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

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

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

Back to VECTOR TRIPLE PRODUCT

Your Progress

0 / 80 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • VECTOR TRIPLE PRODUCT

Vector triple product practice · Mixed difficulty

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 1-30 of 80 questions

Preparing practice and saved progress…

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

View All →

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.

Revision notes and key takeaways

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

Key Takeaways

✓ This set contains 80 single-correct-answer MCQs on VECTOR TRIPLE PRODUCT.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

Showing 1-30 of 80 questions

Your Progress

Vector triple product practice · Mixed difficulty

Explore Other Chapters in MATHEMATICS

Key Takeaways

Continue Your Practice

Your practice