JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

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Analyzing +4 | -1

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

are any three vectors forming a linearly independent system, then

\forall \theta \in R

\left[\mathbf{a} \cos \theta+\mathbf{b} \sin \theta+\mathbf{c o s} 2 \theta, \mathbf{a} \cos \left(\frac{2 \pi}{3}+\theta\right)\right.

+ b\sin \left( {\frac{{2\pi }}{3} + \theta } \right) + c\cos 2\left( {\frac{{2\pi }}{3} + \theta } \right){\text{, }}

acos\left. {\left( {\theta - \frac{{2\pi }}{3}} \right) + b\sin \left( {\theta - \frac{{2\pi }}{3}} \right) + c\cos 2\left( {\theta - \frac{{2\pi }}{3}} \right)} \right]

equals

View answer and explanation

Correct option: D

None of the above

Option D (

\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+\frac{(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{b} \times \mathbf{c})}{|b|^2} \mathbf{b}

) is correct because it represents the component of

\mathbf{a} \times(\mathbf{b} \times \mathbf{c})

perpendicular to

\mathbf{b}

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}