$\frac{{{\bf{\vec a}}\cdot\left( {\;{\bf{\vec b}} \times \;{\bf{\vec c}}} \right)}}{{{\bf{\vec b}}\cdot\left( {\;{\bf{\vec c}} \times {\bf{\vec a}}} \right)}} + \frac{{{\bf{\vec b}}\cdot\left( {\;{\bf{\vec a}} \times {\bf{\vec b}}} \right)}}{{{\bf{\vec a}}\cdot\left( {{\bf{\vec b}} \times \;{\bf{\vec c}}} \right)}}\;{\rm{is\;equal\;to}}$

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Understanding +4 | -1

\frac{{\mathbf{\vec a}\cdot\left( {\;\mathbf{\vec b} \times \;\mathbf{\vec c}} \right)}}{{\mathbf{\vec b}\cdot\left( {\;\mathbf{\vec c} \times \mathbf{\vec a}} \right)}} + \frac{{\mathbf{\vec b}\cdot\left( {\;\mathbf{\vec a} \times \mathbf{\vec b}} \right)}}{{\mathbf{\vec a}\cdot\left( {\mathbf{\vec b} \times \;\mathbf{\vec c}} \right)}}\;\mathrm{is\;equal\;to}

View answer and explanation

Correct option: A

Option A (1) is correct because the expression simplifies to 1 using properties of scalar triple products.

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}