If ${{{\bf{\vec u}}}_1}\;{\rm{and\;}}{{{\bf{\vec u}}}_2}$ be vectors of unit length and ${\rm{\theta }}$ be the angle between them, then $\frac{1}{2}\left| {{{{\bf{\vec u}}}_2} - {{{\bf{\vec u}}}_1}} \right|{\rm{is}}$

$\sin \frac{{\rm{\theta }}}{2}$

$\cos \frac{{\rm{\theta }}}{2}$

The projection vector of $\vec{a}$ on $\vec{b}$ is

$\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\right) \vec{b}$

$\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}\right)$

$\left(\frac{\vec{a} \cdot \vec{b}}{|\overrightarrow{\mathrm{a}}|^2}\right) \hat{b}$

Let ${\bf{\vec a}} = \widehat {\bf{i}} - \widehat {\bf{j}}$, ${\bf{\vec b}} = \widehat {\bf{j}} - {\bf{\hat k}},\;{\bf{\vec c}} = {\bf{\hat k}} - \widehat {\bf{i}}$. If ${{\bf{\vec d}}}$ is a unit vectors such that ${\bf{\vec a}}\cdot{\rm{\;}}{\bf{\vec d}} = 0 = \left[ {{\rm{\;}}{\bf{\vec b}}\;{\rm{\;}}{\bf{\vec c}}\;{\rm{\;}}{\bf{\vec d}}} \right],$ then ${{\bf{\vec d}}}$ is (are)

$ \pm \frac{{\widehat {\bf{i}} + \widehat {\bf{j}} - 2{\bf{\hat k}}}}{{\sqrt 6 }}$

$ \pm \frac{{\widehat {\bf{i}} + \widehat {\bf{j}} - {\bf{\hat k}}}}{{\sqrt 3 }}$

$ \pm \frac{{\widehat {\bf{i}} + \widehat {\bf{j}} + {\bf{\hat k}}}}{{\sqrt 3 }}$

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JEE • MATHEMATICS • DOT PRODUCT OF VECTORS

Dot product of vectors practice · Easy

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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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121 Easy

121 of 241

Application +4 | -1

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

and three vectors such that

\mathbf{\vec a} = \mathrm{\;}\mathbf{\vec b} + \mathbf{\vec c}

and the angle between

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

\frac{\pi }{2}\mathrm{then\;}

View answer and explanation

Correct option: A

{a^2} = {b^2} + {c^2}

Option A (

{a^2} = {b^2} + {c^2}

) is correct because it represents the Pythagorean theorem, which applies when two vectors are orthogonal.

122 Easy

122 of 241

Application +4 | -1

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

are unit vectors such that

\mathbf{a}+\mathbf{b}-\mathbf{c}=0

, then the angle between

\mathbf{a}

and

\mathbf{b}

View answer and explanation

Correct option: D

2 \pi / 3

Option D (

2 \pi / 3

) is correct because using the dot product and the given condition

\mathbf{a}+\mathbf{b}-\mathbf{c}=0

, we can find the angle between a and b.

123 Easy

123 of 241

Application +4 | -1

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

are mutually perpendicular unit vectors, then

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

is equal to

View answer and explanation

Correct option: A

\sqrt 3

Option A (

\sqrt 3

) is correct because the magnitude of the sum of three mutually perpendicular unit vectors is

\sqrt 3

124 Easy

124 of 241

Application +4 | -1

Given that

\left| {\,\mathbf{\vec a}\,} \right| = 3,\,\,\left| {\,\mathbf{\vec b}\,\left| { = 4,\,} \right|\,\mathbf{\vec a} \times \mathbf{\vec b}\,} \right| = 10

, then

|\mathbf{\vec a} \cdot \mathbf{\vec b}{|^2}

equals

View answer and explanation

Correct option: B

Option B (44) is correct because it is derived from the relationship between the dot product, cross product, and magnitudes of the vectors, using the given values.

125 Easy

125 of 241

Application +4 | -1

\vec a = 4\hat i + 6\hat j

and

\vec b = 3\hat j + 4\hat k

, then the vector form of component of

{\vec a}

along

{\vec b}

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents the projection of vector

\vec a

onto vector

\vec b

in vector form.

126 Easy

126 of 241

Application +4 | -1

The position vector of a point

P

{r}=x {i}+y {j}+z {k}

, where

x, y, z \in {N}

and

{u}={i}+{j}+{k}

. If

{r} \cdot {u}=10

, then the number of possible positions of

P

View answer and explanation

Correct option: B

Option B (

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

) is correct because it is collinear with vector a and satisfies the condition a . b = 2.

127 Easy

127 of 241

Application +4 | -1

\mathbf{\vec a} = \widehat \mathbf{i} + 2\widehat \mathbf{j} - 3\mathbf{\hat k}

and

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

then the angle between the vectors

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

and

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

View answer and explanation

Correct option: B

90^\circ

Option B (

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.

128 Easy

128 of 241

Application +4 | -1

\bar{a}_1, \bar{a}_2, \bar{a}_3, \ldots . \bar{a}_{20}

are 20 unit vectors such that

\sum\limits_{{i=1}}^{20} \bar{a}_i=0

then the value of

\sum\limits_{{0 \leq I < j \leq 20}} \overline{a_i} \cdot \overline{a_j}

View answer and explanation

Correct option: A

-10

Option A (-10) is correct. The sum of dot products of all pairs of unit vectors is calculated using the given condition that the sum of all vectors is zero.

129 Easy

129 of 241

Application +4 | -1

If the angle between

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

and

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

\frac{\pi }{3}

, then the value of

a

View answer and explanation

Correct option: B

- 4

or 0

Option B (-4 or 0) is correct because the values of 'a' are found by using the dot product formula for the angle between two vectors and solving the resulting quadratic equation.

130 Easy

130 of 241

Understanding +4 | -1

\left( {\mathbf{\vec a} \cdot \widehat {\mathbf{\dot i}}} \right)\,\,\widehat {\mathbf{\dot i}} + \left( {\mathbf{\vec a} \cdot \widehat {\mathbf{\dot j}}} \right)\,\,\widehat {\mathbf{\dot j}} + \left( {\mathbf{\vec a} \cdot \mathbf{\hat k}} \right)\,\,\mathbf{\hat k}

̂ is equal to

View answer and explanation

Correct option: A

{\mathbf{\vec a}}

Option A (

{\mathbf{\vec a}}

) is correct because the expression is the vector projection of

\vec a

onto the i, j, and k axes, which sums back to the original vector.

131 Easy

131 of 241

Application +4 | -1

{{\mathbf{\vec u}}_1}\;\mathrm{and\;}{{\mathbf{\vec u}}_2}

be vectors of unit length and

\mathrm{\theta }

be the angle between them, then

\frac{1}{2}\left| {{{\mathbf{\vec u}}_2} - {{\mathbf{\vec u}}_1}} \right|\mathrm{is}

View answer and explanation

Correct option: B

\sin \frac{\mathrm{\theta }}{2}

Option B (

\sin \frac{\mathrm{\theta }}{2}

) is correct because the expression simplifies to

\sin \frac{\mathrm{\theta }}{2}

using trigonometric identities and the properties of unit vectors.

132 Easy

132 of 241

Remembering +4 | -1

The projection vector of

\vec{a}

\vec{b}

View answer and explanation

Correct option: B

\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\right)

The correct answer is B. The projection of vector

\vec{a}

\vec{b}

is given by the formula

\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

133 Easy

133 of 241

Analyzing +4 | -1

\hat{\mathbf{a}}

is a unit vector and projection of

\mathbf{x}

along

\hat{\mathbf{a}}

is 2 units and

(\hat{\mathbf{a}} \times \mathbf{x})+\mathbf{b}=\mathbf{x}

, then

\mathbf{x}

is equal to

View answer and explanation

Correct option: B

\frac{1}{2}[2 \hat{\mathbf{a}}-\mathbf{b}+(\hat{\mathbf{a}} \times \mathbf{b})]

Option B (

\frac{1}{2}[2 \hat{\mathbf{a}}-\mathbf{b}+(\hat{\mathbf{a}} \times \mathbf{b})]

) is correct because it is obtained by manipulating the given vector equation and using the projection information to solve for

\mathbf{x}

134 Easy

134 of 241

Analyzing +4 | -1

The vector

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

is rotated through a right angle, passing through the

y

-axis in its way and the resulting vector is

\vec{b}

. Then the projection of

3 \vec{a}+\sqrt{2 \vec{b}}

\vec{c}=5 \hat{i}+4 \hat{j}+3 \hat{k}

View answer and explanation

Correct option: D

3 \sqrt{2}

The correct answer is D (

3 \sqrt{2}

) because after rotating the vector and finding the resulting vector, the projection is calculated correctly.

135 Easy

135 of 241

Application +4 | -1

The angle between the vectors

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

and

2 i-2 j+4 k

View answer and explanation

Correct option: B

\sin ^{-1} \frac{2}{\sqrt{7}}

Option B (

\sin ^{-1} \frac{2}{\sqrt{7}}

) is correct because it is the angle derived from calculating the dot product and magnitudes of the given vectors.

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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

136 Easy

136 of 241

Application +4 | -1

Let

\mathbf{u}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathbf{v}=\hat{\mathbf{i}}-\hat{\mathbf{j}}

and

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

. If

\mathbf{n}

is a unit vector such that

\mathbf{u} \cdot \mathbf{n}=0

and

\mathbf{v} \cdot \mathbf{n}=0

, then

|\mathbf{w} \cdot \mathbf{n}|

is equal to

View answer and explanation

Correct option: D

Option D (3) is correct because finding a unit vector

\mathbf{n}

perpendicular to both

\mathbf{u}

and

\mathbf{v}

and then calculating

|\mathbf{w} \cdot \mathbf{n}|

yields 3.

137 Easy

137 of 241

Application +4 | -1

A particle moves along a curve so that its coordinates at time

t

are

x=t, y=\frac{1}{2} t^2, z=\frac{1}{3} t^3

. The acceleration at

t=1

View answer and explanation

Correct option: A

{j}+2 {k}

Option B (

{j}+2 {k}

) is correct because the acceleration vector at t=1 is found by taking the second derivative of the position vector and evaluating it at t=1.

138 Easy

138 of 241

Application +4 | -1

Let

\mathbf{\vec a} = \widehat \mathbf{i} - \widehat \mathbf{j}

\mathbf{\vec b} = \widehat \mathbf{j} - \mathbf{\hat k},\;\mathbf{\vec c} = \mathbf{\hat k} - \widehat \mathbf{i}

. If

{\mathbf{\vec d}}

is a unit vectors such that

\mathbf{\vec a}\cdot\mathrm{\;}\mathbf{\vec d} = 0 = \left[ {\mathrm{\;}\mathbf{\vec b}\;\mathrm{\;}\mathbf{\vec c}\;\mathrm{\;}\mathbf{\vec d}} \right],

then

{\mathbf{\vec d}}

is (are)

View answer and explanation

Correct option: B

\pm \frac{{\widehat \mathbf{i} + \widehat \mathbf{j} - 2\mathbf{\hat k}}}{{\sqrt 6 }}

Option B (

\pm \frac{{\widehat \mathbf{i} + \widehat \mathbf{j} - 2\mathbf{\hat k}}}{{\sqrt 6 }}

) is correct because it satisfies the given conditions:

\mathbf{\vec a}\cdot\mathrm{\;}\mathbf{\vec d} = 0

and

\left[ {\mathrm{\;}\mathbf{\vec b}\;\mathrm{\;}\mathbf{\vec c}\;\mathrm{\;}\mathbf{\vec d}} \right] = 0

139 Easy

139 of 241

Analyzing +4 | -1

If vectors

\vec{a}, \vec{b}, \vec{c}

satisfy the condition

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

, then

(\vec{b}-\vec{a}) \cdot\left(\vec{c}-\frac{\vec{a}+\vec{b}}{2}\right)

is equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the dot product simplifies to zero based on the given condition

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

140 Easy

140 of 241

Analyzing +4 | -1

Let the unit vectors

{\vec a}

and

{\vec b}

be perpendicular to each other and the unit vector

{\vec c}

be inclined at an angle

\theta

to both

{\vec a}

and

{\vec b}

. If

\vec c = x\vec a + y\vec b + \vec c\left( {\vec a \times \vec b} \right)

, then

View answer and explanation

Correct option: D

x = y = \cos \theta ,\;{z^2} = - \cos 2\theta

Option D (

x = y = \cos \theta ,\;{z^2} = - \cos 2\theta

) is correct because by taking dot products with

\vec a

and

\vec b

, we find

x = y = \cos \theta

, and using the magnitude of

\vec c

, we derive

{z^2} = -\cos 2\theta

141 Easy

141 of 241

Application +4 | -1

Let

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

and

\vec{b}=3 \hat{i}+5 \hat{j}+4 \hat{k}

be two vectors, such that

\vec{a} \times \vec{b}=-\hat{i}+9 \hat{i}+12 \hat{k}

. Then the projection of

\vec{b}-2 \vec{a}

\vec{b}+\vec{a}

is equal to

View answer and explanation

Correct option: D

\frac{46}{5}

Option D (

\frac{46}{5}

) is correct because it is the result of calculating the projection of

\vec{b}-2\vec{a}

\vec{b}+\vec{a}

after finding the values of

\alpha

and

\beta

142 Easy

142 of 241

Application +4 | -1

Let

\mathbf{\vec a},\mathbf{\vec b}

and

\mathbf{\vec c}\;

be vectors with magnitude 3,4 and 5 respectively and

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

, then the value of

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

View answer and explanation

Correct option: D

- 25

Option D (

- 25

) is correct because it is derived from squaring the equation

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

and using the given magnitudes.

143 Easy

143 of 241

Application +4 | -1

Two forces

\overrightarrow{{P}}=3 \hat{{i}}-2 \hat{{j}}+\hat{{k}}

and

\overrightarrow{{Q}}=\hat{{i}}+3 \hat{{j}}-5 \hat{{k}}

acting on a particle

A

move it to

B

. If the position vectors of

A

and

B

are respectively

-2 \hat{{i}}+5 \hat{{k}}

and

3 \hat{{i}}-7 \hat{{j}}+2 \hat{{k}}

, then the total work done is

View answer and explanation

Correct option: D

25 unit

Option D (25 unit) is correct because the work done is calculated by finding the dot product of the total force and the displacement vector.

144 Easy

144 of 241

Application +4 | -1

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

are three vectors, such that

\mathbf{a}+\mathbf{b}+\mathbf{c}=0,|\mathbf{a}|=1

|\mathbf{b}|=2,|\mathbf{c}|=3

, then

\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}

is equal to

View answer and explanation

Correct option: B

-7

Option B (-7) is correct because squaring the equation

\mathbf{a} + \mathbf{b} + \mathbf{c} = 0

and using the given magnitudes leads to

\mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{a} = -7

145 Easy

145 of 241

Application +4 | -1

If the scalar projection of the vector

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

on the vector

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

\frac{1}{{\sqrt {30} }}\mathrm{\;then\;the\;value\;of\;}x\;\mathrm{is\;}

View answer and explanation

Correct option: A

- 3/2

Option A (

- 3/2

) is correct because the scalar projection formula is used, and solving for x gives -3/2.

146 Easy

146 of 241

Application +4 | -1

The value of

c

so that for all real

x

, then vectors o

cx\;\widehat \mathbf{i} - 6\widehat \mathbf{j} + 3\mathbf{\hat k},\;x\widehat \mathbf{i} + 2\widehat \mathbf{j} + 2cx\mathbf{\hat k}

make an obtuse angle are

View answer and explanation

Correct option: C

- \frac{4}{3} < c < 0

Option C (

- \frac{4}{3} < c < 0

) is correct because for the angle to be obtuse, the dot product must be negative, leading to a quadratic inequality whose solution gives the range of c.

147 Easy

147 of 241

Application +4 | -1

If the vectors

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

and

\mathbf{\vec c} = \mathrm{\lambda }\widehat \mathbf{i} + \widehat \mathbf{j} + \mathrm{\mu }\mathbf{\hat k}

are mutually orthogonal, then

\left( {\lambda ,\mathrm{\mu }} \right)

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the values of

\lambda

and

\mu

are found by using the orthogonality condition (dot product equals zero) between the vectors.

148 Easy

148 of 241

Application +4 | -1

\mathrm{\theta }

is the angle between the lines

AB

and

AC

where

A,B

and

C

are the three points with coordinates

\left( {1,2, - 1} \right),\left( {2,0,3} \right),\left( {3, - 1,2} \right)

respectively, then

\sqrt {462} \cos \mathrm{\theta }

is equal to

View answer and explanation

Correct option: A

Option A (20) is correct because using the dot product formula to find the cosine of the angle between the vectors AB and AC, and then multiplying by

\sqrt{462}

, yields 20.

149 Easy

149 of 241

Application +4 | -1

The position vector of a point

P

r = x\widehat i + y\hat j + z\widehat k

, where

x,y, z \in N

and

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

. If

\mathbf{r} \cdot \mathbf{a}=20

, then the number of possible position of

P

View answer and explanation

Correct option: A

Option A (81) is correct because it represents the number of possible positions of point P, calculated by finding the number of positive integer solutions to the given equation.

150 Easy

150 of 241

Application +4 | -1

\theta

be the angle between the unit vectors a and

\mathbf{b}

, then

\cos \frac{\theta}{2}=

View answer and explanation

Correct option: B

\frac{1}{2}|a+b|

Option B (

\frac{1}{2}|a+b|

) is correct because it directly follows from the formula relating the magnitude of the sum of two unit vectors to the cosine of half the angle between them.

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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