Let ${{\bf{\hat u}}}$ and ${{\bf{\hat v}}}$ are unit vectors such that ${\bf{\hat u}}\cdot{\bf{\hat v}} = 0$ If ${{\bf{\hat r}}}$ is any vector coplanar with ${{\bf{\hat u}}}$ and ${{\bf{\hat v}}}$, then the magnitude of the vector ${\bf{\vec r}} \times \left( {{\bf{\hat u}} \times {\bf{\hat v}}} \right)$ is

$2\left| {{\bf{\vec r}}} \right|$

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

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

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

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

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

If ${{\bf{\vec a}}}$ and ${{\bf{\vec b}}}$ are unit vectors, then the vectors $\left( {{\bf{\vec a}} + {\bf{\vec b}}} \right) \times \left( {{\bf{\vec a}} \times {\bf{\vec b}}} \right)$ is parallel to the vector

${\bf{\vec a}} - {\bf{\vec b}}$

${\bf{\vec a}} + {\bf{\vec b}}$

$2\;{\bf{\vec a}} - {\bf{\vec b}}$

$2\;{\bf{\vec a}} + {\bf{\vec b}}$

IF ${\bf{\vec a}} \times \left( {{\bf{\vec b}} \times {\bf{\vec c}}} \right) = \left( {{\bf{\vec a}} \times {\bf{\vec b}}} \right) \times {\bf{\vec c}}$, where ${{\bf{\vec a}}}$, ${{\bf{\vec b}}}$ and ${{\bf{\vec c}}}$ are any three vectors such that ${\bf{\vec a}}.{\bf{\vec b}} \ne 0,\;\;\;{\bf{\vec b}}\cdot{\bf{\vec c}} \ne 0,$ then ${\bf{\vec a}}{\rm{\;and\;}}{\bf{\vec c}}$ are

inclined at angle of $\frac{{\rm{\pi }}}{6}{\rm{\;\;}}$ between them

inclined at an angle of $\frac{{\rm{\pi }}}{3}$ between them

Let ${\bf{\vec a}} = \widehat {\bf{j}} - {\bf{\hat k}}$ and ${\bf{\vec c}} = \widehat {\bf{i}} - \widehat {\bf{j}} - {\bf{\hat k}}$. Then the vectors ${{\bf{\vec b}}}$ satisfying ${\bf{\vec a}} \times {\bf{\vec b}} + {\bf{\vec c}} = \vec 0$ and ${\bf{\vec a}} \cdot {\bf{\vec b}} = 3$ is

$ - \widehat {\bf{i}} + \widehat {\bf{j}} - 2{\bf{\hat k}}$

$2\widehat {\bf{i}} - \widehat {\bf{j}} + 2{\bf{\hat k}}$

$\widehat {\bf{i}} - \widehat {\bf{j}} - 2{\bf{\hat k}}$

$\widehat {\bf{i}} + \widehat {\bf{j}} - 2{\bf{\hat k}}$

If ${\bf{\vec a}} = \left( {\widehat {\bf{i}} + \widehat {\bf{j}} + {\bf{\hat k}}} \right),\;{\bf{\vec a}}\cdot{\bf{\vec b}} = 1\;$ and ${\bf{\vec a}} \times {\bf{\vec b}} = \widehat {\bf{j}} - {\bf{\hat k}},$ then ${{\bf{\vec b}}}$ is

$\widehat {\bf{i}} - \widehat {\bf{j}} + {\bf{\hat k}}$

$2\widehat {\bf{j}} - {\bf{\hat k}}$

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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 1-30 of 80 questions

1 Easy

1 of 80

Application +4 | -1

Let

{\mathbf{\hat u}}

and

{\mathbf{\hat v}}

are unit vectors such that

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

{\mathbf{\hat r}}

is any vector coplanar with

{\mathbf{\hat u}}

and

{\mathbf{\hat v}}

, then the magnitude of the vector

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

View answer and explanation

Correct option: C

\left| {\mathbf{\vec r}} \right|

Option C (

\left| {{\bf{{\vec r}}}} \right|

) is correct because the magnitude of the vector product simplifies to the magnitude of the original vector

{\bf{{\vec r}}}

2 Medium

2 of 80

Analyzing +4 | -1

Let

\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}

then

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

is equal to

View answer and explanation

Correct option: C

140

Option C (140) is correct because it is obtained by simplifying the vector expression, finding the value of lambda, and then calculating the magnitude squared.

3 Medium

3 of 80

Analyzing +4 | -1

Let

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

such that

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

and

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

then

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

equals

View answer and explanation

Correct option: A

3(\hat{i}+\hat{j}+\hat{k})

Option A (3(\hat{i}+\hat{j}+\hat{k})) is correct because it is the result of the vector operation and simplification as shown in the solution.

4 Easy

4 of 80

Analyzing +4 | -1

Let

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

be any three vectors. Then, vectors

\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)

are such that they are

View answer and explanation

Correct option: C

coplanar

Option C (coplanar) is correct because the sum of the given vectors is zero, implying they lie in the same plane.

5 Easy

5 of 80

Application +4 | -1

a

and

b

are perpendicular, then

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

is equal to

View answer and explanation

Correct option: D

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

The correct answer is D (

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

) because it correctly applies the vector triple product and uses the perpendicularity condition.

6 Easy

6 of 80

Application +4 | -1

For any vector

\vec{a}

, the value of

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

, is

View answer and explanation

Correct option: B

2 \vec{a}

Option B (

2 \vec{a}

) is correct because the expression simplifies to

2 \vec{a}

using vector triple product identities and dot product properties.

7 Easy

7 of 80

Application +4 | -1

Let

{a},{b}, {c}

be the three vectors such that

{a} \cdot({b}+{c})={b} \cdot({c}+{a})={c} \cdot({a}+{b})=0

and

|{a}|=1,|{b}|=4

|{c}|=8

, then

|{a}+{b}+{c}|=

View answer and explanation

Correct option: C

Option C (9) is correct because the magnitude of the sum of the vectors is calculated using the given dot product conditions and magnitudes.

8 Easy

8 of 80

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 is correct because it uses the vector triple product identity correctly and simplifies the expression.

9 Easy

9 of 80

Application +4 | -1

\mathbf{a} \times[\mathbf{a} \times(\mathbf{a} \times \mathbf{b})]

is equal to

View answer and explanation

Correct option: D

(a . a)(b \times a)

Option D (

(a . a)(b \times a)

) is correct because by applying the vector triple product formula twice and simplifying, the expression

\mathbf{a} \times[\mathbf{a} \times(\mathbf{a} \times \mathbf{b})]

reduces to

(a . a)(b \times a)

10 Easy

10 of 80

Application +4 | -1

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

are unit vectors, then the vectors

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

is parallel to the vector

View answer and explanation

Correct option: A

\mathbf{\vec a} - \mathbf{\vec b}

Option A (

\mathbf{\vec a} - \mathbf{\vec b}

) is correct because expanding the vector expression and using the properties of unit vectors simplifies the result to a vector parallel to

\mathbf{\vec a} - \mathbf{\vec b}

11 Medium

11 of 80

Application +4 | -1

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

are coplanar vectors, then

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

View answer and explanation

Correct option: D

Option D (0) is correct because if vectors

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

are coplanar, the scalar triple products

[\,\mathbf{a} \,\mathbf{b} \,\mathbf{d}

and

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

are zero, leading to

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

12 Easy

12 of 80

Understanding +4 | -1

\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}

, then which relation is correct

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because when

\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}

, the cross products between pairs of vectors are equal.

13 Easy

13 of 80

Application +4 | -1

Let

\vec a = \hat i + \hat j - \hat k,\vec b = \hat i - \hat j + \hat k

and

{\vec c}

be a unit vector perpendicular to

{\vec a}

and coplanar with

{\vec a}

and

\vec b,

then it is given by

View answer and explanation

Correct option: A

\frac{1}{{\sqrt 6 }}\left( {2\hat i - \hat j + \hat k} \right)

Option A is correct because the unit vector perpendicular to

\vec{a}

and coplanar with

\vec{a}

and

\vec{b}

is found using the vector triple product and normalization.

14 Easy

14 of 80

Application +4 | -1

Let

P, Q, R

be points with position vectors

{r}_1=3 {i}-2 {j}-{k}, {r}_2={i}+3 {j}+4{k}

and

{r}_3=2 {i}+{j}-2 {k}

relative to an origin 0. The distance of

P

from the plane

O Q R

is (magnitude)

View answer and explanation

Correct option: B

Option B (3) is correct because the distance of point P from the plane OQR is calculated using the formula involving the scalar triple product and equals 3.

15 Easy

15 of 80

Analyzing +4 | -1

(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})

is not equal to

View answer and explanation

Correct option: B

\{(\vec{a} \times \vec{b}) \times \vec{c}\} \,\vec{d}

Option B (

\{(\vec{a} \times \vec{b}) \times \vec{c}\} \cdot \vec{d}

) is correct because it is not generally equal to the other expressions, which are equivalent representations of the scalar product of two cross products.

Explore Other Chapters in MATHEMATICS

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SEQUENCE OF SERIES 1764 MCQs DEFINITE INTEGRATION 1443 MCQs MATRICES 1363 MCQs INDEFINITE INTEGRATION 1179 MCQs PROBABILITY 1146 MCQs COMPLEX NUMBERS 980 MCQs

16 Easy

16 of 80

Application +4 | -1

\mathbf{\vec a} = \widehat \mathbf{i} + 2\widehat \mathbf{j} + 3\mathbf{\hat k},\;

and

\mathrm{\;}\mathbf{\vec b} = \widehat \mathbf{i} \times \left( {\mathbf{\vec a} \times \widehat \mathbf{i}} \right) + \widehat \mathbf{j} \times \left( {\mathbf{\vec a} \times \widehat \mathbf{j}} \right) + \mathbf{\hat k} \times \left( {\mathbf{\vec a} \times \mathbf{\hat k}} \right)

Then length of

{\mathbf{\vec b}}

is equal to

View answer and explanation

Correct option: D

2\sqrt {14}

Option D (

2\sqrt {14}

) is correct because after simplifying the expression for

\mathbf{\vec b}

using vector triple product identities, its magnitude is found to be

2\sqrt{14}

17 Medium

17 of 80

Analyzing +4 | -1

Let

a = \widehat i - \widehat j,\,\,\,b = \widehat j - \widehat k

and

c = \widehat k - \widehat i.

If d is a unit vectors such that

a \cdot d = 0 = (bcd),

then d is equal to

View answer and explanation

Correct option: A

\pm \frac{{(\widehat i + \widehat j - 2\widehat k)}}{{\sqrt 6 }}

The correct answer is A (

\pm \frac{{(\widehat i + \widehat j - 2\widehat k)}}{{{\sqrt 6 }}}

) because it is derived by finding a vector perpendicular to a and coplanar with b and c, then normalizing it to obtain the unit vector d.

18 Medium

18 of 80

Understanding +4 | -1

The value of

[\mathbf{a}-\mathbf{b} \mathbf{b}-\mathbf{c} \mathbf{c}-\mathbf{a}]

, where

|\mathbf{a}|=1,|\mathbf{b}|=5

and

|c|=3

View answer and explanation

Correct option: A

\mathrm{O}

Option A is correct because the scalar triple product of (a-b), (b-c), and (c-a) is always zero.

19 Easy

19 of 80

Analyzing +4 | -1

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

, where

{\mathbf{\vec a}}

{\mathbf{\vec b}}

and

{\mathbf{\vec c}}

are any three vectors such that

\mathbf{\vec a}.\mathbf{\vec b} \ne 0,\;\;\;\mathbf{\vec b}\cdot\mathbf{\vec c} \ne 0,

then

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

are

View answer and explanation

Correct option: C

Parallel

Option C (Parallel) is correct because the given condition implies that vectors

\mathbf{\vec a}

and

\mathbf{\vec c}

are parallel.

20 Medium

20 of 80

Application +4 | -1

\vec{a}

is perpendicular to both

\vec{b}

and

\vec{c}

such that

|\vec{a}|=2,|\vec{b}|=3,|\vec{c}|=4

and the angle between

\vec{b}

and

\vec{c}

\frac{2 \pi}{3}

, then

|[\vec{a} \vec{b} \vec{c}]|

is equal to

View answer and explanation

Correct option: C

12 \sqrt{3}

Option C (

12 \sqrt{3}

) is correct because the scalar triple product

|[\vec{a} \vec{b} \vec{c}]|

is equal to

|\vec{a}| |\vec{b}| |\vec{c}| \sin \theta

, where

\theta

is the angle between

\vec{b}

and

\vec{c}

, given that

\vec{a}

is perpendicular to both

\vec{b}

and

\vec{c}

21 Medium

21 of 80

Analyzing +4 | -1

Let

\vec{a}, \vec{b}

and

\vec{c}

be three non zero vectors such that

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

and

\vec{a} \times(\vec{b} \times \vec{c})=\frac{\vec{b}-\vec{c}}{2}

. If

\vec{d}

be a vector such that

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

, then

(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})

is equal to

View answer and explanation

Correct option: A

\frac{1}{4}

Option A (

\frac{1}{4}

) is correct because by using vector triple product and given conditions, we can find the value of the expression.

22 Easy

22 of 80

Application +4 | -1

Let

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

and

\mathbf{\vec c} = \widehat \mathbf{i} - \widehat \mathbf{j} - \mathbf{\hat k}

. Then the vectors

{\mathbf{\vec b}}

satisfying

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

and

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

View answer and explanation

Correct option: A

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

Option A (-i + j - 2k) is correct because it satisfies both the cross product and dot product conditions given in the problem.

23 Easy

23 of 80

Application +4 | -1

Let

\vec{a}, \vec{b}

and

\vec{c}

be three vectors having magnitudes 1,1 and 2 respectively such that

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

then the acute angle between

\vec{a}

and

\vec{c}

View answer and explanation

Correct option: C

\frac{\pi}{6}

Option C (

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

) is correct because by using the vector triple product identity and equating magnitudes, we find that the angle between

\vec{{a}}

and

\vec{{c}}

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

24 Easy

24 of 80

Application +4 | -1

Let a, b and c be non-zero vectors such that

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

\theta

is the acute angle between the vector b and c, then

\sin \theta

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is the value of

\sin \theta

derived from the given vector equation and the condition that

\theta

is an acute angle.

25 Easy

25 of 80

Analyzing +4 | -1

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

and

\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right] \neq 0

, then

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

is equal to

View answer and explanation

Correct option: A

\overrightarrow{0}

Option A (

\overrightarrow{0}

) is correct because the given condition and the non-coplanarity imply that

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

must be the zero vector.

26 Medium

26 of 80

Application +4 | -1

\mathbf{\vec a} = \left( {\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}} \right),\;\mathbf{\vec a}\cdot\mathbf{\vec b} = 1\;

and

\mathbf{\vec a} \times \mathbf{\vec b} = \widehat \mathbf{j} - \mathbf{\hat k},

then

{\mathbf{\vec b}}

View answer and explanation

Correct option: C

\widehat \mathbf{i}

Option C (

\widehat \mathbf{i}

) is the correct answer because it is derived by using the vector triple product identity and solving for

\mathbf{\vec b}

27 Medium

27 of 80

Application +4 | -1

A vector

c

perpendicular to the vectors

2 {i}+3 {j}-{k}

and

{i}-2 {j}+3 {k}

satisfying

{c}(2 {i}-{j}+{k})=-6

View answer and explanation

Correct option: C

-3 {i}+3 {j}+3 {k}

Option C (-3i + 3j + 3k) is correct because it is perpendicular to both given vectors and satisfies the dot product condition.

28 Easy

28 of 80

Application +4 | -1

Let a, b and c be three vectors mutually perpendicular to each other and have same magnitude. If a vector r satisfies.

a \times \{ (r - b) \times a\} + b \times \{ (r - c) \times b\} + c \times \{ (r - a) \times c\} = 0,

then r is equal to

View answer and explanation

Correct option: C

\frac{1}{2}(a + b + c)

Option C (

\frac{1}{2}(a + b + c)

) is correct because it is the solution to the given vector equation after applying vector triple product and simplifying.

29 Medium

29 of 80

Application +4 | -1

\vec{a}=4 \hat{i}+6 \hat{j}

and

\vec{b}=3 \hat{j}+4 \hat{k}

, the component of

\vec{a}

along

\vec{b}

\mathrm{K}

then the value of

[\mathrm{K}]

(where [.] represents G.I.F.)

View answer and explanation

Correct option: A

Option A (3) is correct because the component of vector a along vector b is calculated using the formula (a.b)/|b| and then taking the greatest integer function.

30 Medium

30 of 80

Application +4 | -1

Given

\vec{A}=2 \hat{i}+3 \hat{j}+6 \hat{k}, \vec{B}=\hat{i}+\hat{j}-2 \hat{k}

and

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

. Compute the value of

|\vec{A} \times[\vec{A} \times(\vec{A} \times \vec{B})] \cdot \vec{C}|

View answer and explanation

Correct option: A

343

Option A (343) is correct because the scalar triple product simplifies to

|-|\vec{{A}}|^2[\vec{{A}} \vec{{B}} \vec{{C}}] |

and the calculation yields 343.

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