JEE MATHEMATICS MCQ: SCALAR TRIPLE PRODUCT

1 Easy

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Analyzing +4 | -1

Two adjacent sides of a parallelogram

A B C D

are given by

\overrightarrow{A B}=2 \hat{i}+10 \hat{j}+11 \hat{k}

and

\overrightarrow{A D}=-\hat{i}+2 \hat{j}+2 \hat{k}.

The side

A D

is rotated by an acute angle

\alpha

in the plane of the parallelogram so that

A D

becomes

A D^{\prime}.

A D^{\prime}

makes a right angle with the side

A B,

then the cosine of the angle

\alpha

is given by

View answer and explanation

Correct option: B

\frac{\sqrt{17}}{9}

Option A (

\hat{j}-\hat{k}

and

-\hat{j}+\hat{k}

) is correct because these vectors are coplanar with the given vectors and perpendicular to the vector

\hat{i}+\hat{j}+\hat{k}

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}