If ${\bf{\vec A}} \times \left( {{\bf{\vec B}} \times {\bf{\vec C}}} \right) = {\bf{\vec B}} \times \left( {{\bf{\vec C}} \times {\bf{\vec A}}} \right)$ and $\left[ {{\bf{\vec A}}\;{\bf{\vec B}}\;{\bf{\vec C}}} \right] \ne 0,$ then ${\bf{\vec A}} \times \left( {{\bf{\vec B}} \times {\bf{\vec C}}} \right)$ is equal to

${\bf{\vec A}} \times {\bf{\vec B}}$

${\bf{\vec B}} \times {\bf{\vec C}}$

${\bf{\vec C}} \times {\bf{\vec A}}$

Back to VECTOR TRIPLE PRODUCT

Your Progress

0 / 66 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • VECTOR TRIPLE PRODUCT

Vector triple product practice · Easy

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

Take a chapter quiz →

Showing 61-66 of 66 questions

Preparing practice and saved progress…

61 Easy

61 of 66

Application +4 | -1

Let

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

be non-zero vectors such that

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

. If

\theta

is an acute angle between the vectors

b

and

\mathbf{c}

, then

\sin \theta

is equal to

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because solving the equation

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

for

\sin \theta

yields

\sin \theta = \frac{2 \sqrt{2}}{3}

62 Easy

62 of 66

Application +4 | -1

Let

\vec{p}, \vec{q}, \vec{r}

be three mutually perpendicular vectors of the same magnitude. If a vector

\vec{x}

satisfies the equation

\vec{p} \times(\vec{x}-\vec{q}) \times \vec{p}\}+\vec{q} \times(\vec{x}-\vec{r}) \times \vec{q}\}+\vec{r} \times\{\vec{x}-\vec{p}) \times \vec{r}\}=\overrightarrow{0},

then

\vec{x}

is given by

View answer and explanation

Correct option: B

\frac{1}{2}(\vec{p}+\vec{q}+\vec{r})

Option B (

\frac{1}{2}(\vec{p}+\vec{q}+\vec{r})

) is correct because substituting this value of

\vec{x}

into the given equation makes it equal to

\overrightarrow{0}

, utilizing the mutually perpendicular nature of

\vec{p}, \vec{q}, \vec{r}

63 Easy

63 of 66

Analyzing +4 | -1

The unit vector which is orthogonal to the vector

3 \hat{i}+2 \hat{j}+6 \hat{k}

and is coplanar with vectors

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

and

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

View answer and explanation

Correct option: C

\frac{1}{\sqrt{10}}(3 \hat{j}-\hat{k})

Option C (

\frac{1}{\sqrt{10}}(3 \hat{j}-\hat{k})

) is correct because it represents the unit vector orthogonal to the given vector and coplanar with the other two, found using vector triple product.

64 Easy

64 of 66

Application +4 | -1

Let

\vec{a}, \vec{b}

and

\vec{c}

be three unit vectors such that

\vec{a} \times(\vec{b} \times \vec{c})=\frac{\sqrt{3}}{2}(\vec{b}+\vec{c})

. If

\vec{b}

is not parallel to

\vec{c}

, then the angle between

\vec{a}

and

\vec{b}

View answer and explanation

Correct option: D

\frac{5 \pi}{6}

Option D (

\frac{{5 \pi}}{{6}}

) is correct because solving the vector equation using the vector triple product and comparing coefficients yields this angle between

\vec{a}

and

\vec{b}

65 Easy

65 of 66

Application +4 | -1

Let

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

and

\vec{c}=\hat{i}-\hat{j}-\hat{k}.

Then the vector

\vec{b}

satisfying

\vec{a} \times \vec{b}+\vec{c}=\overrightarrow{0}

and

\vec{a} \cdot \vec{b}=3,

View answer and explanation

Correct option: C

-\hat{i}+\hat{j}-2 \hat{k}

Option C (

-\hat{i}+\hat{j}-2 \hat{k}

) is correct because it satisfies both given conditions:

\vec{a} \times \vec{b}+\vec{c}=\overrightarrow{0}

and

\vec{a} \cdot \vec{b}=3

66 Easy

66 of 66

Analyzing +4 | -1

\mathbf{\vec A} \times \left( {\mathbf{\vec B} \times \mathbf{\vec C}} \right) = \mathbf{\vec B} \times \left( {\mathbf{\vec C} \times \mathbf{\vec A}} \right)

and

\left[ {\mathbf{\vec A}\;\mathbf{\vec B}\;\mathbf{\vec C}} \right] \ne 0,

then

\mathbf{\vec A} \times \left( {\mathbf{\vec B} \times \mathbf{\vec C}} \right)

is equal to

View answer and explanation

Correct option: A

{\vec 0}

The correct answer is

\vec 0

because given the condition

\mathbf{\vec A} \times ({{\mathbf{\vec B} \times \mathbf{\vec C}}}) = \mathbf{\vec B} \times ({{\mathbf{\vec C} \times \mathbf{\vec A}}})

and

\left[ {{{\mathbf{\vec A}}}\;{{\mathbf{\vec B}}}\;{{\mathbf{\vec C}}}} \right] \ne 0

, it implies that

\mathbf{\vec A} \times ({{\mathbf{\vec B} \times \mathbf{\vec C}}}) = \vec 0

Revision notes and key takeaways

Review VECTOR TRIPLE PRODUCT in your textbook or class notes, then use this set to check your understanding.

Showing 61-66 of 66 questions

Your Progress

Vector triple product practice · Easy

Continue Your Practice

Your practice