${\bf{\vec a}} \times \left[ {{\bf{\vec a}} \times \left( {\;{\bf{\vec a}} \times {\bf{\vec b}}} \right)} \right]$ is equal to

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

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

${\bf{\vec a}}\cdot\left( {\;{\bf{\vec b}} \times {\bf{\vec a}}} \right) - {\bf{\vec b}}\cdot\left( {\;{\bf{\vec a}} \times {\bf{\vec b}}} \right)$

$\left[ {{\bf{\vec a}}\cdot\left( {\;{\bf{\vec a}} \times {\bf{\vec b}}} \right)} \right]\overrightarrow {\;{\bf{a}}} $

If $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=0$, then

$|\mathbf{a}|=|\mathbf{b}| \cdot|\mathbf{c}|=1$

Let ${\bf{\vec a}} = 2\widehat {\bf{i}} + \widehat {\bf{j}} + {\bf{\hat k}},\;{\bf{\vec b}} = \widehat {\bf{i}} + 2\widehat {\bf{j}} - {\bf{\hat k}}$ and a unit vector ${{\bf{\vec c}}}$ be coplanar. If ${{\bf{\vec c}}}$ is perpendicular to ${\bf{\vec a}}\;{\rm{and\;}}{\bf{\vec c}}$ is equal to

$ \pm \frac{1}{{\sqrt 2 }}\left( { - \widehat {\bf{j}} + {\bf{\hat k}}} \right)$

$ \pm \frac{1}{{\sqrt 3 }}\left( { - \widehat {\bf{i}} - \widehat {\bf{j}} - {\bf{\hat k}}} \right)$

$ \pm \frac{1}{{\sqrt 5 }}\left( {\widehat {\bf{i}} - 2\;\widehat {\bf{j}}} \right)$

$ \pm \frac{1}{{\sqrt 3 }}\left( {\widehat {\bf{i}} - \widehat {\bf{j}} - {\bf{\hat k}}} \right)$

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

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

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

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JEE • MATHEMATICS • VECTOR TRIPLE PRODUCT

Vector triple product practice · Easy

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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 31-60 of 66 questions

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31 Easy

31 of 66

Application +4 | -1

\mathbf{\vec a} \times \left[ {\mathbf{\vec a} \times \left( {\;\mathbf{\vec a} \times \mathbf{\vec b}} \right)} \right]

is equal to

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because the vector triple product can be simplified to this expression.

32 Easy

32 of 66

Application +4 | -1

Let

\vec{a}=3 \hat{i}+\hat{j}

and

\vec{b}=\hat{i}+2 \hat{j}+\hat{k}

. Let

\vec{c}

be a vector satisfying

\vec{a} \times(\vec{b} \times \vec{c})=\vec{b}+\lambda \vec{c}

. If

\vec{b}

and

\vec{c}

are nonparallel, then the value of

\lambda

View answer and explanation

Correct option: A

-5

Option A (-5) is correct because by applying the vector triple product identity and comparing coefficients, we find that

\lambda = -5

33 Easy

33 of 66

Analyzing +4 | -1

\mathbf{a} \times(\mathbf{b} \times \mathbf{c})=0

, then

View answer and explanation

Correct option: B

\| \mathbf{c}

Option B (b

\|\|

c) is correct because

\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = 0

implies that

\mathbf{a}

is parallel to

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

\mathbf{b} \times \mathbf{c} = 0

, which means

\mathbf{b}

and

\mathbf{c}

are parallel.

34 Easy

34 of 66

Application +4 | -1

\bar{a}, \bar{b}, \bar{c}

are three unit vectors.

\bar{b}

is not parallel to

\bar{c}

and

\bar{a} \times(\bar{b} \times \bar{c})=\frac{1}{2} \bar{b}

then

(\bar{a}, \bar{b})=\ldots

;

(\bar{a}, \bar{c})=

View answer and explanation

Correct option: A

90^{\circ}, 60^{\circ}

Option A (

90^{{\circ}}, 60^{{\circ}}

) is correct because the dot product of

\bar{{a}}

and

\bar{{b}}

is 0, indicating they are orthogonal, and the dot product of

\bar{{a}}

and

\bar{{c}}

is 1/2, leading to an angle of 60 degrees.

35 Easy

35 of 66

Application +4 | -1

For any vector r ⃗, the value of

\hat i \times \left( {\vec r \times \hat i} \right) + \hat j \times \left( {\vec r \times \hat j} \right) + \hat k \times \left( {\vec r \times \hat k} \right),

View answer and explanation

Correct option: B

2\;\vec r

Option B (

2\;\vec r

) is correct because it is the simplified result of the given vector expression using vector triple product identities.

36 Easy

36 of 66

Analyzing +4 | -1

Let

\mathbf{\vec a} = 2\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k},\;\mathbf{\vec b} = \widehat \mathbf{i} + 2\widehat \mathbf{j} - \mathbf{\hat k}

and a unit vector

{\mathbf{\vec c}}

be coplanar. If

{\mathbf{\vec c}}

is perpendicular to

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

is equal to

View answer and explanation

Correct option: A

\pm \frac{1}{{\sqrt 2 }}\left( { - \widehat \mathbf{j} + \mathbf{\hat k}} \right)

The correct answer is A (

\pm \frac{1}{{\sqrt 2 }}\left( { - \widehat \mathbf{j} + \mathbf{\hat k}} \right)

) because it represents a unit vector perpendicular to

\vec a

and coplanar with

\vec a

and

\vec b

37 Easy

37 of 66

Analyzing +4 | -1

\left( {\vec a \times \vec b} \right) \times \vec c = \vec a \times \left( {\vec b \times \vec c} \right),

then

View answer and explanation

Correct option: A

\vec b \times \left( {\vec c \times \vec a} \right) = \vec 0

Option A (

\vec b \times (\vec c \times \vec a) = \vec 0

) is correct because the given condition leads to the conclusion that

\vec b \times (\vec c \times \vec a)

must be the zero vector.

38 Easy

38 of 66

Application +4 | -1

The distance of the point

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

from the plane

\mathbf{r} .(\mathbf{i}-2 \mathbf{j}+4 \mathbf{k})=9

View answer and explanation

Correct option: A

\frac{13}{\sqrt{21}}

Option A,

\frac{13}{\sqrt{21}}

, is correct because it represents the perpendicular distance of the given point from the plane, calculated using the formula.

39 Easy

39 of 66

Application +4 | -1

\theta

be the angle between the vectors

a

and

b

and

|\mathbf{a} \times \mathbf{b}|=\mathbf{a} \cdot \mathbf{b}

, then

\theta=

View answer and explanation

Correct option: C

\frac{\pi}{4}

Option C (

\frac{{\pi}}{{4}}

) is correct because when the magnitudes of the cross product and dot product are equal, the angle between the vectors is 45 degrees.

40 Easy

40 of 66

Analyzing +4 | -1

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

, where

\vec{a}, \vec{b}

and

\vec{c}

are any thee iectors such that

\vec{a} \cdot \vec{b} \neq 0, \vec{b} \cdot \vec{c} \neq 0

, then

\vec{a}

and

\vec{c}

are

View answer and explanation

Correct option: D

parallel

Option D (parallel) is correct because the given condition implies that

\vec{a}

is a scalar multiple of

\vec{c}

, hence they are parallel.

41 Easy

41 of 66

Application +4 | -1

\mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{b}=\mathbf{i}+\mathbf{j}, \mathbf{c}=\mathbf{i}

and

(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}=\lambda \mathbf{a}+\mu \mathbf{b}

, then

\lambda+\mu=

View answer and explanation

Correct option: A

Option A (0) is correct because when

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

is expressed as

\lambda \mathbf{a} + \mu \mathbf{b}

, the sum of

\lambda

and

\mu

equals 0.

42 Easy

42 of 66

Analyzing +4 | -1

Let

\mathbf{\vec a},\mathbf{\vec b}

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}{4}\left| {\mathbf{\vec b}} \right|\left| {\mathbf{\vec c}} \right|\mathbf{\vec a}.

\theta

is the acute angle between vectors

{\mathbf{\vec b}}

and

{\mathbf{\vec c}}

, then the angle between

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

is equal to

View answer and explanation

Correct option: D

\frac{\pi }{2}

Option D (

\frac{\pi}{2}

) is correct because the given condition implies that the dot product of vectors

\vec{a}

and

\vec{c}

is zero, meaning they are orthogonal.

43 Easy

43 of 66

Analyzing +4 | -1

Let

\vec{a}

and

\vec{b}

be two non-collinear unit vectors. If

\vec{u}=\vec{a}-(\vec{a} \cdot \vec{b}) \,\vec{b}

and

\vec{v}=\vec{a} \times \vec{b},

then

|\vec{v}|

View answer and explanation

Correct option: C

|\vec{u}|+|\vec{u} \cdot \vec{b}|

Option C (

|\vec{u}|+|\vec{u} \cdot \vec{b}|

) is correct because

\vec{u}

is perpendicular to

\vec{b}

, making

\vec{u} \cdot \vec{b}=0

, and thus

|\vec{v}| = |\vec{u}|+|\vec{u} \cdot \vec{b}|

44 Easy

44 of 66

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

45 Easy

45 of 66

Application +4 | -1

Given three unit vector

\mathbf{a}, \mathbf{b}, \mathbf{c}

such that

\mathbf{a} \perp \mathbf{b}

and

\mathbf{a} \| \mathbf{c}

, then

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

View answer and explanation

Correct option: B

Option B (b) is correct because, given the conditions, the expression simplifies to b using vector triple product identities and the properties of dot products.

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46 Easy

46 of 66

Application +4 | -1

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

are three non-zero vectors and

\hat{n}

is a unit vector perpendicular to

\vec{c}

such that

\vec{a}=\alpha \vec{b}-\hat{n},(\alpha \neq 0)

and

\vec{b} \cdot \vec{c}=12

, then

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

is equal to

View answer and explanation

Correct option: C

Option C (12) is correct because it is the magnitude of the vector triple product, simplified using vector identities and the given conditions.

47 Easy

47 of 66

Analyzing +4 | -1

Let

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

and

\vec{b}=\hat{i}+\hat{j}.

\vec{c}

is a vector such that

\vec{a} \cdot \vec{c}=|\vec{c}|,|\vec{c}-\vec{a}|=2 \sqrt{2}

and the angle between

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

and

\vec{c}

30^{\circ},

then

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

View answer and explanation

Correct option: B

3 / 2

Option B (

3 / 2

) is correct because after finding

|\vec{c}|=1

and

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

, we can calculate

|(\vec{a} \times \vec{b}) \times \vec{c})| = \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{3}{2}

48 Easy

48 of 66

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.

49 Easy

49 of 66

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.

50 Easy

50 of 66

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.

51 Easy

51 of 66

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.

52 Easy

52 of 66

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

53 Easy

53 of 66

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.

54 Easy

54 of 66

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.

55 Easy

55 of 66

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.

56 Easy

56 of 66

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}

57 Easy

57 of 66

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.

58 Easy

58 of 66

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

59 Easy

59 of 66

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}

60 Easy

60 of 66

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

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