If ${\bf{\vec a}} = 2{\bf{\hat i}} - 3{\bf{\hat j}} + 5{\bf{\hat k}}$, ${\bf{\vec b}} = 3{\bf{\hat i}} - 4{\bf{\hat j}} + 5{\bf{\hat k}}$ and ${\bf{\vec c}} = 5{\bf{\hat i}} - 3{\bf{\hat j}} - 2{\bf{\hat k}}$, then the volume of the parallelopiped with coterminous edges ${\bf{\vec a}} + {\bf{\vec b}},{\bf{\vec b}} + {\bf{\vec c}},{\bf{\vec c}} + {\bf{\vec a}}$ is

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

\mathbf{\vec a} = 2\mathbf{\hat i} - 3\mathbf{\hat j} + 5\mathbf{\hat k}

\mathbf{\vec b} = 3\mathbf{\hat i} - 4\mathbf{\hat j} + 5\mathbf{\hat k}

and

\mathbf{\vec c} = 5\mathbf{\hat i} - 3\mathbf{\hat j} - 2\mathbf{\hat k}

, then the volume of the parallelopiped with coterminous edges

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

View answer and explanation

Correct option: D

Option D (8) is correct because it is the volume of the parallelepiped calculated using the scalar triple product of the given vectors.

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}