If ${\bf{\vec a}} = 4\,{\bf{\hat i}} + 6\,{\bf{\hat j}}{\rm{\;}}\,{\rm{and}}\,{\rm{\;}}{\bf{\vec b}} = 3\,{\bf{\hat j}} + 4\,{\bf{\hat k}}$, the vector form of the component of ${\bf{\vec a}}{\rm{\;}}\,\,{\rm{along}}\,{\rm{\;}}{\bf{\vec b}}$ is

$\frac{{18}}{{25}}\,\,\left( {3\,{\bf{\hat j}} + 4\,{\bf{\hat k}}{\rm{\;}}} \right)$

$\frac{{18}}{5}\,\,\left( {3\,{\bf{\hat i}} + 4\,{\bf{\hat k}}{\rm{\;}}} \right)$

$\frac{{36}}{{25}}\,\,\left( {3\,{\bf{\hat j}}\, + 4\,{\bf{\hat k}}{\rm{\;}}} \right)$

$\frac{{19}}{{18}}\,\,\left( {2\,{\bf{\hat i}} + 3\,{\bf{\hat j}}{\rm{\;}}} \right)$

If $\left| {\vec a + \vec b} \right| = \left| {\vec a - \vec b} \right|$, then

${\vec a}$ is parallel to ${\vec b}$

Let ${\bf{\vec a}} = \widehat {{\bf{\dot i}}} + \widehat {{\bf{\dot j}}} - {\bf{\hat k}},\,\;{\bf{\vec b}} = \widehat {{\bf{\dot i}}} - \widehat {{\bf{\dot j}}} + 3\,{\bf{\hat k}}\,\;{\rm{and\;}}\,{\bf{\vec c}}$ be a unit victor perpendicular to ${{\bf{\vec a}}}$ and coplanar with ${{\bf{\vec a}}}$ and ${{\bf{\vec b}}}$, then ${{\bf{\vec c}}}$ is

$\frac{1}{{\sqrt 6 }}\,\,\left( {2\widehat {\,{\bf{\dot i}}} - \widehat {{\bf{\dot j}}} + {\bf{\hat k}}} \right)$

$\frac{1}{{\sqrt 2 }}\,\,\left( {\widehat {{\bf{\dot j}}} + {\bf{\hat k}}} \right)$

$\frac{1}{{\sqrt 2 }}\,\,\left( {\widehat {{\bf{\dot j}}} - {\bf{\hat k}}} \right)$

$\frac{1}{{\sqrt 6 }}\,\,\left( {\widehat {{\bf{\dot i}}} - 2\widehat {\,{\bf{\dot j}}} + {\bf{\hat k}}} \right)$

If $\left| {\vec a} \right| = \left| {\vec b} \right|$, then

$\vec a + \vec b$ is $ \bot $ to $\vec a - \vec b$

$\left( {\vec a + \vec b} \right)$ is parallel to $\vec a - \vec b$

$\left( {\vec a + \vec b} \right).\left( {\vec a - \vec b} \right) = 2{\left| {\vec a} \right|^2}$

Back to DOT PRODUCT OF VECTORS

Your Progress

0 / 241 Answered

Correct

Incorrect

Accuracy

JEE • MATHEMATICS • DOT PRODUCT OF VECTORS

Dot product of vectors practice · Easy

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 61-90 of 241 questions

Preparing practice and saved progress…

61 Easy

61 of 241

Application +4 | -1

\left| {\;\mathbf{\vec a} + \mathbf{\vec b}\;\left| = \right|\mathbf{\vec a} - \mathbf{\vec b}\mathrm{\;}} \right|

, then angle between

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

\left( {\;\mathbf{\vec a}\; \ne \vec 0,\mathbf{\vec b} \ne \vec 0} \right)

View answer and explanation

Correct option: D

\frac{\pi }{2}

Option D (

\frac{{\pi }}{{2}}

) is correct because if the magnitudes of the sum and difference of two vectors are equal, the angle between them is 90 degrees.

62 Easy

62 of 241

Application +4 | -1

Let

\overrightarrow{\mathrm{a}}

and

\overrightarrow{\mathrm{b}}

be two unit vectors such that

|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}|=\sqrt{3}

.
If

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

, then

2|\vec{c}|

is equal to :

View answer and explanation

Correct option: A

\sqrt{55}

Option D (

12 \sqrt{2}

) is correct because it is the value of the dot product

\overrightarrow{\mathrm{c}} \cdot(-7 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})

calculated using the given conditions.

63 Easy

63 of 241

Application +4 | -1

Let

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

be unit vectors inclined at an angle

2\alpha \,\,\left( {0 \le \alpha \le \pi } \right)

each other, then

\left| {\;\,\mathbf{\vec a} + \mathbf{\vec b}\,} \right| < 1

, if

View answer and explanation

Correct option: D

\frac{\pi }{3} < \alpha < \frac{{2\,\pi }}{3}

Option D (

\frac{\pi}{3} < \alpha < \frac{{2\pi}}{3}

) is correct because it represents the range of

\alpha

for which the magnitude of the sum of the unit vectors is less than 1.

64 Easy

64 of 241

Understanding +4 | -1

The vector equation of the sphere whose centre is the point (1,0,1)and radius is 4, is

View answer and explanation

Correct option: A

\left| {\mathbf{\vec r} - \left( {\widehat \mathbf{i} + \mathbf{\hat k}} \right)} \right| = 4

Option A (

\left| {\mathbf{\vec r} - \left( {\widehat {\mathbf{i}} + \mathbf{\hat k}} \right)} \right| = 4

) is correct because the vector equation of a sphere with center

\vec a

and radius R is given by

\left| {\mathbf{\vec r} - \mathbf{\vec a}} \right| = R

65 Easy

65 of 241

Application +4 | -1

If a has magnitude 5 and points North-East and vector

\mathbf{b}

has magnitude 5 and points North-West, then

|\mathbf{a}-\mathbf{b}|

is equal to

View answer and explanation

Correct option: D

5 \sqrt{2}

Option D (

5 \sqrt{2}

) is correct because it is the magnitude of the difference between the two vectors, calculated using the Pythagorean theorem since the vectors are perpendicular.

66 Easy

66 of 241

Application +4 | -1

If the vectors

a \hat{i}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}

and

-\hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\mathrm{a} \hat{\mathrm{k}}

are perpendicular to each other then

a

is equal to:

View answer and explanation

Correct option: C

-5

Option C (-5) is correct because the dot product of two perpendicular vectors is zero, leading to the equation -a + 10 + 3a = 0, which simplifies to a = -5.

67 Easy

67 of 241

Application +4 | -1

\vec a + \vec b + \vec c = 0,\;\left| {\vec a} \right| = 3,\;\left| {\vec b} \right| = 5,\;\left| {\vec c} \right| = 7

, then the angle between

{\vec a}

and

{\vec b}

View answer and explanation

Correct option: A

\pi /3

Option A (

\pi /3

) is correct because the angle between the vectors is found using the cosine rule derived from the given vector equation.

68 Easy

68 of 241

Analyzing +4 | -1

\mathbf{a}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+5 \hat{\mathbf{k}}

and

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

are given vector.

A

vector

\mathrm{w}

which is perpendicular to

Z

-axis satisfying

\mathbf{c} \cdot \mathbf{a}=9

and

\mathbf{c} \cdot \mathbf{b}=-4

. If inclination of

\mathbf{c}

with

X

-axis and

Y

-axis is

\alpha

and

\beta

respectively, then which of the following is not true?

View answer and explanation

Correct option: C

\alpha>\frac{\pi}{2}

The correct answer is C (

\alpha>\frac{\pi}{2}

) because based on the calculated vector c, the angle alpha is less than

\frac{\pi}{2}

69 Easy

69 of 241

Application +4 | -1

If the constant force

2\widehat \mathbf{i} - 5\widehat \mathbf{j} + 6\mathbf{\hat k}

and

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

act on a particle due to which it is displaced from a point

A\left( {4, - 3, - 2} \right)

to a point

B\left( {6,1, - 3} \right)

then the work done by the force is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the work done is calculated as -15 units, which is not among the given options.

70 Easy

70 of 241

Application +4 | -1

Let

\mathbf{u}, \mathbf{v}, \mathbf{w}

be such that

|\mathbf{u}|=1,|\mathbf{v}|=2,|\mathbf{w}|=3

. If the projection

\mathbf{v}

along

\mathbf{u}

is equal to that of

\mathbf{w}

along

\mathbf{u}

and

\mathbf{v}, \mathbf{w}

are perpendicular to each other, then

|\mathbf{u}-\mathbf{v}+\mathbf{w}|

equal to

View answer and explanation

Correct option: C

\sqrt{14}

Option C (

\sqrt{14}

) is correct because using the given conditions and the formula for the magnitude of the sum of vectors, we find

|\mathbf{u}-\mathbf{v}+\mathbf{w}| = \sqrt{14}

71 Easy

71 of 241

Application +4 | -1

Let

\vec{a}=5 \hat{t}-\hat{j}-3 \hat{k}

and

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

be two vectors. Then which one of the following statements is True?

View answer and explanation

Correct option: B

Projection of

\vec{a}

\vec{b}

\frac{17}{\sqrt{35}}

and the direction of the projection vector is opposite to the direction of

\vec{b}

The correct answer is D because the projection of

\vec{a}

\vec{b}

is negative, indicating the projection vector is opposite to the direction of

\vec{b}

72 Easy

72 of 241

Application +4 | -1

If the vectors

\hat i - 2x\;\hat j - 3y\;\hat k

and

\hat i + 3x\;\hat j + 2y\;\hat k

are orthogonal to each other, then the locus of the point

\left( {x,\;y} \right)

View answer and explanation

Correct option: A

A circle

Option A (A circle) is correct because the equation derived from the orthogonality condition represents a circle.

73 Easy

73 of 241

Application +4 | -1

Let

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

be two unit vectors such that angle between them is

60^\circ

. Then,

\left| {\mathbf{\vec a} - \mathbf{\vec b}} \right|

is equal to

View answer and explanation

Correct option: D

Option D (1) is correct because using the formula for the magnitude of the difference of two vectors, and substituting the given angle and unit vector magnitudes, the result is 1.

74 Easy

74 of 241

Application +4 | -1

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

are two vectors such that

\left| {\,\mathbf{\vec a}\,} \right| + 3\,\sqrt 3

\mathbf{\vec b} = 4

and

\left| {\,\mathbf{\vec a} + \mathbf{\vec b}\,} \right| = \sqrt 7

, then the angle between

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

View answer and explanation

Correct option: D

150^\circ

Option D (

150^\circ

) is correct because it is the angle that satisfies the given magnitudes using the law of cosines.

75 Easy

75 of 241

Understanding +4 | -1

\mathbf{a}

and

\mathbf{b}

are adjacent sides of a rhombus, then

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because in a rhombus, all sides are equal in length, so the dot product of a vector with itself is equal to the square of its magnitude.

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

76 Easy

76 of 241

Application +4 | -1

Let

\mathbf{a}

and

\mathbf{b}

be two unit vectors inclined at an angle

\theta

, then

\sin (\theta / 2)

is equal to

View answer and explanation

Correct option: A

\frac{1}{2}|\mathbf{a}-\mathbf{b}|

Option A (

\frac{1}{2}|\mathbf{a}-\mathbf{b}|

) is correct because it is derived from the law of cosines and the properties of unit vectors.

77 Easy

77 of 241

Application +4 | -1

\mathbf{\vec a} = 4\,\mathbf{\hat i} + 6\,\mathbf{\hat j}\mathrm{\;}\,\mathrm{and}\,\mathrm{\;}\mathbf{\vec b} = 3\,\mathbf{\hat j} + 4\,\mathbf{\hat k}

, the vector form of the component of

\mathbf{\vec a}\mathrm{\;}\,\,\mathrm{along}\,\mathrm{\;}\mathbf{\vec b}

View answer and explanation

Correct option: B

\frac{{18}}{{25}}\,\,\left( {3\,\mathbf{\hat j} + 4\,\mathbf{\hat k}\mathrm{\;}} \right)

Option B (

\frac{{18}}{{25}}\left( {3\mathbf{\hat j} + 4{\mathbf{\hat k}\}} \right)

) is correct because it represents the vector component of

\vec{a}

along

\vec{b}

78 Easy

78 of 241

Application +4 | -1

\left| {\vec a + \vec b} \right| = \left| {\vec a - \vec b} \right|

, then

View answer and explanation

Correct option: B

\vec a \bot \vec b

Option B (

\vec a \bot \vec b

) is correct because if the magnitudes of the sum and difference of two vectors are equal, the vectors must be perpendicular.

79 Easy

79 of 241

Application +4 | -1

A unit vector

\mathbf{c}

perpendicular to

\mathbf{a}

and coplanar with

\mathbf{a}

and

\mathbf{b}, \mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{b}=\mathbf{i}+2 \mathbf{j}

is given by

View answer and explanation

Correct option: D

\frac{1}{\sqrt{2}}(-\mathbf{j}+\mathbf{k})

Option D (

\frac{1}{\sqrt{2}}(-\mathbf{j}+\mathbf{k})

) is correct because it represents a unit vector perpendicular to

\mathbf{a}

and coplanar with

\mathbf{a}

and

\mathbf{b}

80 Easy

80 of 241

Application +4 | -1

Let

\mathbf{\vec a} = \widehat {\mathbf{\dot i}} + \widehat {\mathbf{\dot j}} - \mathbf{\hat k},\,\;\mathbf{\vec b} = \widehat {\mathbf{\dot i}} - \widehat {\mathbf{\dot j}} + 3\,\mathbf{\hat k}\,\;\mathrm{and\;}\,\mathbf{\vec c}

be a unit victor perpendicular to

{\mathbf{\vec a}}

and coplanar with

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

, then

{\mathbf{\vec c}}

View answer and explanation

Correct option: D

\frac{1}{{\sqrt 6 }}\,\,\left( {2\widehat {\,\mathbf{\dot i}} - \widehat {\mathbf{\dot j}} + \mathbf{\hat k}} \right)

Option D is correct. The vector

\mathbf{\vec c}

is found by using the condition that it is perpendicular to

\mathbf{\vec a}

and coplanar with

\mathbf{\vec a}

and

\mathbf{\vec b}

81 Easy

81 of 241

Analyzing +4 | -1

Let

a b>0

and

\alpha=\frac{\hat{\mathbf{i}}}{a}+\frac{4 \hat{\mathbf{j}}}{b}+b \hat{\mathbf{k}}

and

\beta=b \hat{\mathbf{i}}+a \hat{\mathbf{j}}+\frac{1}{b} \hat{\mathbf{k}}

then the maximum value of

\frac{10}{5+\alpha \cdot \beta}

View answer and explanation

Correct option: A

Option D (None of the above) is correct because the given expression simplifies to zero due to the linear dependence of the vectors.

82 Easy

82 of 241

Application +4 | -1

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

and

|\mathbf{a} \times \mathbf{b}|=8

, then

\mathbf{a}

. b is equal to

View answer and explanation

Correct option: D

Option D (6) is correct because it follows from using the magnitudes of the vectors and the magnitude of their cross product to find the dot product.

83 Easy

83 of 241

Application +4 | -1

If the position vectors

A

and

B

6 {i}+{j}-3 {k}

and

4 {i}-3 {j}-2 {k}

, then the work done by the force

F={i}-3 {j}

+5 {k}

in displacing a particle from

{A}

{B}

View answer and explanation

Correct option: A

15 units

Option A (15 units) is correct because it is the work done calculated by the dot product of the force and the displacement vector.

84 Easy

84 of 241

Application +4 | -1

If the vector

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

makes angles

\alpha, \beta

and

\gamma

with vectors

\hat{\mathbf{i}}, \hat{\mathbf{j}}

and

\hat{\mathbf{k}}

respectively, then

View answer and explanation

Correct option: D

\alpha=\beta=\gamma

Option D (

\alpha=\beta=\gamma

) is correct because the vector

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

makes equal angles with each of the coordinate axes.

85 Easy

85 of 241

Application +4 | -1

The vectors

{\mathbf{\vec a}}

and

\mathbf{\vec b}\;

of equal magnitude 5 originating from a point and directs respectively towards north-east and north-west. Then, the magnitude of

\mathbf{\vec a} - \mathbf{\vec b}

View answer and explanation

Correct option: D

5\sqrt 2

The correct answer is D. The magnitude of the difference between two vectors of equal magnitude is calculated using the law of cosines, considering the angle between them is 90 degrees.

86 Easy

86 of 241

Application +4 | -1

The angle between the vectors

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

and

\mathbf{\vec a} - \mathbf{\vec b}

, where

\mathbf{\vec a} = \left( {1,1,4} \right)

and

\mathbf{\vec b} = \left( {1, - 1,\mathrm{\;}4} \right)

View answer and explanation

Correct option: A

90^\circ

Option A (

90^\circ

) is correct because the dot product of the vectors

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

and

\mathbf{\vec a} - \mathbf{\vec b}

is zero, indicating they are orthogonal.

87 Easy

87 of 241

Application +4 | -1

If the moduli of

\mathbf{a}

and

\mathbf{b}

are equal and angle between them is

120^{\circ}

and

\mathbf{a} \cdot \mathbf{b}=-8

, then

|\mathbf{a}|

is equal to

View answer and explanation

Correct option: C

Option C (4) is correct because the magnitude of vector a is found to be 4 using the given dot product and angle between the vectors.

88 Easy

88 of 241

Understanding +4 | -1

\mathbf{\vec a}\cdot\mathbf{\vec b} = - \left| {\mathbf{\vec a}} \right|\left| {\mathbf{\vec b}} \right|,

then the angle between

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

View answer and explanation

Correct option: B

180^\circ

Option B (

180^\circ

) is correct because the dot product is equal to the negative product of magnitudes when the angle between the vectors is 180 degrees.

89 Easy

89 of 241

Application +4 | -1

{\vec a}

and

{\vec b}

are two unit vectors inclined at an angle

\theta

such that

\vec a + \vec b

is a unit vector, then

\theta

is equal to

View answer and explanation

Correct option: D

\frac{{2\pi }}{3}

Option D (

\frac{{2\pi }}{{3}}

) is correct because the angle between the two unit vectors is found using the condition that their sum is also a unit vector.

90 Easy

90 of 241

Understanding +4 | -1

\left| {\vec a} \right| = \left| {\vec b} \right|

, then

View answer and explanation

Correct option: B

\vec a + \vec b

\bot

\vec a - \vec b

Option B (

\vec a + \vec b

\bot

\vec a - \vec b

) is correct because the dot product of

(\vec a + \vec b)

and

(\vec a - \vec b)

is zero when

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

Revision notes and key takeaways

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

Showing 61-90 of 241 questions

Your Progress

Dot product of vectors practice · Easy

Explore Other Chapters in MATHEMATICS

Continue Your Practice

Your practice