If ${\bf{\vec a}}\cdot\widehat {\bf{i}} = {\bf{\vec a}}\cdot\left( {\widehat {\bf{i}} + \widehat {\bf{j}}} \right) = {\bf{\vec a}}\cdot\left( {\widehat {\bf{i}} + \widehat {\bf{j}} + {\bf{\hat k}}} \right) = 1$, then ${{\bf{\vec a}}}$ is equal to

$\widehat {\bf{i}} + \widehat {\bf{j}}$

$\widehat {\bf{i}} - {\bf{\hat k}}$

$\widehat {\bf{i}} + \widehat {\bf{j}} - {\bf{\hat k}}$

A unit vector coplanar with ${\bf{\hat i}} + {\bf{\hat j}} + 2{\bf{\hat k}}$ and ${\bf{\hat i}} + 2{\bf{\hat j}} + {\bf{\hat k}}$ and perpendicular to ${\bf{\hat i}} + {\bf{\hat j}} + {\bf{\hat k}}$ is

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

$\left( {\frac{{{\bf{\hat i}} + {\bf{\hat j}} + {\bf{\hat k}}}}{{\sqrt 3 }}} \right)$

$\left( {\frac{{{\bf{\hat i}} + {\bf{\hat j}} + 2{\bf{\hat k}}}}{{\sqrt 6 }}} \right)$

$\left( {\frac{{{\bf{\hat i}} + 2{\bf{\hat j}} + {\bf{\hat k}}}}{{\sqrt 6 }}} \right)$

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JEE • MATHEMATICS • DOT PRODUCT OF VECTORS

Dot product of vectors 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 1-30 of 241 questions

1 Easy

1 of 241

Analyzing +4 | -1

A vector which makes equal angles with the vectors

\frac{1}{3}\left( {\hat i - 2\hat j + 2\hat k} \right),\frac{1}{5}\left( { - 4\hat i - 3\hat k} \right)

, and

{\hat j}

, is

View answer and explanation

Correct option: B

- 5\hat i + \hat j + 5\hat k

Option B (

- 5\hat i + \hat j + 5\hat k

) is correct because it makes equal angles with the given vectors.

2 Easy

2 of 241

Application +4 | -1

If vector

\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}

and vector

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

, then

\frac{\text { Projection of vector } \mathbf{a} \text { on vector } \mathbf{b}}{\text { Projection of vector } \mathbf{b} \text { on vector } \mathbf{a}}

is equal to

View answer and explanation

Correct option: B

\frac{7}{3}

Option B (

\frac{7}{3}

) is correct because it is the ratio of the magnitudes of the two vectors, as the projection of a on b divided by the projection of b on a equals |a|/|b|.

3 Easy

3 of 241

Application +4 | -1

The angle between

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

\frac{{5\pi }}{6}

and the projection of

{\mathbf{\vec a}}

in the direction of

{\mathbf{\vec b}}

\frac{{ - 6}}{{\sqrt 3 }}

then

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

is equal to

View answer and explanation

Correct option: D

The correct answer is D (4) because it is obtained by using the projection formula and the given angle between the vectors.

4 Easy

4 of 241

Application +4 | -1

A velocity

\frac{1}{4} \mathrm{~m} / \mathrm{s}

is resolved into two components along

O A

and

O B

making angles

30^{\circ}

and

45^{\circ}

respectively with the given velocity. Then the component along

O B

View answer and explanation

Correct option: A

\frac{1}{8}(\sqrt{6}-\sqrt{2}) \mathrm{m} / \mathrm{s}

Option A (

\frac{1}{8}(\sqrt{6}-\sqrt{2}) \mathrm{~m} / \mathrm{s}

) is correct because it is the component of the velocity along OB, calculated using the sine rule for resolving velocities.

5 Easy

5 of 241

Application +4 | -1

A plane meets the co-ordinate axes at

P, Q

and

R

such that the position vector of the oentroid of

\triangle P Q R

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

. Then the equation of the plane is

View answer and explanation

Correct option: A

\vec{r} \cdot(20 \hat{i}-8 \hat{j}+5 \hat{k})=120

Option A (

\vec{r} \cdot(20 \hat{i}-8 \hat{j}+5 \hat{k})=120

) is correct because it represents the equation of a plane in vector form, derived from the intercepts on the coordinate axes.

6 Easy

6 of 241

Application +4 | -1

The angle between the vector

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

and

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

View answer and explanation

Correct option: A

\pi / 2

Option A (

\pi / 2

) is correct because the dot product of the two vectors is zero, indicating they are perpendicular.

7 Easy

7 of 241

Analyzing +4 | -1

\vec a = \hat i + \hat j - \hat k,\vec b = \hat i - \hat j + \hat k

and

{\vec c}

is a unit vector perpendicular to the vector

\vec a

and coplanar with

\vec a

and

\vec b,

then a unit vector

{\vec d}

perpendicular to both

\vec a

and

{\vec c}

View answer and explanation

Correct option: B

\frac{1}{{\sqrt 2 }}\left( {\hat j + \hat k} \right)

Option B is correct. The unit vector perpendicular to both

\vec a

and

\vec c

is calculated using the cross product and normalization, resulting in

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

8 Easy

8 of 241

Application +4 | -1

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

and

\left| {\mathbf{\vec a}} \right| = 5,\;\mathrm{\;}\left| {\mathbf{\vec b}} \right| = 4\;

and

\left| {\mathbf{\vec c}} \right| = 3,\;

then the value of

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

View answer and explanation

Correct option: C

- 25

Option C (

-25

) is correct because it is derived from the given condition

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

and the magnitudes of the vectors.

9 Easy

9 of 241

Application +4 | -1

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

are three unit vectors such that

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

, where

{\vec 0}

is null vector, then

\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: C

- \frac{3}{2}

Option C (

- \frac{3}{2}

) is correct because it is the result of calculating the dot product sum given the condition that the sum of the three unit vectors is zero.

10 Easy

10 of 241

Analyzing +4 | -1

Let

S

be the set of all

a \in R

for which the angle between the vectors

\vec{u}=a\left(\log _e b\right) \hat{i}-6 \hat{j}+3 \hat{k}

and

\vec{v}=\left(\log _e b\right) \hat{i}+2 \hat{j}+2 a\left(\log _e b\right) \hat{k},(b>1)

is acute. Then

S

is equal to :

View answer and explanation

Correct option: B

\Phi

Option B (

\Phi

) is correct because there is no real value of 'a' that satisfies the condition for an acute angle between the vectors given the constraint b > 1.

11 Easy

11 of 241

Application +4 | -1

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

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

and

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

, then the angle between

2\mathbf{\vec a} - \mathbf{\vec c}

and

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

View answer and explanation

Correct option: B

\frac{\pi }{3}

Option B (

\frac{\pi}{3}

) is correct because the angle between the vectors

2\mathbf{\vec a} - \mathbf{\vec c}

and

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

is found using the dot product formula.

12 Easy

12 of 241

Application +4 | -1

A particle is acted upon by constant forces

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

and

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

which displace it from a point

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

to the point

5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\hat{\mathbf{k}}

. The work done in standard units by the forces is given by

View answer and explanation

Correct option: A

40 units

Option A (40 units) is correct because the work done is the dot product of the total force and the displacement vector, which equals 40.

13 Easy

13 of 241

Application +4 | -1

Let

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

and

\vec{\beta}=\hat{i}+2 \hat{j}-4 \hat{k}

. Let

\vec{\beta}_{1}

be parallel to

\vec{\alpha}

and

\vec{\beta}_{2}

be perpendicular to

\vec{\alpha}

. If

\vec{\beta}=\vec{\beta}_{1}+\vec{\beta}_{2}

, then the value of

5 \vec{\beta}_{2} \cdot(\hat{i}+\hat{j}+\hat{k})

View answer and explanation

Correct option: A

Option A (7) is correct because it involves finding the component of a vector parallel to another and then calculating a dot product.

14 Easy

14 of 241

Application +4 | -1

Given

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

are vectors such that

[\vec{a} \vec{b} \vec{c}]=\frac{1}{3}

. If the vector

\overrightarrow{{V}}

can be expressed as linear combination of

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

and

\vec{a} \times \vec{b}

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

z(\vec{a} \times \vec{b})

then

(x+y+z)

has the value equal to

View answer and explanation

Correct option: B

3 \overrightarrow{{V}} \cdot(\overrightarrow{{a}}+\overrightarrow{{b}}+\overrightarrow{{c}})

Option B (

3 \overrightarrow{{V}} \cdot(\overrightarrow{{a}}+\overrightarrow{{b}}+\overrightarrow{{c}})

) is correct because it correctly relates the scalar triple product to the dot product of V with the sum of the vectors a, b, and c.

15 Easy

15 of 241

Application +4 | -1

|\mathbf{a}|=3,|\mathbf{b}|=4

and

|\mathbf{a}+\mathbf{b}|=5

, then

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

View answer and explanation

Correct option: B

Option B (5) is correct because it is derived using the parallelogram law of vector addition and subtraction.

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16 Easy

16 of 241

Application +4 | -1

The magnitude of the projection of the vector

2\hat i + 3\hat j + \hat k

on the vector perpendicular to the plane containing the vectors

\hat i + \hat j + \hat k\,

and

\,\hat i\, + \,2\hat j\, + \,3\hat k,\,

View answer and explanation

Correct option: D

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

Option D is correct because it accurately calculates the magnitude of the projection of the vector onto the vector perpendicular to the plane.

17 Easy

17 of 241

Understanding +4 | -1

\left( {\mathbf{\vec a} \cdot \mathbf{\hat i}} \right)\mathbf{\hat i} + \left( {\mathbf{\vec a} \cdot \mathbf{\hat j}} \right)\mathbf{\hat 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 represents the decomposition of a vector into its components along the standard basis vectors.

18 Easy

18 of 241

Application +4 | -1

\mathbf{\vec A} = \widehat \mathbf{i} + 2\widehat \mathbf{j} + 3\mathbf{\hat k}\;,\;\mathbf{\vec B} = \widehat \mathbf{i} + 2\widehat \mathbf{j} + \mathbf{\hat k}

and

\mathbf{\vec C} = 3\widehat \mathbf{i} + \widehat \mathbf{j},

evaluate

t

, if the vector

\left( {\mathbf{\vec A} + t\mathbf{\vec B}} \right)

and

\mathbf{\vec C}\;

are mutually perpendicular.

View answer and explanation

Correct option: A

Option A (5) is correct because when the dot product of the two vectors

\left( {\mathbf{\vec A} + t\mathbf{\vec B}} \right)

and

\mathbf{\vec C}\;

is set to zero, solving for t yields 5.

19 Easy

19 of 241

Application +4 | -1

The vectors

{\vec X}

and

{\vec Y}

satisfy the equations

2\vec X + \vec Y = \vec p

and

\vec X + 2\vec Y = \vec q

, where

\vec p = \vec i + \vec j

and

\vec q = \hat i - \hat j

. If

\theta

is the angle between

{\vec X}

and

{\vec Y}

, then

View answer and explanation

Correct option: C

\cos \theta = - \frac{4}{5}

Option C (cos θ = -4/5) is correct because it calculates the cosine of the angle between the two vectors X and Y after solving the system of equations.

20 Easy

20 of 241

Application +4 | -1

\mathbf{\vec a}\cdot\widehat \mathbf{i} = \mathbf{\vec a}\cdot\left( {\widehat \mathbf{i} + \widehat \mathbf{j}} \right) = \mathbf{\vec a}\cdot\left( {\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}} \right) = 1

, then

{\mathbf{\vec a}}

is equal to

View answer and explanation

Correct option: C

\widehat \mathbf{i}

Option C (

\widehat \mathbf{i}

) is correct because it satisfies all the given dot product conditions.

21 Easy

21 of 241

Application +4 | -1

The unit vector in

XOY

plane and making angles

45^\circ

and

60^\circ

respectively with

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

and

b = 0\hat i + \hat j - \hat k

, is

View answer and explanation

Correct option: B

\frac{1}{{\sqrt 2 }}\hat i - \frac{1}{{\sqrt 2 }}\hat k

Option D (None of these) is incorrect, the correct answer is B (

\frac{1}{{\sqrt 2 }}\hat i - \frac{1}{{\sqrt 2 }}\hat k

) because it satisfies the given conditions of making angles 45° and 60° with the given vectors.

22 Easy

22 of 241

Application +4 | -1

A unit vector coplanar with

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

and

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

and perpendicular to

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

View answer and explanation

Correct option: A

\left( {\frac{{\mathbf{\hat j} - \mathbf{\hat k}}}{{\sqrt 2 }}} \right)

Option A (

\left( {\frac{{\mathbf{\hat j} - \mathbf{\hat k}}}{{\sqrt 2 }}} \right)

) is correct because it satisfies the coplanarity and perpendicularity conditions.

23 Easy

23 of 241

Analyzing +4 | -1

If in a

\triangle A B C, \mathbf{B C}=\frac{\mathbf{e}}{|\mathrm{e}|}-\frac{\mathbf{f}}{|\mathbf{f}|}

and

\mathbf{A C}=\frac{2 \mathbf{e}}{|\mathbf{e}|} ;|\mathbf{e}| \neq|\mathbf{f}|

, then the value of

\cos 2 A+\cos 2 B+\cos 2 C

must be

View answer and explanation

Correct option: A

-1

Option A (-1) is correct because the triangle is right-angled at B, making cos 2B = -1 and cos 2A + cos 2C = 0.

24 Easy

24 of 241

Application +4 | -1

If the position vectors of the points

A

and

B

are

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

and

\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}

respectively, then the equation of the plane through

B

and perpendicular to

A B

View answer and explanation

Correct option: A

2 x+3 y+6 z+28=0

Option A (

2 x+3 y+6 z+28=0

) is the correct equation of the plane because it passes through point B and is perpendicular to vector AB.

25 Easy

25 of 241

Application +4 | -1

Vectors

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

are inclined at an angle

\mathrm{\theta } = 120^\circ

. If

\left| {\mathbf{\vec a}} \right| = 1,\;\left| {\mathbf{\vec b}} \right| = 2,

then

[\left( {\mathbf{\vec a} + 3\mathbf{\vec b}} \right) \times {(3\mathbf{\vec a} + \mathbf{\vec b}]^2}

is equal to

View answer and explanation

Correct option: D

192

Option D (192) is correct because the magnitude of the cross product is calculated correctly using the given magnitudes and angle.

26 Easy

26 of 241

Analyzing +4 | -1

Let a vector

\vec{a}

has magnitude 9 . Let a vector

\vec{b}

be such that for every

(x, y) \in \mathbf{R} \times \mathbf{R}-\{(0,0)\}

, the vector

(x \vec{a}+y \vec{b})

is perpendicular to the vector

(6 y \vec{a}-18 \times \vec{b})

. Then the value of

|\vec{a} \times \vec{b}|

is equal to

View answer and explanation

Correct option: B

27 \sqrt{3}

Option B (

27 \sqrt{{3}}

) is correct because it is the magnitude of the cross product of vectors \vec{{a}} and \vec{{b}} calculated using the given perpendicularity condition.

27 Easy

27 of 241

Application +4 | -1

If the position vectors of the points

A, B, C, D

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

and

\mathbf{i}-6 \mathbf{j}-\mathbf{k}

, then the angle between the vectors

\overrightarrow{A B}

and

\overrightarrow{C D}

View answer and explanation

Correct option: D

\pi

Option D (

\pi

) is correct because the vectors AB and CD are anti-parallel, resulting in an angle of 180 degrees.

28 Easy

28 of 241

Application +4 | -1

If the vector

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

is perpendicular to

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

then

c

is equal to

View answer and explanation

Correct option: B

Option B (4) is correct because the dot product of two perpendicular vectors is zero, leading to the equation 18 + 2 - 5c = 0, which solves to c = 4.

29 Easy

29 of 241

Application +4 | -1

The cartesian from of the plane

\mathbf{\vec r} = \left( {s - 2t} \right)\widehat {\mathbf{\dot i}} + \left( {3 - t} \right)\widehat {\mathbf{\dot j}} + \left( {2s + t} \right)\mathbf{\hat k}

View answer and explanation

Correct option: C

2x - 5y - z + 15 = 0

Option C (

2x - 5y - z + 15 = 0

) is correct because it is derived by finding the normal vector to the plane and using the point-normal form of the plane equation.

30 Easy

30 of 241

Application +4 | -1

Let

|\mathbf{a}|=3

and

|\mathbf{b}|=4

. The value of

\mu

for which the vectors

\mathbf{a}+\mu \mathbf{b}

and

\mathbf{a}-\mu \mathbf{b}

will be perpendicular is

View answer and explanation

Correct option: A

3 / 4

Option A (

3 / 4

) is correct because it satisfies the condition for perpendicularity, which is that the dot product of the two vectors is zero.

Revision notes and key takeaways

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

Key Takeaways

✓ This set contains 241 single-correct-answer MCQs on DOT PRODUCT OF VECTORS.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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