expression.

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]

a \cdot(a \times b)=

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

\mathbf{b}=-\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}

\mathbf{c}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}

\mathbf{a}

a

[\mathbf{a} \mathbf{b} \mathbf{c}]