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 $\vec a - \vec b$

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

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JEE • MATHEMATICS • DOT PRODUCT OF VECTORS

Dot product of vectors 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 91-120 of 296 questions

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91 Easy

91 of 296

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}

92 Easy

92 of 296

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}

93 Easy

93 of 296

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.

94 Medium

94 of 296

Application +4 | -1

\vec a,\;\,\vec b,\;\,\vec c

are three mutually perpendicular vectors of equal magnitude, then the angle

\theta

which

\vec a + \vec b + \vec c

makes with any one of three given vectors is given by

View answer and explanation

Correct option: A

{\cos ^{ - 1}}\,\,\frac{1}{{\sqrt 3 }}

Option A (

{\cos ^{ - 1}}\frac{1}{{\sqrt 3 }}

) is correct because it is derived by finding the cosine of the angle between the vector sum and one of the mutually perpendicular vectors.

95 Easy

95 of 296

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.

96 Medium

96 of 296

Application +4 | -1

\vec{a}=-\hat{\imath}+2 \hat{\jmath}+\hat{k}

is rotated by

90^{\circ}

about origin passing through

y

-axis. If new vector is

\vec{b}

then projection of

\vec{b}

\vec{c}=5 \hat{\imath}+4 \hat{\jmath}+3 \hat{k}

is equal to:

View answer and explanation

Correct option: A

\frac{6}{5}

Option A (6/5) is correct because the projection of the rotated vector b onto vector c is calculated to be 6/5.

97 Easy

97 of 296

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.

98 Easy

98 of 296

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.

99 Easy

99 of 296

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.

100 Easy

100 of 296

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}

101 Easy

101 of 296

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.

102 Easy

102 of 296

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}

103 Easy

103 of 296

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}

104 Easy

104 of 296

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.

105 Easy

105 of 296

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.

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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

106 Medium

106 of 296

Application +4 | -1

If a unit vector lies in

y z

-plane and makes angles of

30^{\circ}

and

60^{\circ}

with the positive

y

-axis and

z

-axis respectively, then its components along the co-ordinate axes will be

View answer and explanation

Correct option: B

0, \frac{\sqrt{3}}{2}, \frac{1}{2}

Option B (0,

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

\frac{1}{2}

) is correct because the components of the unit vector are found using the given angles with the y and z axes and the fact that it lies in the yz-plane.

107 Easy

107 of 296

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.

108 Easy

108 of 296

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.

109 Easy

109 of 296

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.

110 Easy

110 of 296

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.

111 Easy

111 of 296

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.

112 Easy

112 of 296

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.

113 Easy

113 of 296

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.

114 Easy

114 of 296

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}|

115 Easy

115 of 296

Analyzing +4 | -1

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

are three unit of vectors, a and

\mathbf{b}

are perpendicular to each other and vector

c

is equally inclined to both

\mathbf{a}

and

\mathbf{b}

at an angle

\theta

. If

\mathbf{c}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma(\mathbf{a} \times \mathbf{b})

, where

\alpha, \beta, \gamma

are constants, then

View answer and explanation

Correct option: C

\alpha=\beta=\cos \theta, \gamma^2=-\cos 2 \theta

Option C (

\alpha=\beta=\cos \theta, \gamma^2=-\cos 2 \theta

) is correct because it satisfies the given conditions when taking the dot product and using trigonometric identities.

116 Easy

116 of 296

Application +4 | -1

The two vectors

\left(x^2-1\right) \hat{i}+(x+2) \hat{j}+x^2 \hat{k}

and

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

are orthogonal

View answer and explanation

Correct option: D

For

x=-\frac{1}{2}

and

x=1

Option D (For

x=-\frac{1}{2}

and

x=1

) is correct because these are the values of x that make the dot product of the two vectors equal to zero, satisfying the orthogonality condition.

117 Easy

117 of 296

Application +4 | -1

A force vector applied on a mass represented by

\vec F = 6\hat i - 8\hat j + 10\hat k

and accelerates with

2\,\,{\text{m/}}{{\text{s}}^2}

the mass of the body is

View answer and explanation

Correct option: D

5\sqrt 2 \,\,{\text{kg}}

Option D (

5\sqrt 2 \,\,{\text{kg}}

) is correct because the mass is calculated by dividing the magnitude of the force vector by the acceleration.

118 Easy

118 of 296

Application +4 | -1

The value of

a

, for which the points,

A, B, C

with position vectors

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

and

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

respectively are the vertices of a right angled triangle with

C=\frac{\pi}{2}

are

View answer and explanation

Correct option: D

2 and 1

Option D (2 and 1) is correct because the dot product of AC and BC must be zero for a right-angled triangle at C, leading to two possible values for a.

119 Easy

119 of 296

Analyzing +4 | -1

\left| {\,\vec a\,} \right| = 4,\;\,\left| {\,\vec b\,} \right| = 4

and

\left| {\,\vec c\,} \right| = 5

such that

\vec a \bot \left( {\vec b + \vec c} \right),\,\;\vec b \bot \left( {\vec c + \vec a} \right)

and

\vec c \bot \left( {\vec a + \vec b} \right)

, then

\left| {\,\vec a + \vec b + \vec c\,} \right|

View answer and explanation

Correct option: D

\sqrt {57}

Option D (

\sqrt {57}

) is correct because it is the magnitude of the vector sum derived from the given orthogonality conditions.

120 Easy

120 of 296

Application +4 | -1

The value of

\lambda

for which the vectors

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

and

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

are perpendicular, is

View answer and explanation

Correct option: A

None

Option A (None) is correct because the dot product of the two vectors is always 1, regardless of the value of lambda, so they can never be perpendicular.

Revision notes and key takeaways

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

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