If ${{\bf{\vec b}}}$ is a unit vector, then $\left( {{\bf{\vec a}}\cdot{\bf{\vec b}}} \right){\bf{\vec b}} + {\bf{\vec b}} \times \left( {{\bf{\vec a}} \times {\bf{\vec b}}} \right)$ is

$|{\bf{\vec a}}{|^2}{\bf{\vec b}}$

$\left| {{\bf{\vec a}}\cdot{\bf{\vec b}}} \right|{\bf{\vec a}}$

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

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

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

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

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JEE • MATHEMATICS • VECTOR TRIPLE PRODUCT

Vector triple product practice · Mixed difficulty

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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.

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Showing 61-80 of 80 questions

61 Easy

61 of 80

Application +4 | -1

{\mathbf{\vec b}}

is a unit vector, then

\left( {\mathbf{\vec a}\cdot\mathbf{\vec b}} \right)\mathbf{\vec b} + \mathbf{\vec b} \times \left( {\mathbf{\vec a} \times \mathbf{\vec b}} \right)

View answer and explanation

Correct option: C

{\mathbf{\vec a}}

Option C (

{\mathbf{\vec a}}

) is correct because the expression simplifies to

\vec a

using the vector triple product identity and the fact that

\vec b

is a unit vector.

62 Easy

62 of 80

Application +4 | -1

\vec a \times \left( {\vec a \times \left( {\vec a \times \vec b} \right)} \right)

equals

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it correctly applies the vector triple product identity twice.

63 Easy

63 of 80

Application +4 | -1

\vec{a}

and

\vec{b}

are perpendicular vectors, then

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

is equal to

View answer and explanation

Correct option: C

|\vec{a}|^4 \vec{b}

Option C (

|\vec{a}|^4 \vec{b}

) is correct because repeated application of the vector triple product, considering the perpendicularity of

\vec{a}

and

\vec{b}

, leads to this result.

64 Easy

64 of 80

Analyzing +4 | -1

Let

a = \hat i + \hat j + 2\hat k

and

b = - \hat i + 2\hat j + 3\hat k

. Then the vector product

(a + b) \times [(a \times \{ (a - b) \times b)\} \times b]\,

is equal to

View answer and explanation

Correct option: B

7\,(34\hat i - 5\hat j + 3\hat k)

Option B (

7(34\hat i - 5\hat j + 3\hat k)

) is correct because it is the result of simplifying the vector triple product and substituting the given vectors.

65 Easy

65 of 80

Application +4 | -1

\vec{a}=-\hat{i}+\hat{j}+\hat{k}, \hat{b}=2 \hat{i}+0 \hat{j}+\hat{k}

, then

a

vector

\vec{X}

satisfying the conditions:
(i) that it is coplanar with

\vec{a}

and

\vec{b}

(ii) that is perpendicular to

\vec{b}

,

(iii) that

\vec{a} \cdot \vec{X}=7

, is

View answer and explanation

Correct option: B

\frac{1}{2}(-3 \hat{i}+5 \hat{j}+6 \hat{k})

Option B (

\frac{1}{2}(-3 \hat{i}+5 \hat{j}+6 \hat{k})

) is correct because it satisfies all the given conditions: coplanarity with

\vec{a}

and

\vec{b}

, perpendicularity to

\vec{b}

, and

\vec{a} \cdot \vec{X}=7

66 Easy

66 of 80

Understanding +4 | -1

If a vector

\alpha

lie in the plane

\beta

and

\gamma

then which is correct

View answer and explanation

Correct option: A

[\alpha \beta \gamma]=0

Option A ([

\alpha \beta \gamma

]=0) is correct because if a vector lies in the plane of two other vectors, then the scalar triple product of the three vectors is zero.

67 Easy

67 of 80

Analyzing +4 | -1

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

View answer and explanation

Correct option: C

\vec{a} \times \vec{b}

Option C (

\vec{a} \times \vec{b}

) is correct because the given expression simplifies to the vector cross product using the properties of scalar triple product and vector representation.

68 Easy

68 of 80

Analyzing +4 | -1

Let a, b and c be three unit vectors, out of which vectors b and c are non-parallel. If

\alpha

and

\beta

are the angles which vector a makes with vectors b and c respectively and

a \times \left( {b \times c} \right) = \frac{1}{2}b,

then

\left| {\alpha - \beta } \right|

is equal to

View answer and explanation

Correct option: A

30^\circ

Option A (

30^\circ

) is correct because by using the vector triple product and comparing coefficients, we can find the angles

\alpha

and

\beta

, and then calculate the difference.

69 Easy

69 of 80

Application +4 | -1

\mathbf{u}=\hat{\mathbf{i}} \times(\mathbf{a} \times \hat{\mathbf{i}})+\hat{\mathbf{j}} \times(\mathbf{a} \times \hat{\mathbf{j}})+\hat{\mathbf{k}} \times(\mathbf{a} \times \hat{\mathbf{k}})

, then

View answer and explanation

Correct option: C

\mathbf{u}=2 \mathrm{a}

Option C (

\mathbf{u}=2 \mathrm{a}

) is correct because using the vector triple product identity simplifies the expression to

2\mathbf{a}

70 Easy

70 of 80

Application +4 | -1

Let

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

be the three vectors having magnitudes 1 , 5 and 3 respectively such that the angle between

\mathbf{a}

and

\mathbf{b}

\theta

and

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

, then

\tan \theta

is equal to

View answer and explanation

Correct option: D

\frac{3}{4}

Option D (

\frac{3}{4}

) is correct because it is the value of

\tan \theta

calculated from the given vector equation and magnitudes.

71 Easy

71 of 80

Application +4 | -1

If the vectors

{\vec a}

and

{\vec b}

are mutually perpendicular, then

\vec a \times \left\{ {\vec a \times \left\{ {\vec a \times \left( {\vec a \times \vec b} \right)} \right\}} \right\}

is equal to

View answer and explanation

Correct option: C

{\left| {\vec a} \right|^4}\vec b

Option C (

{\left| {\vec a} \right|^4}\vec b

) is correct because repeated application of the vector triple product identity, along with the perpendicularity of

\vec{a}

and

\vec{b}

, simplifies the expression to

{\left| {\vec a} \right|^4}\vec b

72 Easy

72 of 80

Understanding +4 | -1

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

is coplanar with

View answer and explanation

Correct option: A

b and c

The correct answer is A (b and c) because the cross product

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

results in a vector that lies in the plane defined by vectors

\mathbf{b}

and

\mathbf{c}

73 Medium

73 of 80

Analyzing +4 | -1

\bar{u}, \bar{m}, \bar{r}

be three mutually perpendicular vectors with same magnitude. If

\bar{e}

satisfies the relation

\bar{u} \times\{(\bar{e}-\bar{m}) \times \bar{u}\}+\bar{m} \times\{(\bar{e}-\bar{r}) \times \bar{m}\}+\bar{r} \times\{(\bar{e}-\bar{u}) \times \bar{r}\}=\overline{0}

then

\bar{e}=

View answer and explanation

Correct option: B

\frac{1}{2}(\bar{u}+\bar{m}+\bar{r})

Option B (

\frac{1}{2}(\bar{u}+\bar{m}+\bar{r})

) is correct because it satisfies the given vector equation after simplification using vector triple product.

74 Easy

74 of 80

Application +4 | -1

Let

\mathbf{\vec a}\cdot\mathbf{\vec b}\;\mathrm{and\;}\mathbf{\vec c}

be non-zero vectors such that

\left( {\mathbf{\vec a} \times \mathbf{\vec b}} \right) \times \mathbf{\vec c} = \frac{1}{3}\left| {\mathbf{\vec b}\left| \; \right|\mathbf{\vec c}} \right|\mathbf{\vec a}

\mathrm{\theta }

is the acute angle between the vectors

{\mathbf{\vec b}}

and

{\mathbf{\vec c}}

then

\sin \mathrm{\theta }\;

equals

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it is the result of using the vector triple product identity and comparing coefficients to find

\cos \theta

, then calculating

\sin \theta

75 Easy

75 of 80

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}

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76 Easy

76 of 80

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}

77 Easy

77 of 80

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.

78 Easy

78 of 80

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}

79 Easy

79 of 80

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

80 Easy

80 of 80

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

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