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

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

$\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 · 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 61-90 of 149 questions

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61 Easy

61 of 149

Understanding +4 | -1

\quad(a-b) \times(a+b)=

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the cross product

(a-b) \times (a+b)

simplifies to

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

62 Easy

62 of 149

Application +4 | -1

|\mathrm{a}| = 2,|\mathrm{b}| = 5

and

\mathrm{a} \times \mathrm{b| = 8},

then

|\mathrm{a}.\mathrm{b}|

is equal to

View answer and explanation

Correct option: A

Option A (6) is correct because it's the absolute value of the dot product calculated using the magnitudes of the vectors and the angle between them, derived from the cross product.

63 Easy

63 of 149

Analyzing +4 | -1

\vec a \times \vec b = \vec c \times \vec d

and

\vec a \times \vec c = \vec b \times \vec d

, then

View answer and explanation

Correct option: A

\left( {\vec a \pm \vec d} \right) = \lambda \left( {\vec b \pm \vec c} \right)

Option A is correct because the given conditions imply that the vectors (

\vec a - \vec d

) and (

\vec b - \vec c

) are parallel, as are (

\vec a + \vec d

) and (

\vec b + \vec c

64 Easy

64 of 149

Application +4 | -1

\mathbf{a}

and

\mathbf{b}

are two vectors, then

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

is equal to

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the correct expression for

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

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

, which can be represented by a determinant but not any of the given options.

65 Easy

65 of 149

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.

66 Easy

66 of 149

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}

67 Easy

67 of 149

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.

68 Easy

68 of 149

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.

69 Easy

69 of 149

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.

70 Easy

70 of 149

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.

71 Easy

71 of 149

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.

72 Easy

72 of 149

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.

73 Easy

73 of 149

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.

74 Easy

74 of 149

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.

75 Easy

75 of 149

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.

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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

76 Easy

76 of 149

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.

77 Easy

77 of 149

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.

78 Easy

78 of 149

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.

79 Easy

79 of 149

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}}}})

80 Easy

80 of 149

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.

81 Easy

81 of 149

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.

82 Easy

82 of 149

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.

83 Easy

83 of 149

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}

84 Easy

84 of 149

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

85 Easy

85 of 149

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.

86 Easy

86 of 149

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.

87 Easy

87 of 149

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.

88 Easy

88 of 149

Application +4 | -1

Let the points

A, B

and

P

(-2,2,4),(2,6,3)

and

(1,2,1)

respectively. The magnitude of the moment of the force represented by

\overrightarrow{A B}

and acting at

A

about

P

View answer and explanation

Correct option: B

3 \sqrt{41}

Option B (

3 \sqrt{41}

) is correct because it represents the magnitude of the moment of the force calculated using the cross product of the position vector and the force vector.

89 Easy

89 of 149

Application +4 | -1

\vec{a}=3 \hat{i}-5 \hat{j}

and

\vec{b}=6 \hat{i}+3 \hat{j}

are two vectors and

\vec{c}

isa vector such that

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

then

|\vec{a}|:|\vec{b}|:|\vec{c}|

View answer and explanation

Correct option: B

\sqrt{34}: \sqrt{45}: 39

Option B (

\sqrt{34}: \sqrt{45}: 39

) is correct because it accurately represents the magnitudes of the vectors a, b, and c, where c is the cross product of a and b.

90 Easy

90 of 149

Analyzing +4 | -1

Let

\vec \alpha = 3\hat i + \hat j

and

\vec \beta = 2\hat i - \hat j + 3\hat k.

\vec \beta = {\vec \beta _1} - {\vec \beta _2}

where

{\vec \beta _1}

is parallel to

\vec \alpha

and

{\vec \beta _2}

is perpendicular to

\vec \alpha

, then

{\vec \beta _1} \times {\vec \beta _2}

is equal to

View answer and explanation

Correct option: B

\frac{1}{2}\left( { - 3\hat i + 9\hat j + 5\hat k} \right)

Option B (

\frac{1}{2}(-3\hat{i} + 9\hat{j} + 5\hat{k})

) is correct because it represents the correct cross product of the components of

\vec{\beta}

parallel and perpendicular to

\vec{\alpha}

Revision notes and key takeaways

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