JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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Analyzing +4 | -1

If the vectors

\vec{a}

and

\vec{b}

are perpendicular to each other, then a vector

\vec{v}

in terms of

\vec{a}

and

\vec{b}

satisfying the equations

\vec{v} \cdot \vec{a}=0, \vec{v} \cdot \vec{b}=1

and

\left[\begin{array}{lll}\vec{v} & \vec{a} & \vec{b}\end{array}\right]=1,

View answer and explanation

Correct option: A

\frac{\vec{b}}{|\vec{b}|^2}+\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|^2}

Option A (

\frac{\vec{{b}}}{{|\vec{{b}}|^2}}+\frac{{\vec{{a}} \times \vec{{b}}}}{{|\vec{{a}} \times \vec{{b}}|^2}}

) is correct because it satisfies the given conditions for the vector

\vec{{v}}

when

\vec{{a}}

and

\vec{{b}}

are perpendicular.

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}