JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

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Let

\hat{\mathbf{a}}

and

\hat{\mathbf{b}}

be two unit vectors such that

\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}=\frac{1}{3}

and

\hat{\mathbf{a}} \times \hat{\mathbf{b}}=\hat{\mathbf{c}}

. Also

\bar{F}=\alpha \hat{\mathbf{a}}+\beta \hat{\mathbf{b}}+\gamma \hat{\mathbf{c}}

, where

\alpha, \beta, \gamma

are scalars. If

\alpha=k_1(\hat{\mathbf{F}} \cdot \hat{\mathbf{a}})-k_2(\hat{\mathbf{F}} \cdot \hat{\mathbf{b}})

, then the value of

2\left(k_1+k_2\right)

View answer and explanation

Correct option: C

The correct answer is C (3) because it is obtained by calculating the scalar triple product and using vector identities to simplify the expression for alpha.

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

for some non-zero vector

\vec r,

then the value of

\left[ {\vec a\vec b\vec c} \right],

is The shortest distance between the lines

\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}}

and

\frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3},

is Let

\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}

and

\mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}.

Then

\left[ {\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}} \right]

depends on If

\vec{a}, \vec{b}

are non-zero vectors, then

((\vec{a} \times \vec{b}) \times \vec{a}) \cdot((\vec{b} \times \vec{a}) \times \vec{b})

equals

a \cdot(a \times b)=

The length of the perpendicular from the origin to the plane passing through the point a and containing the line

\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}