Let $\mathbf{b}=-\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ and $\mathbf{c}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}$. If $\mathbf{a}$ be $a$ unit vector and the scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$ has the greatest value, then a is equal to

$\frac{1}{\sqrt{3}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$

$\frac{1}{\sqrt{5}}(\sqrt{2} \hat{i}-\hat{j}-\sqrt{2} \hat{k})$

$\frac{1}{\sqrt{59}}(3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}})$

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JEE • MATHEMATICS • SCALAR TRIPLE PRODUCT

Scalar triple product practice · Mixed difficulty

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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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1 Easy

1 of 230

Understanding +4 | -1

\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0

for some non-zero vector

\vec r,

then the value of

\left[ {\vec a\vec b\vec c} \right],

View answer and explanation

Correct option: C

Option C (0) is correct because if

\vec r

is perpendicular to

\vec a

\vec b

, and

\vec c

, then

\vec a

\vec b

, and

\vec c

are coplanar, making their scalar triple product zero.

2 Easy

2 of 230

Application +4 | -1

The shortest distance between the lines

\frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}}

and

\frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3},

View answer and explanation

Correct option: A

\frac{{18}}{{\sqrt 5 }}

Option A (

\frac{{18}}{{\sqrt 5 }}

) is correct because the shortest distance between the two lines is calculated to be

\frac{{18}}{{\sqrt 5 }}

using the formula.

3 Easy

3 of 230

Analyzing +4 | -1

Let

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

and

\mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}.

Then

\left[ {\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}} \right]

depends on

View answer and explanation

Correct option: A

neither

x

nor

y

The correct answer is A (neither

x

nor

y

) because the scalar triple product simplifies to 1, which is independent of both

x

and

y

4 Easy

4 of 230

Application +4 | -1

\vec{a}, \vec{b}

are non-zero vectors, then

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

equals

View answer and explanation

Correct option: A

-(\vec{a} \cdot \vec{b})|\vec{a} \times \vec{b}|

Option D ((

\vec{a} \cdot \vec{b})|\vec{a} \times \vec{b}|^2

) is correct because expanding both triple vector products and simplifying using dot product properties results in this expression.

5 Easy

5 of 230

Understanding +4 | -1

a \cdot(a \times b)=

View answer and explanation

Correct option: C

\mathrm{O}

Option C is correct because the dot product of a vector with the cross product of itself and another vector is always zero, since a x b is perpendicular to a.

6 Medium

6 of 230

Application +4 | -1

The length of the perpendicular from the origin to the plane passing through the point a and containing the line

\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}

View answer and explanation

Correct option: C

\frac{[a \mathbf{b c}]}{|\mathbf{b} \times \mathbf{c}+\mathbf{c \times a}|}

The correct answer is

\frac{{[a \mathbf{b c}]}}{{|\mathbf{b} \times \mathbf{c}+\mathbf{c \times a}|}}

because it represents the formula for the perpendicular distance from the origin to the plane.

7 Medium

7 of 230

Application +4 | -1

Let

\mathbf{b}=-\hat{\mathbf{i}}+4 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}

and

\mathbf{c}=2 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-10 \hat{\mathbf{k}}

. If

\mathbf{a}

a

unit vector and the scalar triple product

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

has the greatest value, then a is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the scalar triple product is maximized when a is in the direction of b x c, and normalizing b x c gives the unit vector a.

8 Easy

8 of 230

Analyzing +4 | -1

If the vectors

p \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+q \hat{\mathbf{j}}+\hat{\mathbf{k}}

and

\hat{\mathbf{i}}+\hat{\mathbf{j}}+r \hat{\mathbf{k}}

(Where,

p \neq q \neq r \neq 1

) are coplanar, then the value of

p q r-(p+q+r)

View answer and explanation

Correct option: A

-2

Option A (-2) is correct because when the given vectors are coplanar, the determinant of their components is zero, leading to the result pqr - (p+q+r) = -2.

9 Easy

9 of 230

Application +4 | -1

y=\sec \left(\tan ^{-1} x\right)

, then

\left.\frac{d y}{d x}\right|_{x=1}

is equal to

View answer and explanation

Correct option: D

\frac{1}{\sqrt{2}}

Option D (

\frac{1}{\sqrt{2}}

) is correct because it is the value of the derivative at x=1 after simplification.

10 Medium

10 of 230

Application +4 | -1

The vector equation of a plane, which is at a distance of 8 unit from the origin and which is normal to the vector

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

, is

View answer and explanation

Correct option: B

\mathbf{r} \cdot(2 \mathbf{i}+\mathbf{j}+2 \mathbf{k})=24

Option B,

\mathbf{r} \cdot(2 \mathbf{i}+\mathbf{j}+2 \mathbf{k})=24

, is correct because it represents the vector equation of a plane at a distance of 8 units from the origin and normal to the given vector.

11 Easy

11 of 230

Analyzing +4 | -1

\left[\begin{array}{lll}\vec{a} \times \vec{b} & \vec{b} \times \vec{c} & \vec{c} \times \vec{a}\end{array}\right]=\lambda\left[\begin{array}{ll}\vec{a} \vec{b} \vec{c}\end{array}\right]^2

then

\lambda

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because the scalar triple product of

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

\vec{b} \times \vec{c}

, and

\vec{c} \times \vec{a}

is equal to the square of the scalar triple product of

\vec{a}

\vec{b}

, and

\vec{c}

12 Medium

12 of 230

Analyzing +4 | -1

Let

\mathbf{r}=(\mathbf{a} \times \mathbf{b}) \sin x+(\mathbf{b} \times \mathbf{c}) \cos y+(\mathbf{c} \times \mathbf{a})

, where

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

are non-zero non-coplanar vectors. If

\mathbf{r}

is orthogonal to

3 \mathbf{a}+5 \mathbf{b}+2 \mathbf{c}

, then the value of

\sec ^2 y+\operatorname{cosec}^2 x+\sec y \operatorname{cosec} x

View answer and explanation

Correct option: A

Option A (3) is correct because substituting

\sin x = -1

and

\cos y = -1

into the expression

\sec^2 y + \csc^2 x + \sec y \csc x

yields 3.

13 Easy

13 of 230

Application +4 | -1

\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}

, then the value of

\left|\begin{array}{ccc}\mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{c} \cdot \mathbf{a} & \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{c}\end{array}\right|

View answer and explanation

Correct option: C

Option C (16) is correct because the determinant can be simplified to the square of the scalar triple product of the vectors a, b, and c, which evaluates to 16.

14 Easy

14 of 230

Understanding +4 | -1

(\mathbf{r} \cdot \mathbf{i})^2+(\mathbf{r} \cdot \mathbf{j})^2+(\mathbf{r} \cdot \mathbf{k})^2=

View answer and explanation

Correct option: B

r^2

Option B (

r^2

) is correct because

(\mathbf{r} \cdot \mathbf{i})^2+(\mathbf{r} \cdot \mathbf{j})^2+(\mathbf{r} \cdot \mathbf{k})^2

simplifies to

x^2 + y^2 + z^2

, which is equal to

r^2

15 Medium

15 of 230

Application +4 | -1

The shortest distance between the lines

\mathbf{r}=(3 \mathbf{i}-2 \mathbf{j}-2 \mathbf{k})+\mathbf{i t}

and

\mathbf{r}=\mathbf{i}-\mathbf{j}+2 \mathbf{k}+\mathbf{j} s

(

t

and

s

being parameters) is

View answer and explanation

Correct option: C

Option C (4) is correct because the shortest distance between the two lines is calculated using the formula and found to be 4.

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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 230

Understanding +4 | -1

Assertion : For any three vectors

\mathrm{a}, \mathrm{b}

and

\mathrm{c}

\left[\begin{array}{lll}a & b & c\end{array}\right]=\left[\begin{array}{lll}b & c & a\end{array}\right]=\left[\begin{array}{lll}c & a & b\end{array}\right]

Reason : Cyclic permutation of three vectors does not change the value of the scalar triple product.

View answer and explanation

Correct option: A

Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.

Assertion is correct, Reason is correct; Reason is a correct explanation for assertion because the scalar triple product is invariant under cyclic permutations.

17 Easy

17 of 230

Application +4 | -1

The volume of the tetrahedron having the edges

\widehat {\mathbf{\dot i}} + 2\,\widehat {\mathbf{\dot j}} - \mathbf{\hat k},\,\;\;\widehat {\mathbf{\dot i}} + \widehat {\mathbf{\dot j}} + \mathbf{\hat k},\widehat {\,\mathbf{\dot i}} - \widehat {\mathbf{\dot j}} + \lambda \,\mathbf{\hat k}

as coterminous is

\frac{2}{3}

cu unit. Then,

\mathrm{\lambda }

equals

View answer and explanation

Correct option: A

The correct answer is A (1) because the volume of the tetrahedron is given by one-sixth of the scalar triple product of the coterminous edges, and solving for lambda gives 1.

18 Easy

18 of 230

Application +4 | -1

A parallelopiped is formed by planes drawn parallel to coordinate axes through the points A = (1, 2, 3) and B = (9, 8, 5). The volume of that parallelopiped is equal to (in cubic units)

View answer and explanation

Correct option: D

Option D (96) is correct because the volume of the parallelepiped is the product of the lengths of its coterminous edges.

19 Easy

19 of 230

Application +4 | -1

If the volume of a parallelopiped whose coterminous edges are given by the vectors

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

\vec{b}=2 \hat{i}+4 \hat{j}-n \hat{k}

and

\vec{c}=\hat{i}+n \hat{j}+3 \hat{k}(n \geq 0)

, is 158 cubic units, then

View answer and explanation

Correct option: C

\vec{b} \cdot \vec{c}=10

The correct answer is C (

\vec{b} \cdot \vec{c}=10

) because after calculating the value of n using the volume of the parallelepiped, we find that

\vec{b} \cdot \vec{c}=10

20 Medium

20 of 230

Application +4 | -1

The shortest distance between the lines

\frac{x-4}{4}=\frac{y+2}{5}=\frac{z+3}{3}

and

\frac{x-1}{3}=\frac{y-3}{4}=\frac{z-4}{2}

View answer and explanation

Correct option: D

3 \sqrt{6}

Option D (

3 \sqrt{6}

) is correct because it is the shortest distance calculated using the formula for the distance between two skew lines.

21 Easy

21 of 230

Application +4 | -1

The acute angle that the vector

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

makes with the plane determined by the vectors

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

and

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

View answer and explanation

Correct option: B

\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)

Option B (

\sin^{{-1}}(\frac{{1}}{{\sqrt{{3}}}})

) is correct because it is found by calculating the sine of the angle between the vector and the normal to the plane.

22 Easy

22 of 230

Application +4 | -1

The points with position vectors

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

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

are coplanar if

View answer and explanation

Correct option: A

(1-\alpha)(1+\beta)=0

The correct answer is A (

(1-\alpha)(1+\beta)=0

) because the points are coplanar if the scalar triple product of the vectors formed by these points is zero.

23 Easy

23 of 230

Application +4 | -1

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

is perpendicular to

{\mathbf{\vec b}}

and

\mathbf{\vec c}\left| {\mathbf{\vec a}} \right| = 2,\;\left| {\mathbf{\vec b}\left| { = 3,\;} \right|\mathbf{\vec c}} \right| = 4

and the angle between

{\mathbf{\vec b}}

and

\mathbf{\vec c}\;\mathrm{is}\frac{{2\mathrm{\pi }}}{3}

, then

\left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]

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 is calculated using the magnitudes of the vectors and the sine of the angle between them.

24 Easy

24 of 230

Analyzing +4 | -1

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

are unit coplanar vectors, then

\left[ {2\mathrm{\;}\mathbf{\vec a} - \mathbf{\vec b}\mathrm{\;}2\mathbf{\vec b}\mathrm{\;} - \mathbf{\vec c}\mathrm{\;}2\mathbf{\vec c} - \mathbf{\vec a}\mathrm{\;}} \right]

is equal to

View answer and explanation

Correct option: B

Option B (0) is correct because if vectors are coplanar, any linear combination of them will also be coplanar, resulting in a zero scalar triple product.

25 Easy

25 of 230

Analyzing +4 | -1

Two adjacent sides of a parallelogram

A B C D

are given by

\overrightarrow{A B}=2 \hat{i}+10 \hat{j}+11 \hat{k}

and

\overrightarrow{A D}=-\hat{i}+2 \hat{j}+2 \hat{k}.

The side

A D

is rotated by an acute angle

\alpha

in the plane of the parallelogram so that

A D

becomes

A D^{\prime}.

A D^{\prime}

makes a right angle with the side

A B,

then the cosine of the angle

\alpha

is given by

View answer and explanation

Correct option: B

\frac{\sqrt{17}}{9}

Option A (

\hat{j}-\hat{k}

and

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

) is correct because these vectors are coplanar with the given vectors and perpendicular to the vector

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

26 Easy

26 of 230

Analyzing +4 | -1

Let

\hat a{\text{ }}\,\,{\text{and}}\,{\text{ }}\hat b

be two vectors such that

\hat a.\,\,\hat b = \frac{1}{3}

and

\hat a \times \,\hat b = \hat c.

Also

\overline F = \alpha \hat a + \beta \hat b + \gamma \hat c,

where

\alpha ,\,\,\beta ,\,\,\gamma

are scalars. If

\alpha = {k_1}(\hat F.\,\hat a)\, - {k_2}(\hat F.\hat b),

then the value of

2({k_1} + {k_2})

View answer and explanation

Correct option: C

The correct answer is 3 because it is found by expressing F in terms of a, b, and c, then using the properties of the scalar triple product to solve for k1 and k2, and finally calculating 2(k1 + k2).

27 Easy

27 of 230

Analyzing +4 | -1

The vectors

\vec u = \left( {al + {a_1}{l_1}} \right)\hat i + \left( {am + {a_1}{m_1}} \right)\hat j + \left( {an + {a_1}{n_1}} \right)\hat k,

\vec \upsilon = \left( {bl + {b_1}{l_1}} \right)\hat i + \left( {bm + {b_1}{m_1}} \right)\hat j + \left( {bn + {b_1}{n_1}} \right)\hat k,

\vec w = \left( {cl + {c_1}{l_1}} \right)\hat i + \left( {cm + {c_1}{m_1}} \right)\hat j + \left( {cn + {c_1}{n_1}} \right)\hat k

View answer and explanation

Correct option: B

Are coplanar

The correct answer is B (Are coplanar) because the scalar triple product of the vectors is zero, indicating that they lie in the same plane.

28 Easy

28 of 230

Application +4 | -1

|a| = 1,\,\,|b| = 3{\text{ }}\,\,{\text{and}}\,\,{\text{ }}|c| = 5

then the value of

[a - b\,\,b - c\,\,c - a]

View answer and explanation

Correct option: A

Option A (0) is correct because the scalar triple product of vectors in the form [a-b b-c c-a] is always zero.

29 Easy

29 of 230

Application +4 | -1

Volume of the parallelopiped whose coterminous edges are

2 \mathbf{i}-3 \mathbf{j}+4 \mathbf{k}, \mathbf{i}+2 \mathbf{j}-2 \mathbf{k}, 3 \mathbf{i}-\mathbf{j}+\mathbf{k}

, is

View answer and explanation

Correct option: C

7 cubic unit

Option C (7 cubic unit) is correct because the volume of the parallelepiped is the absolute value of the determinant formed by the given vectors, which equals 7.

30 Easy

30 of 230

Application +4 | -1

If vectors

\vec{A}=2 i+3 j+4 \mathbf{k}, \vec{B}=i+j+5 \mathbf{k}

, and

\vec{C}

form a left handed system, then

\vec{C}

View answer and explanation

Correct option: B

-11 \mathbf{i}+6 \mathbf{j}+\mathbf{k}

Option D (-11i+6j-k) is incorrect. Option B (-11 \mathbf{i}+6 \mathbf{j}+ \mathbf{k}) is correct because for a left-handed system, the scalar triple product [A B C] must be negative.

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Key Takeaways

✓ This set contains 230 single-correct-answer MCQs on SCALAR TRIPLE PRODUCT.
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✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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