Let $a = 3\hat i + 2\hat j + x\hat k{\rm{ }}\,{\rm{\,\,and}}\,\,{\rm{ }}b = \hat i - \hat j + \hat k$ for some real $x.$ Then $\left| {a \times b} \right| = r$ is possible if

$0 < r \le \sqrt {\frac{3}{2}} $

$\sqrt {\frac{3}{2}} < r \le 3\sqrt {\frac{3}{2}} $

$3\sqrt {\frac{3}{2}} < r < 5\sqrt {\frac{3}{2}} $

A unit vector perpendicular to the plane of ${\bf{\vec a}} = 2\widehat {\bf{i}} - 6\widehat {\bf{j}} - 3{\bf{\hat k}}$ and ${\bf{\vec b}} = 4\widehat {\bf{i}} + 3\widehat {\bf{j}} - {\bf{\hat k}}$ is

$\frac{{3\widehat {\bf{i}} - 2\widehat {\bf{j}} + 6{\bf{\hat k}}}}{7}$

$\frac{{4\widehat {\bf{i}} + 3\widehat {\bf{j}} - {\bf{\hat k}}}}{{\sqrt {26} }}$

$\frac{{2\widehat {\bf{i}} - 6\widehat {\bf{j}} - 3{\bf{\hat k}}}}{7}$

$\frac{{2\widehat {\bf{i}} - 3\widehat {\bf{j}} - 6{\bf{\hat k}}}}{7}$

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JEE • MATHEMATICS • CROSS PRODUCT

Cross product practice · Easy

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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 149 questions

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1 Easy

1 of 149

Understanding +4 | -1

A solution of the vector equation

\vec{r} \times \vec{b}=\vec{a} \times \vec{b}

, where

\vec{a}, \vec{b}

are two given vectors, is where

\lambda

is a parameter.

View answer and explanation

Correct option: B

\vec{r}=\vec{a}+\lambda \vec{b}

Option B (

\vec{r}=\vec{a}+\lambda \vec{b}

) is the correct solution because it satisfies the given vector equation after simplification and considering parallel vectors.

2 Easy

2 of 149

Application +4 | -1

Let

a = 3\hat i + 2\hat j + x\hat k\mathrm{ }\,\mathrm{\,\,and}\,\,\mathrm{ }b = \hat i - \hat j + \hat k

for some real

x.

Then

\left| {a \times b} \right| = r

is possible if

View answer and explanation

Correct option: D

r \ge 5\sqrt {\frac{3}{2}}

Option D (

r \ge 5\sqrt{\frac{3}{2}}

) is correct because the minimum value of

|a \times b|

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

, thus

|a \times b|

must be greater than or equal to this value.

3 Easy

3 of 149

Application +4 | -1

A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(-1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is

View answer and explanation

Correct option: C

{\cos ^{ - 1}}\left( {\frac{{19}}{{35}}} \right)

Option D (303) is correct. After simplifying the expression and using the formula for the sum of squares, the value of A is found to be 303.

4 Easy

4 of 149

Analyzing +4 | -1

Let

a = 2\widehat i + \widehat j - 2\widehat k{\text{ }}\,\,{\text{and }}\,\,b = \widehat i + \widehat j.

If c is a vector such that

a \cdot c = |c|,\,\,|c - a\mid = 2\sqrt 2

and the angle between

a \times b

and

c

30^\circ ,

then

|(a \times b) \times c|

equal to

View answer and explanation

Correct option: B

03-Feb

The correct answer is 3/2 because it is calculated by finding the magnitude of vector c, then using the cross product formula and the given angle to find the magnitude of the triple vector product.

5 Easy

5 of 149

Application +4 | -1

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

, then

(\vec{A} \times \vec{B}) \times \vec{C}

View answer and explanation

Correct option: A

5(-i+3 j+4 k)

Option A (

5(-\mathbf{i}+3 \mathbf{j}+4 \mathbf{k})

) is correct because it is the result of calculating the vector triple product

(\vec{A} \times \vec{B}) \times \vec{C}

6 Easy

6 of 149

Application +4 | -1

Angle between the line

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

and the normal to the plane

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

View answer and explanation

Correct option: A

\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)

Option A (

\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)

) is correct because it is the angle between the line and the normal to the plane, calculated using the dot product.

7 Easy

7 of 149

Application +4 | -1

Let

a = 2\hat i + \hat j - 2\hat k,b = \hat i + \hat j

and

c

be a vector such that

\left| {c - a} \right| = 3,\,\,\left| {\left( {a \times b} \right) \times c} \right| = 3

and the angle between

c

and

a \times b

30^\circ

. Then

a \cdot c

is equal to

View answer and explanation

Correct option: B

Option B (2) is correct because by using the given vector conditions and properties of dot and cross products, we can solve for the dot product of a and c.

8 Easy

8 of 149

Application +4 | -1

\mathbf{\vec u} = \widehat {\mathbf{\dot i}}\,\, \times \,\,\left( {\mathbf{\vec a} \times \widehat {\mathbf{\dot i}}} \right) + \widehat {\mathbf{\dot j}}\,\, \times \,\,\left( {\mathbf{\vec a} \times \widehat {\mathbf{\dot j}}} \right) + \mathbf{\hat k}\,\, \times \,\,\left( {\mathbf{\vec a} \times \mathbf{\hat k}} \right)

is equal

View answer and explanation

Correct option: B

2\,\mathbf{\vec a}

Option B (

2\,\mathbf{\vec a}

) is correct because the expression simplifies to

2\vec{a}

using vector triple product properties.

9 Easy

9 of 149

Application +4 | -1

If the area of the parallelogram with

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

as two adjacent side is 15 sq units, then the area of the parallelogram having

3\mathbf{\vec a} + 2\mathbf{\vec b}

and

\mathbf{\vec a} + 3\mathbf{\vec b}

as two adjacent sides in sq units is

View answer and explanation

Correct option: B

105

Option B (105) is correct because the area of the new parallelogram is 7 times the area of the original parallelogram.

10 Easy

10 of 149

Application +4 | -1

The vector

\mathbf{c}

is perpendicular to the vectors

\mathbf{a}=(2,-3,1)

\mathbf{b}=(1,-2,3)

and satisfies the condition

\mathbf{c} \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-7 \hat{\mathbf{k}})=10

. Then, the vector

\mathbf{c}

is equal to

View answer and explanation

Correct option: B

(-7,-5,-1)

Option B ((-7,-5,-1)) is correct because it is the cross product of vectors a and b, scaled to satisfy the given condition.

11 Easy

11 of 149

Application +4 | -1

Let

\quad \vec{a}=\alpha \hat{i}+3 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}-\beta \hat{j}+4 \hat{k} \quad

and

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

where

\alpha, \beta \in \mathbf{R}

, be three vectors. If the projection of

\vec{a}

\vec{c}

\frac{10}{3}

and

\vec{b} \times \vec{c}=-6 \hat{i}+10 \hat{j}+7 \hat{k}

, then the value of

\alpha+\beta

is equal to:

View answer and explanation

Correct option: A

Option A (3) is correct because it is found by using the projection formula and the cross product to solve for alpha and beta, then summing them.

12 Easy

12 of 149

Analyzing +4 | -1

Let

\vec{a}

be a non-zero vector parallel to the line of intersection of the two planes described by

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

and

\hat{i}-\hat{j}, \hat{j}-\hat{k}

. If

\theta

is the angle between the vector

\vec{a}

and the vector

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

and

\vec{a} \cdot \vec{b}=6

, then the ordered pair

(\theta,|\vec{a} \times \vec{b}|)

is equal to

View answer and explanation

Correct option: D

\left(\frac{\pi}{4}, 6\right)

Option D is correct. By finding the direction vector of the line of intersection, then using the dot product formula, we can find the angle and the magnitude of the cross product.

13 Easy

13 of 149

Application +4 | -1

|\mathbf{a}|=3,|\mathbf{b}|=1,|\mathbf{c}|=4

and

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

, then

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

View answer and explanation

Correct option: A

-13

Option A (-13) is correct because it is obtained by squaring the equation a+b+c=0 and using the given magnitudes to solve for the required dot product sum.

14 Easy

14 of 149

Application +4 | -1

A unit vector which is perpendicular to

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

and

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

View answer and explanation

Correct option: D

\frac{1}{\sqrt{5}}(2 \mathrm{i}+\mathbf{k})

Option D (

\frac{{1}}{{\sqrt{{5}}}}(2 \mathbf{{i}} + \mathbf{{k}})

) is correct because it represents a unit vector that is perpendicular to both given vectors, found by normalizing their cross product.

15 Easy

15 of 149

Application +4 | -1

The area of a parallelogram whose adjacent sides are

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

and

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

(in sq. units) is

View answer and explanation

Correct option: B

5 \sqrt{6}

Option B (

5 \sqrt{6}

) is correct because the area of the parallelogram is the magnitude of the cross product of the adjacent sides.

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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 149

Application +4 | -1

The line of intersection of the planes
r.

(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})=1

and

\vec{r}_{.}(\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})=2

, is:

View answer and explanation

Correct option: C

\frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{-7}=\frac{z}{-13}

Option C (

\frac{{x-\frac{{6}}{{13}}}}{{2}}=\frac{{y-\frac{{5}}{{13}}}}{{-7}}=\frac{{z}}{{-13}}

) is correct because it represents the line of intersection of the two planes, found by solving the system of equations and finding the direction vector.

17 Easy

17 of 149

Application +4 | -1

If the vector

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

is written as the sum of a vector

\vec{b}_1

, parallel to

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

and a vector

\vec{b}_2

, perpendicular to

\vec{a}

, then

\overrightarrow{b_1} \times \overrightarrow{b_2}

is equal to:

View answer and explanation

Correct option: B

6 \hat{i}-6 \hat{j}+\frac{9}{2} \hat{k}

Option B (

6 \hat{i}-6 \hat{j}+\frac{9}{2} \hat{k}

) is correct because it is the cross product of the vector component parallel to

\vec{a}

and the vector component perpendicular to

\vec{a}

18 Easy

18 of 149

Application +4 | -1

Consider the parallelopiped with side

\vec{a}=3 \hat{i}+2 \hat{j}+\hat{k}, \, \vec{b}=\hat{i}+\hat{j}+2 \hat{k}

and

\vec{c}=\hat{i}+3 \hat{j}+3 \hat{k}

then the angle between

\overrightarrow{{a}}

and the plane containing the face determined by

\vec{b}

and

\vec{c}

View answer and explanation

Correct option: C

\sin ^{-1} \frac{9}{14}

Option C is correct because it involves finding the angle between a vector and a plane using the dot product of the vector and the normal to the plane.

19 Easy

19 of 149

Application +4 | -1

For any two vectors

\mathbf{a}

and

\mathbf{b},(\mathbf{a} \times \mathbf{b})^2

is equal to

View answer and explanation

Correct option: C

a^2 b^2-(\mathbf{a} \cdot \mathbf{b})^2

Option C (

a^2 b^2 - (\mathbf{{a}} \cdot \mathbf{{b}})^2

) is correct because it represents the squared magnitude of the cross product in terms of the magnitudes of the vectors and their dot product.

20 Easy

20 of 149

Application +4 | -1

A unit vector perpendicular to the plane of

\mathbf{\vec a} = 2\widehat \mathbf{i} - 6\widehat \mathbf{j} - 3\mathbf{\hat k}

and

\mathbf{\vec b} = 4\widehat \mathbf{i} + 3\widehat \mathbf{j} - \mathbf{\hat k}

View answer and explanation

Correct option: C

\frac{{3\widehat \mathbf{i} - 2\widehat \mathbf{j} + 6\mathbf{\hat k}}}{7}

The correct answer is C (

\frac{{3\widehat \mathbf{i} - 2\widehat \mathbf{j} + 6\mathbf{\hat k}}}{7}

) because it represents a unit vector perpendicular to both given vectors, obtained by normalizing their cross product.

21 Easy

21 of 149

Application +4 | -1

The position vectors of the points

A, B

and

C

are

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

and

\mathbf{k}+\mathbf{i}

respectively. The vector area of the

\triangle A B C= \pm \frac{1}{2} \vec{\alpha}

where

\vec{\alpha}=

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because the vector area of the triangle is half the cross product of two vectors formed by the points.

22 Easy

22 of 149

Application +4 | -1

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

and

\mathbf{b}=6 \mathbf{i}-3 \mathbf{j}+2 \mathbf{k}

, then the value of

\mathbf{a} \times \mathbf{b}

View answer and explanation

Correct option: C

i-10 j-18 k

Option C (

i-10j-18k

) is correct because it is the result of calculating the cross product of the two given vectors.

23 Easy

23 of 149

Application +4 | -1

\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{c}=\hat{j}-\hat{k}, \vec{a} \cdot \vec{b}=3

and

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

, then

\vec{b}

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it is the vector \vec{b} that satisfies the given conditions.

24 Easy

24 of 149

Application +4 | -1

Let

\hat a

\hat b

and

\hat c

be three unit vectors such that

\hat a \times \left( {\hat b \times \hat c} \right) = \frac{{\sqrt 3 }}{2}\left( {\hat b + \hat c} \right)

. If

\hat b

is not parallel to

\hat c

, then the angle between

\hat a

and

\hat b

View answer and explanation

Correct option: D

\frac{{5\pi }}{6}

Option D (

\frac{{5\pi }}{6}

) is correct because it is the angle between

\hat a

and

\hat b

calculated from the given vector equation.

25 Easy

25 of 149

Analyzing +4 | -1

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

then

\vec{a}+\vec{b}+\vec{c}=

View answer and explanation

Correct option: C

The correct answer is C (0) because the given condition implies that the vectors are coplanar and their sum must be zero.

26 Easy

26 of 149

Analyzing +4 | -1

Let

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

and

\vec c = 2\hat i - 3\hat j + 2\hat k

Then the number of vectors

\overrightarrow b

such that

\overrightarrow b \times \overrightarrow c = \overrightarrow a

and

\left| {\overrightarrow b } \right| \in \{ 1,2,....,10\}

is:

View answer and explanation

Correct option: A

Option A (0) is correct because the dot product of the given vectors

\vec c

and

\vec a

is not zero, which contradicts the vector triple product property.

27 Easy

27 of 149

Application +4 | -1

For any vector

\mathbf{a}

, the value of

(\mathbf{a} \times \hat{\mathbf{i}})^2+(\mathbf{a} \times \hat{\mathbf{j}})^2+(\mathbf{a} \times \hat{\mathbf{k}})^2

is equal to

View answer and explanation

Correct option: B

2 \mathrm{a}^2

Option B (

2 \mathrm{a}^2

) is correct because expanding the expression and simplifying using vector identities results in

2(a_1^2 + a_2^2 + a_3^2) = 2a^2

28 Easy

28 of 149

Application +4 | -1

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

and

|\mathbf{a} \times \mathbf{b}|=4

, then the angle between

\mathbf{a}

and

\mathbf{b}

View answer and explanation

Correct option: B

\cos ^{-1} \frac{3}{5}

Option B (

\cos^{{-1}} \frac{{3}}{{5}}

) is correct because it represents the angle between the vectors, derived from the ratio of the dot product to the magnitude of the cross product.

29 Easy

29 of 149

Application +4 | -1

A vector perpendicular to the plane containing the points

A\left( {1. - 1,2} \right),\;B\left( {2,0, - 1} \right),\;C\left( {0,2,1} \right)

View answer and explanation

Correct option: B

8\widehat \mathbf{i} + 4\widehat \mathbf{j} + 4\mathbf{\hat k}

Option B (

8\widehat{\mathbf{i}} + 4\widehat{\mathbf{j}} + 4\mathbf{\hat k}

) is correct because it represents a vector perpendicular to the plane containing the points A, B, and C, found by taking the cross product of vectors formed by these points.

30 Easy

30 of 149

Application +4 | -1

The area of the parallelogram whose diagonals are the vectors

2 a-b

and

4 a-5 b

, where

a

and

b

are the unit vectors forming an angle of

45^{\circ}

, is

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because the area of the parallelogram is half the magnitude of the cross product of its diagonals, considering the angle between the unit vectors.

Revision notes and key takeaways

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

Key Takeaways

✓ This set contains 149 single-correct-answer MCQs on CROSS 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.

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