${\rm{The\;value\;of\;}}\frac{{{{\left( {{\bf{\vec a}} \times {\bf{\vec b}}} \right)}^2} + {{\left( {{\bf{\vec a}} \cdot {\bf{\vec b}}} \right)}^2}}}{{2{{{\bf{\vec a}}}^2}{{{\bf{\vec b}}}^2}}}{\rm{is\;}}$

${\bf{\vec a}} \cdot {\bf{\vec b}}$

$\left( {{\bf{\vec x}} - {\bf{\vec y}}} \right) \times \left( {{\bf{\vec x}} + {\bf{\vec y}}} \right) = $……. where ${\bf{\vec x}},{\bf{\vec y}}\; \in {\rm{\;}}{R^3}$

$2\left( {{\bf{\vec x}} \times {\bf{\vec y}}} \right)$

$\left| {{\bf{\vec x}}{|^2} - } \right|{\bf{\vec y}}{|^2}$

$\frac{1}{2}\left( {{\bf{\vec x}} \times {\bf{\vec y}}} \right)$

If $2{\bf{\vec a}} + 3{\bf{\vec b}} + {\bf{\vec c}} = \vec 0$, then ${\bf{\vec a}} \times {\bf{\vec b}} + {\bf{\vec b}} \times {\bf{\vec c}} + {\bf{\vec c}} \times {\bf{\vec a}}$ is equal to

$3\left( {{\bf{\vec b}} \times {\bf{\vec c}}} \right)$

$6\left( {{\bf{\vec b}} \times {\bf{\vec c}}} \right)$

$2\left( {{\bf{\vec b}} \times {\bf{\vec c}}} \right)$

The area of a parallelogram whose adjacent sides are given by the vectors ${\bf{\hat i}} + 2{\bf{\hat j}} + 3{\bf{\hat k}}$ and $ - 3{\bf{\hat i}} - 2{\bf{\hat j}} + {\bf{\hat k}}$ (in sq unit), is

$\sqrt {180} \;{\rm{sq\;unit}}$

$\sqrt {140} \;{\rm{sq\;unit}}$

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JEE • MATHEMATICS • CROSS PRODUCT

Cross product practice · Mixed difficulty

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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 91-120 of 195 questions

91 Easy

91 of 195

Application +4 | -1

The value of

m

for which straight line

3 x-2 y+z+3=0=4 x-3 y+4 z+1

is parallel to the plane

2 x-y+m z-2=0

View answer and explanation

Correct option: A

-2

Option A (-2) is correct because it satisfies the condition for the line to be parallel to the plane, which involves the scalar triple product being zero.

92 Easy

92 of 195

Analyzing +4 | -1

\bar{a}

and

\bar{b}

are two non-collinear unit vectors such that

|\bar{a} \times \bar{b}|=\left|\frac{\bar{a}-\bar{b}}{2}\right|

, then the value of

|\bar{a} \times(\bar{a} \times \bar{b})|^2

View answer and explanation

Correct option: A

\frac{3}{4}

Option A (

\frac{3}{4}

) is correct because by using the given condition and vector triple product, the expression simplifies to

\frac{3}{4}

93 Easy

93 of 195

Analyzing +4 | -1

\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c} \neq 0

, then the correct statement is

View answer and explanation

Correct option: C

(a+c)|| b$

Option C ((a+c)|| b) is correct because the given condition implies that the cross product of (a+c) and b is zero, which means they are parallel.

94 Easy

94 of 195

Application +4 | -1

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

are unit vectors such that

\left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec a} \times \mathbf{\vec b}} \right] = \frac{1}{4}

, then angle between

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

View answer and explanation

Correct option: C

\frac{\mathrm{\pi }}{6}

Option C (

\frac{\mathrm{\pi }}{6}

) is correct because the angle is found by simplifying the scalar triple product and using the unit vector property.

95 Easy

95 of 195

Application +4 | -1

A vector of length

\sqrt{7}

which is perpendicular to

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

and

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

and makes an obtuse angle with

y

-axis is

View answer and explanation

Correct option: B

(1 / \sqrt{3})(4 \mathbf{i}-\mathbf{j}-2 \mathbf{k})

Option B (

(1 / \sqrt{3})(4 \mathbf{i}-\mathbf{j}-2 \mathbf{k})

) is correct because it satisfies the conditions of being perpendicular to the given vectors and making an obtuse angle with the y-axis.

96 Medium

96 of 195

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.

97 Easy

97 of 195

Application +4 | -1

If vertices of a triangle are

A(1,-1,2), B(2,0,-1)

and

C(0,2,1)

, then the area of a triangle is

View answer and explanation

Correct option: B

2 \sqrt{6}

Option B (

2 \sqrt{{6}}

) is correct because it is half the magnitude of the cross product of vectors AB and AC.

98 Easy

98 of 195

Application +4 | -1

The value of

\widehat \mathbf{i} \times \left( {\widehat \mathbf{j} \times \mathbf{\hat k}} \right) + \widehat \mathbf{j} \times \left( {\mathbf{\hat k} \times \widehat \mathbf{i}} \right) + \mathbf{\hat k} \times \left( {\widehat \mathbf{i} \times \widehat \mathbf{j}} \right)

View answer and explanation

Correct option: A

{\vec 0}

Option A (

{\vec 0}

) is correct because the cross product of a vector with itself is always the zero vector.

99 Easy

99 of 195

Understanding +4 | -1

\mathrm{The\;value\;of\;}\frac{{{{\left( {\mathbf{\vec a} \times \mathbf{\vec b}} \right)}^2} + {{\left( {\mathbf{\vec a} \cdot \mathbf{\vec b}} \right)}^2}}}{{2{{\mathbf{\vec a}}^2}{{\mathbf{\vec b}}^2}}}\mathrm{is\;}

View answer and explanation

Correct option: D

\frac{1}{2}

Option D (

\frac{1}{2}

) is correct because the expression simplifies to 1/2 using trigonometric identities and the definitions of dot and cross products.

100 Medium

100 of 195

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}

101 Easy

101 of 195

Application +4 | -1

\mathbf{\vec a} = \mathbf{\hat i} + \mathbf{\hat j} + \mathbf{\hat k},\mathbf{\vec b} = \mathbf{\hat i} + 3\mathbf{\hat j} + 5\mathbf{\hat k}

and

\mathbf{\vec c} = 7\mathbf{\hat i} + 9\mathbf{\hat j} + 11\mathbf{\hat k}

, then the area of the parallelogram having diagonals

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

and

\mathbf{\vec b} + \mathbf{\vec c}

View answer and explanation

Correct option: A

4\sqrt 6

Option A (

4\sqrt 6

) is correct because it is calculated by finding half the magnitude of the cross product of the diagonals of the parallelogram.

102 Easy

102 of 195

Application +4 | -1

\mathbf{a}=2 \mathbf{i}+4 \mathbf{j}-5 \mathbf{k}

and

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

, then

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

View answer and explanation

Correct option: A

11 \sqrt{5}

Option A (

11 \sqrt{{5}}

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

103 Medium

103 of 195

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.

104 Easy

104 of 195

Application +4 | -1

The unit vector perpendicular to the vectors

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

and

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

View answer and explanation

Correct option: C

\frac{-2 i+3 \mathbf{i}+5 \mathbf{k}}{\sqrt{38}}

Option C (

\frac{{-2 i+3 \mathbf{j}+5 \mathbf{k}}}{{\sqrt{{38}}}}

) is correct because it is the unit vector obtained by dividing the cross product of the two vectors by its magnitude.

105 Easy

105 of 195

Understanding +4 | -1

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

are position vector of vertices of a triangle

A B C

, then unit vector perpendicular to its plane is

View answer and explanation

Correct option: B

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

Option B is correct because it provides the normalized vector perpendicular to the plane formed by the vertices of the triangle using cross products.

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106 Easy

106 of 195

Application +4 | -1

\vec{a} . \hat{i}=4

, then

(\vec{a} \times \hat{j}) \cdot(2 \hat{j}-3 \hat{k})

View answer and explanation

Correct option: A

Option A (12) is correct because using vector triple product properties, the expression simplifies to 3 times the dot product of vector a and i, which is given as 4.

107 Easy

107 of 195

Application +4 | -1

\left( {\mathbf{\vec x} - \mathbf{\vec y}} \right) \times \left( {\mathbf{\vec x} + \mathbf{\vec y}} \right) =

……. where

\mathbf{\vec x},\mathbf{\vec y}\; \in \mathrm{\;}{R^3}

View answer and explanation

Correct option: A

2\left( {\mathbf{\vec x} \times \mathbf{\vec y}} \right)

Option A (

2\left( {\mathbf{\vec x} \times \mathbf{\vec y}} \right)

) is correct because expanding the cross product and using the properties of cross products results in this expression.

108 Easy

108 of 195

Application +4 | -1

\vec{a}+2 \vec{b}+3 \vec{c}=\overrightarrow{0}

, then

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

is equal to

View answer and explanation

Correct option: A

6(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})

Option A (

6(\overrightarrow{{\mathrm{{b}}}} \times \overrightarrow{{\mathrm{{c}}}})

) is correct because by manipulating the given equation and using properties of the cross product, we find that

\vec{{a}} \times \vec{{b}} + \vec{{b}} \times \vec{{c}} + \vec{{c}} \times \vec{{a}} = 6(\overrightarrow{{\mathrm{{b}}}} \times \overrightarrow{{\mathrm{{c}}}})

109 Easy

109 of 195

Application +4 | -1

\overrightarrow {\mathbf{AB}} \times \overrightarrow {\mathbf{AC}} = 2\widehat \mathbf{i} - 4\widehat \mathbf{j} + 4\mathbf{\hat k},

then the area of

\Delta \;ABC

View answer and explanation

Correct option: A

3 sq units

Option A (3 sq units) is correct because the area of the triangle is half the magnitude of the cross product of the vectors representing two of its sides.

110 Easy

110 of 195

Application +4 | -1

2\mathbf{\vec a} + 3\mathbf{\vec b} + \mathbf{\vec c} = \vec 0

, then

\mathbf{\vec a} \times \mathbf{\vec b} + \mathbf{\vec b} \times \mathbf{\vec c} + \mathbf{\vec c} \times \mathbf{\vec a}

is equal to

View answer and explanation

Correct option: B

3\left( {\mathbf{\vec b} \times \mathbf{\vec c}} \right)

The correct answer is B (

3\left( {\mathbf{\vec b} \times \mathbf{\vec c}} \right)

) because it is derived by manipulating the given equation using cross products and vector properties.

111 Medium

111 of 195

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.

112 Easy

112 of 195

Application +4 | -1

The area of a parallelogram whose adjacent sides are given by the vectors

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

and

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

(in sq unit), is

View answer and explanation

Correct option: A

\sqrt {180} \;\mathrm{sq\;unit}

Option A (

\sqrt {180} \;\mathrm{sq\\;unit}

) is correct because it represents the magnitude of the cross product of the two given vectors, which equals the area of the parallelogram.

113 Medium

113 of 195

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.

114 Easy

114 of 195

Application +4 | -1

The area (in sq. units) of the parallelogram whose diagonals are along the vectors

8 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}

and

3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}

, is :

View answer and explanation

Correct option: B

Option C (14) is correct because it is the magnitude of the projection of

\vec{x} \times \vec{y}

\vec{z}

115 Medium

115 of 195

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.

116 Easy

116 of 195

Application +4 | -1

\vec{a}=\hat{i}+2 \hat{k}, \vec{b}=\hat{i}+\hat{j}+\hat{k}, \vec{c}=7 \hat{i}-3 \hat{j}+4 \hat{k}

, then the vector

\vec{d}

satisfies the relation

\vec{d} \times \vec{b}=\vec{c} \times \vec{b}, \vec{a} \cdot \vec{d}

=0

is given by

View answer and explanation

Correct option: B

2 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}-\hat{\mathrm{k}}

Option B (

2 \hat{{\mathrm{{i}}}} - 8 \hat{{\mathrm{{j}}}} - \hat{{\mathrm{{k}}}}

) is correct because it satisfies both the conditions

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

and

\vec{{a}} \cdot \vec{{d}} = 0

117 Easy

117 of 195

Application +4 | -1

Let

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

and a vector

\vec{b}

be such that

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

and

\vec{a} \cdot \vec{b}=3

. Then

|\vec{b}|

equals?

View answer and explanation

Correct option: A

\sqrt{\frac{11}{3}}

Option A (

\sqrt{\frac{11}{3}}

) is correct because it is the magnitude of vector

\vec{b}

calculated using the given cross product and dot product conditions.

118 Easy

118 of 195

Application +4 | -1

The unit vector perpendicular to

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

and

12 i+5 \mathbf{j}-5 \mathbf{k}

, is

View answer and explanation

Correct option: C

\frac{-5 i+3 j-9 k}{\sqrt{115}}

Option C (

\frac{{-5 \mathbf{{i}} + 3 \mathbf{{j}} - 9 \mathbf{{k}}}}{{\sqrt{{115}}}}

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

119 Medium

119 of 195

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.

120 Easy

120 of 195

Application +4 | -1

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

and

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

represents the adjacent sides of a parallelogram, then the area of this parallelogram is

View answer and explanation

Correct option: C

8 \sqrt{3}

Option C (

8 \sqrt{3}

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

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