Which one of the following vectors is of magnitude 6 and perpendicular to both ${\bf{\vec a}} = 2\widehat {\bf{i}} + 2\widehat {\bf{j}} + {\bf{\hat k}}$ and ${\bf{\vec b}} = \widehat {\bf{i}} - 2\widehat {\bf{j}} + 2{\bf{\hat k}}$?

$2\left( {2\widehat {\bf{i}} - \widehat {\bf{j}} - 2{\bf{\hat k}}} \right)$

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

$2\left( {2\widehat {\bf{i}} - \widehat {\bf{j}} + 2{\bf{\hat k}}} \right)$

$3\left( {2\widehat {\bf{i}} - \widehat {\bf{j}} - 2{\bf{\hat k}}} \right)$

Back to CROSS PRODUCT

Your Progress

0 / 41 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • CROSS PRODUCT

Cross product practice · Medium

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 41 questions

Preparing practice and saved progress…

1 Medium

1 of 41

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.

2 Medium

2 of 41

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.

3 Medium

3 of 41

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.

4 Medium

4 of 41

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.

5 Medium

5 of 41

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.

6 Medium

6 of 41

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.

7 Medium

7 of 41

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}}

8 Medium

8 of 41

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}}

9 Medium

9 of 41

Analyzing +4 | -1

Let

\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{b}=\hat{i}-2 \hat{j}-2 \hat{k}

and

\vec{c}=-\hat{i}+4 \hat{j}+3 \hat{k}

. If

\vec{d}

is a vector perpendicular to both

\vec{b}

and

\vec{c}

, and

\vec{a} \cdot \vec{d}=18

, then

|\vec{a} \times \vec{d}|^{2}

is equal to

View answer and explanation

Correct option: C

720

Option C (720) is correct because it is the result of calculating the magnitude squared of the cross product of vector a and d, given the conditions.

10 Medium

10 of 41

Application +4 | -1

\vec{a}

and

\vec{b}

are two vectors such that

|\vec{a}|=1

|\vec{b}|=4, \vec{a} \cdot \vec{b}=2

. If

\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}

then angle between

\vec{b}

and

\vec{c}

is :

View answer and explanation

Correct option: D

\frac{5 \pi}{6}

Option D (

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

) is correct because the angle between the vectors

\vec{{b}}

and

\vec{{c}}

is calculated using the dot product and magnitudes of the vectors.

11 Medium

11 of 41

Application +4 | -1

Let

\vec{a}

and

\vec{b}

be two vectors such that

|\vec{b}|=1

and

|\vec{b} \times \vec{a}|=2

. Then

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

is equal to

View answer and explanation

Correct option: A

5

The correct answer is A (5) because by expanding the square and using the properties of the cross product, we can find the value of the expression.

12 Medium

12 of 41

Application +4 | -1

a \times(b \times c)+b \times(c \times a)+c \times(a \times b)=

View answer and explanation

Correct option: A

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

Option D (0) is correct because the expression simplifies to 0 due to the properties of the scalar triple product when the vectors are coplanar.

13 Medium

13 of 41

Application +4 | -1

Which one of the following vectors is of magnitude 6 and perpendicular to both

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

and

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

View answer and explanation

Correct option: D

2\left( {2\widehat \mathbf{i} - \widehat \mathbf{j} - 2\mathbf{\hat k}} \right)

Option D (

2(2\widehat{\mathbf{i}} - \widehat{\mathbf{j}} - 2\mathbf{\hat k})

) is correct because it represents a vector with magnitude 6 that is perpendicular to both given vectors, found by taking the cross product and scaling.

14 Medium

14 of 41

Analyzing +4 | -1

Let

\vec{a}=2 i-j+5 k

and

\vec{b}=\alpha i+\beta j+2 k

. If

((\vec{a} \times \vec{b}) \times i) \cdot k=\frac{23}{2}

, then

|\vec{b} \times 2 j|

is equal to

View answer and explanation

Correct option: B

Option B (5) is correct because by calculating the cross products and using the given dot product, we can find the values of

\alpha

and

\beta

, and then calculate

|\vec{b} \times 2 \vec{j}|

15 Medium

15 of 41

Analyzing +4 | -1

A non-zero vector

{\mathbf{\vec a}}

is parallel to the line of intersection of the plane determined by the vectors

\widehat \mathbf{i},\;\widehat \mathbf{i} + \widehat \mathbf{j}

and the plane determined by the vectors

\widehat \mathbf{i} - \widehat \mathbf{j},\;\widehat \mathbf{i} + \mathbf{\hat k}.

The angle between

{\mathbf{\vec a}}

and

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

View answer and explanation

Correct option: D

\frac{\pi }{4}

Option D is correct because the vector

\vec a

is parallel to the intersection of two planes, and the angle between

\vec a

and the given vector can be found using the dot product formula.

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 Medium

16 of 41

Application +4 | -1

A unit vector perpendicular to the vector

4 i-j+3 k

and

-2 i+j-2 k

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents the unit vector obtained by taking the cross product of the two given vectors and normalizing it.

17 Medium

17 of 41

Application +4 | -1

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

then

\vec{b}

is equal to

View answer and explanation

Correct option: C

\hat{\mathrm{i}}

Option C (

\hat{\mathrm{i}}

) is correct. By setting

\vec{b}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}

and using the given dot and cross product conditions, we can solve for the components of

\vec{b}

18 Medium

18 of 41

Application +4 | -1

The scalars

l

and

m

such that

l \mathbf{a}+m \mathbf{b}=\mathbf{c}

, where

\mathbf{a}, \mathbf{b}

and c are given vectors, are equal to

View answer and explanation

Correct option: A

I=\frac{(\mathbf{c} \times \mathbf{b}) .(\mathbf{a} \times \mathbf{b})}{(\mathbf{a} \times \mathbf{b})^2}, m=\frac{(\mathbf{c} \times \mathbf{a}) .(\mathbf{b} \times \mathbf{a})}{(\mathbf{b} \times \mathbf{a})^2}

Option A is correct because it provides the correct formulas for calculating scalars l and m using cross and dot products of the given vectors.

19 Medium

19 of 41

Application +4 | -1

\mathrm{A}(3,1,-1), \mathrm{B}\left(\frac{5}{3}, \frac{7}{3}, \frac{1}{3}\right), \mathrm{C}(2,2,1)

and

\mathrm{D}\left(\frac{10}{3}, \frac{2}{3}, \frac{-1}{3}\right)

are the vertices of a quadrilateral

\mathrm{ABCD}

, then its area is

View answer and explanation

Correct option: A

\frac{4 \sqrt{2}}{3}

Option A (

\frac{{4 \sqrt{{2}}}}{{3}}

) is correct because the area of the quadrilateral is calculated using the cross product of the vectors AC and BD.

20 Medium

20 of 41

Application +4 | -1

Let

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

and the angle between the vectors

\vec{a}

and

\vec{b}

\frac{\pi}{4}

. Then

|(\vec{a}+2 \vec{b}) \times(2 \vec{a}-3 \vec{b})|^{2}

is equal to

View answer and explanation

Correct option: D

882

Option D (882) is correct because after simplifying the cross product and using the given magnitudes and angle, the final result is 882.

21 Medium

21 of 41

Application +4 | -1

Let

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

and

\mathrm{\;}\mathbf{\vec c} = 4\widehat \mathbf{i} - 3\widehat \mathbf{j} + 7\mathbf{\hat k}.\;\mathrm{If}\;\mathrm{\;}\overrightarrow {\mathbf{r}\;}

is a vector such that

\mathbf{\vec r} \times \mathbf{\vec b} = \mathbf{\vec c} \times \mathbf{\vec b}\;\mathrm{and\;}\mathbf{\vec r}\cdot\mathbf{\vec a} = 0,

then value of

\mathbf{\vec r}\cdot\mathbf{\vec b}

View answer and explanation

Correct option: B

- 7

Option B (-7) is correct because solving the vector equations

\mathbf{\vec r} \times \mathbf{\vec b} = \mathbf{\vec c} \times \mathbf{\vec b}

and

\mathbf{\vec r}\cdot\mathbf{\vec a} = 0

yields

\mathbf{\vec r}\cdot\mathbf{\vec b} = -7

22 Medium

22 of 41

Application +4 | -1

\mathbf{d}=\lambda(\mathbf{a} \times \mathbf{b})+\mu(\mathbf{b} \times \mathbf{c})+v(\mathbf{c} \times \mathbf{a})

and

[\mathbf{a} \mathbf{b} \mathbf{c}]=\frac{1}{8}

, then

\lambda+\mu+v

is equal to

View answer and explanation

Correct option: A

8 d \cdot(a+b+c)

Option A (

8 d \cdot(a+b+c)

) is correct because it is derived by taking the dot product of d with c, a, and b, and then summing the results.

23 Medium

23 of 41

Application +4 | -1

P

and

Q

are points of intersection of the line

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

with X-axis and the plane

x+2z=0

respectively, then the area of the triangle

OPQ

is (where

O

is the origin)

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the area of the triangle formed by the origin and the points of intersection of the line with the X-axis and the given plane.

24 Medium

24 of 41

Application +4 | -1

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

, then the vector

\vec{c}

such that

\vec{a} \cdot \vec{c}=2

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

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it satisfies both the dot product and cross product conditions with vectors

\vec{a}

and

\vec{b}

25 Medium

25 of 41

Application +4 | -1

A unit vector perpendicular to the plane determinal by the points

{P}(1,-1,2), {Q}(2,0,-1)

and

{R}(0,2,1)

View answer and explanation

Correct option: A

\frac{2 {i}+{j}+{k}}{\sqrt{6}}

Option A (

\frac{2 {i}+{j}+{k}}{\sqrt{6}}

) is correct because it represents the unit vector perpendicular to the plane determined by the given points, found by taking the cross product of two vectors in the plane and normalizing.

26 Medium

26 of 41

Application +4 | -1

A rigid body rotates about an axis through the origin with an angular velocity

10 \sqrt{3} \mathrm{rad} / \mathrm{s}

. If

\omega

points in the direction of

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

, then the equation to the locus of the points having tangential speed

20 \mathrm{~m} / \mathrm{s}

View answer and explanation

Correct option: C

x^2+y^2+z^2-x y-y z-z x-2=0

Option C (

x^2+y^2+z^2-x y-y z-z x-2=0

) is correct because it represents the locus of points with the given tangential speed, derived from the cross product of angular velocity and position vector.

27 Medium

27 of 41

Application +4 | -1

A force

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

acts at a point

A

, whose position vector is

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

. The moment of

\mathbf{F}

about the origin is

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the result of the cross product of the position vector and the force vector.

28 Medium

28 of 41

Analyzing +4 | -1

a^2+b^2+c^2=1

where

a, b, c \in R

, then the maximum value of

(4 a-3 b)^2+(5 b-4 c)^2+(3 c-5 a)^2

View answer and explanation

Correct option: B

Option B (50) is correct. By using vectors and the Cauchy-Schwarz inequality, we can find the maximum value of the given expression to be 50.

29 Medium

29 of 41

Application +4 | -1

A unit vector perpendicular to vector

\mathbf{c}

and coplanar with vectors

a

and

b

View answer and explanation

Correct option: C

\frac{\mathbf{c} \times(\mathbf{a} \times \mathbf{b})}{|\mathbf{c} \times(\mathbf{a} \times \mathbf{b})|}

Option C (

\frac{\mathbf{c} \times(\mathbf{a} \times \mathbf{b})}{|\mathbf{c} \times(\mathbf{a} \times \mathbf{b})|}

) is correct because it represents a vector perpendicular to c and coplanar with a and b, normalized to a unit vector.

30 Medium

30 of 41

Understanding +4 | -1

Unit vector perpendicular to the plane of

\triangle A B C

with position vectors

a, b, c

of the vertices

A, B, C

View answer and explanation

Correct option: B

\frac{\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}}{2 \Delta}

Option B (

\frac{\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}}{2 \Delta}

) is correct because the unit vector perpendicular to the plane is given by the cross product of two vectors in the plane, divided by twice the area of the triangle.

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 41 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.

Showing 1-30 of 41 questions

Your Progress

Cross product practice · Medium

Explore Other Chapters in MATHEMATICS

Key Takeaways

Continue Your Practice

Your practice