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 · 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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Showing 1-30 of 195 questions

1 Easy

1 of 195

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 195

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 195

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 195

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 195

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 Medium

6 of 195

Application +4 | -1

Let

\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=12 \vec{a}+4 \vec{b}

, where

O

is the origin, If

S

is the parallelogram with adjacent sides

O A

and

O C

, then

\frac{\text { area of the quadrilateral } O A B C}{\text { area of } S}

is equal to

View answer and explanation

Correct option: D

The correct answer is D (8) because the area of the quadrilateral OABC is 8 times the area of the parallelogram S.

7 Easy

7 of 195

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.

8 Easy

8 of 195

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.

9 Easy

9 of 195

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.

10 Easy

10 of 195

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.

11 Easy

11 of 195

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.

12 Easy

12 of 195

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.

13 Easy

13 of 195

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.

14 Medium

14 of 195

Analyzing +4 | -1

Let

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

{\mathbf{\vec c}}

is a vector such that

\mathbf{\vec a}\cdot\mathbf{\vec b}\;\left| {\mathbf{\vec c}} \right|\;

and

\left| {\mathbf{\vec c} - \mathbf{\vec a}} \right| = 2\sqrt 2

and angle between

\mathbf{\vec a} \times \;\mathbf{\vec b}\mathrm{\;and}\;\mathbf{\vec c}\mathrm{\;is\;}30^\circ ,

then

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

View answer and explanation

Correct option: A

\frac{3}{2}

The correct answer is A (3/2) because, after finding the magnitude of vector c and the cross product of a and b, we use the given angle to calculate the magnitude of the final cross product.

15 Easy

15 of 195

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.

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16 Easy

16 of 195

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.

17 Medium

17 of 195

Application +4 | -1

Let

\vec{a}=-\hat{i}-\hat{j}+\hat{k}, \vec{a} \cdot \vec{b}=1

and

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

. Then

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

is equal to

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it is the result of the vector subtraction

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

after solving for vector

\vec{{b}}

using the given dot and cross product conditions.

18 Easy

18 of 195

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.

19 Easy

19 of 195

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.

20 Medium

20 of 195

Analyzing +4 | -1

Let

\triangle A B C

be a given triangle. If

|\mathbf{B A}-t \mathbf{B C}| \geq|\mathbf{A C}|

for any

t \in R

, then

\triangle A B C

View answer and explanation

Correct option: B

Right angled

Option B (Right angled) is correct because the condition

|\mathbf{B A}-t \mathbf{B C}| \geq|\mathbf{A C}|

for all t implies that angle C is 90 degrees, making the triangle right-angled.

21 Easy

21 of 195

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}

22 Easy

22 of 195

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.

23 Medium

23 of 195

Analyzing +4 | -1

A=\left[\begin{array}{lll}l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3\end{array}\right]

and

B=\left[\begin{array}{lll}p_1 & q_1 & r_1 \\ p_2 & q_2 & r_2 \\ p_3 & q_3 & r_3\end{array}\right]

, where

p_i, q_i, r_i

are the cofactors of the elements

l_i, m_i, n_i

for

i=1,2,3

. If

\left(l_1, m_1, n_1\right),\left(l_2, m_2, n_2\right)

and

\left(l_3, m_3, n_3\right)

are the direction cosines of three mutually perpendicular lines, then

\left(p_1, q_1, r_1\right),\left(p_2, q_2, r_2\right)

and

\left(p_3, q_3, r_3\right)

are

View answer and explanation

Correct option: A

the direction cosines of three mutually perpendicular lines

The correct answer is A (

\left{\theta \in(0,2 \pi): \frac{{1+2 i \sin \theta}}{{1-i \sin \theta}} \text { is purely imaginary}\right}.

) because the sum of the elements in A is 4

\pi

after finding the values of

\theta

that make the expression purely imaginary.

24 Medium

24 of 195

Analyzing +4 | -1

Let

\vec{u}=\hat{i}-\hat{j}-2 \hat{k}, \vec{v}=2 \hat{i}+\hat{j}-\hat{k}, \vec{v} \cdot \vec{w}=2

and

\vec{v} \times \vec{w}=\vec{u}+\lambda \vec{v}

. Then

\vec{u} \cdot \vec{w}

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because it is derived by using the properties of dot and cross products of vectors, and solving for the unknown dot product.

25 Easy

25 of 195

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.

26 Medium

26 of 195

Application +4 | -1

If the position vector of the points

A, B, C, D

are

3 \hat{i}-2 \hat{j}-\hat{k}, 2 \hat{i}+3 \hat{j}-4 \hat{k},-\hat{i}+\hat{j}+2 \hat{k}, 4 \hat{i}+5 \hat{j}+\lambda \hat{k}

respectively. If the points A, B, C, D lie on a plane then the value of

\lambda

equals

View answer and explanation

Correct option: A

\frac{-146}{17}

Option A (

\frac{{-146}}{{17}}

) is correct because when points A, B, C, and D lie on a plane, the scalar triple product of the vectors formed by these points is zero, leading to

\lambda = \frac{{-146}}{{17}}

27 Easy

27 of 195

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.

28 Easy

28 of 195

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.

29 Easy

29 of 195

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.

30 Medium

30 of 195

Application +4 | -1

\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}

are such that

[\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}}]=1, \overrightarrow{\mathrm{c}}=\lambda(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})

\overrightarrow{\mathrm{a}}{ }^{\wedge} \overrightarrow{\mathrm{b}}<\frac{2 \pi}{3}

, and

|\overrightarrow{\mathrm{a}}|=\sqrt{2},|\overrightarrow{\mathrm{b}}|=\sqrt{3},|\overrightarrow{\mathrm{c}}|=\frac{1}{\sqrt{3}}

, then the angle between

\vec{a}

and

\vec{b}

View answer and explanation

Correct option: B

\frac{\pi}{4}

Option B (

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

) is correct because given the conditions on the scalar triple product, the magnitudes of the vectors, and the angle between

\vec{{a}}

and

\vec{{b}}

, the angle can be calculated as

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

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 195 single-correct-answer MCQs on CROSS PRODUCT.
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