JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

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

are non-coplanar vectors and

\vec{p}, \vec{q}, \vec{r}

are reciprocal vectors, then

(l \vec{a}+m \vec{b}+n \vec{c}) \cdot(l \vec{p}+m \vec{q}+n \vec{r})

is equal to

View answer and explanation

Correct option: A

l^2+m^2+n^2

Option A (

l^2+m^2+n^2

) is correct because the dot product simplifies to the sum of the squares of the coefficients due to the properties of reciprocal vectors.

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}