Let $\vec{a}, \vec{b}$ and $\vec{c}$ be non-zero vectors such that no two are collinear and $(\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}$ If $\theta$ is the acute angle between the vectors $\vec{b}$ and $\vec{c},$ then $\sin \theta$ equals

JEE MATHEMATICS MCQ: VECTOR TRIPLE PRODUCT

1 Easy

1 of 1

Analyzing +4 | -1

Let

\vec{a}, \vec{b}

and

\vec{c}

be non-zero vectors such that no two are collinear and

(\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b}||\vec{c}| \vec{a}

\theta

is the acute angle between the vectors

\vec{b}

and

\vec{c},

then

\sin \theta

equals

View answer and explanation

Correct option: A

\frac{2 \sqrt{2}}{3}

Option A (

\frac{2 \sqrt{2}}{3}

) is correct because it is derived from the given condition using the properties of vector triple product and solving for

\sin \theta

Practice All Chapter MCQs →

{\mathbf{\hat u}}

{\mathbf{\hat v}}

\mathbf{\hat u}\cdot\mathbf{\hat v} = 0

{\mathbf{\hat r}}

{\mathbf{\hat u}}

{\mathbf{\hat v}}

\mathbf{\vec r} \times \left( {\mathbf{\hat u} \times \mathbf{\hat v}} \right)

\lambda \in \mathrm{R}, \vec{a}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}, \vec{b}=\hat{i}-\lambda \hat{j}+2 \hat{k}

((\vec{a}+\vec{b}) \times(\vec{a} \times \vec{b})) \times(\vec{a}-\vec{b})=8 \hat{i}-40 \hat{j}-24 \hat{k}

|\lambda(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|^2

\vec{a}=-\hat{i}-\hat{j}+\hat{k}

\vec{a} \cdot \vec{b}=1

\vec{a} \times \vec{b}=\hat{i}-\hat{j}

\vec{a}-6 \vec{b}

\vec{a}, \vec{b}, \vec{c}

\vec u = \vec a \times (\vec b \times \vec c),\vec v = \vec b \times (\vec c \times \vec a)\,\,\,{\text{and}}\,\,\,\vec w = \vec c \times (\vec a \times \vec b)

a

b

a\, \times (a\, \times (a\, \times (a\, \times b)))

\vec{a}

\hat{i} \times(\vec{a} \times \vec{i})+\hat{j} \times(\vec{a} \times \hat{j})+\hat{k} \times(\vec{a} \times \vec{k})