Let ${\bf{\vec v}} = 2{\bf{\hat i}} + {\bf{\hat j}} - {\bf{\hat k}}{\rm{\;and\;}}{\bf{\vec w}} = {\bf{\hat i}} + 3{\bf{\hat k}},{\rm{\;If\;}}{\bf{\vec u}}$ is a unit vector, then maximum value of the scalar triple product $\left[ {\;{\bf{\vec u}}\;{\bf{\vec v}}\;{\bf{\vec w}}{\rm{\;}}} \right]$ is

JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

1 of 1

Application +4 | -1

Let

\mathbf{\vec v} = 2\mathbf{\hat i} + \mathbf{\hat j} - \mathbf{\hat k}\mathrm{\;and\;}\mathbf{\vec w} = \mathbf{\hat i} + 3\mathbf{\hat k},\mathrm{\;If\;}\mathbf{\vec u}

is a unit vector, then maximum value of the scalar triple product

\left[ {\;\mathbf{\vec u}\;\mathbf{\vec v}\;\mathbf{\vec w}\mathrm{\;}} \right]

View answer and explanation

Correct option: C

\sqrt {59}

Option C (

\sqrt {59}

) is correct because the maximum value of the scalar triple product is the magnitude of the vector resulting from the cross product of

\vec v

and

\vec w

, since

\vec u

is a unit vector.

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}