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 · Mixed difficulty

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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 296 questions

1 Medium

1 of 296

Application +4 | -1

The distance between the planes given by

\mathbf{r} .(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})+5=0

and

\mathbf{r}-(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})-8=0

View answer and explanation

Correct option: B

\frac{13}{3}

unit

Option B is correct because the distance between two parallel planes is calculated using the formula involving the constants and the magnitude of the normal vector.

2 Easy

2 of 296

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.

3 Easy

3 of 296

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|.

4 Easy

4 of 296

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.

5 Easy

5 of 296

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.

6 Medium

6 of 296

Application +4 | -1

Find the equation of the perpendicular drown from the origin to the plane

2x + 4y - 5z = 10

View answer and explanation

Correct option: B

\mathbf{\vec r} = \left( {2k,4k, - 5k} \right)k \in R

Option B (

\mathbf{\vec r} = \left( {2k,4k, - 5k} \right)k \in R

) is correct because it represents a line passing through the origin and perpendicular to the given plane, with direction ratios matching the coefficients of x, y, and z in the plane's equation.

7 Easy

7 of 296

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.

8 Easy

8 of 296

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.

9 Easy

9 of 296

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)

10 Easy

10 of 296

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.

11 Easy

11 of 296

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.

12 Easy

12 of 296

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.

13 Medium

13 of 296

Application +4 | -1

Let

a, b>0

and

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

and

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

then the maximum value of

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

View answer and explanation

Correct option: A

Option A (1) is correct because the maximum value of the expression is found by minimizing the denominator using AM-GM inequality.

14 Easy

14 of 296

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.

15 Easy

15 of 296

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.

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16 Medium

16 of 296

Application +4 | -1

The projection of the vector

\hat i - 2\hat j + \hat k

on the vector

4\hat i - 4\hat j + 7\hat k

, is

View answer and explanation

Correct option: B

\frac{{19}}{9}

Option B (

\frac{{19}}{9}

) is correct because it is the result of projecting the vector

\hat i - 2\hat j + \hat k

onto the vector

4\hat i - 4\hat j + 7\hat k

using the projection formula.

17 Easy

17 of 296

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.

18 Easy

18 of 296

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.

19 Medium

19 of 296

Analyzing +4 | -1

The vector

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

lying in the plane of the vectors

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

and

\vec{c}

bisects the angle between

\vec{b}

and

\vec{c}

. If

|\vec{c}|=\sqrt{2}

, then

\vec{c} \cdot \vec{a}

View answer and explanation

Correct option: C

Option C (3) is correct because by expressing

\vec{a}

as a linear combination of

\frac{\vec{b}}{|\vec{b}|}

and

\vec{c}

, and using the given conditions, we find

\vec{c} = \hat{j} + \hat{k}

, so

\vec{c} \cdot \vec{a} = 3

20 Easy

20 of 296

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.

21 Easy

21 of 296

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.

22 Easy

22 of 296

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.

23 Easy

23 of 296

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.

24 Easy

24 of 296

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.

25 Medium

25 of 296

Application +4 | -1

If the plane

\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1

, cuts the coordinate axes in

A, B, C

, then the area of

\triangle A B C

View answer and explanation

Correct option: C

\sqrt{61} \mathrm{sq}

. Units

Option C (

\sqrt{{61}}

sq. Units) is correct because it is the area of the triangle formed by the intercepts of the plane with the coordinate axes.

26 Easy

26 of 296

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.

27 Easy

27 of 296

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.

28 Easy

28 of 296

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.

29 Medium

29 of 296

Application +4 | -1

\overrightarrow{\mathrm{d}}=\lambda(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})+\mu(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})+v(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})

and

[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=\frac{1}{8}

, then

\lambda+\mu+v

is equal to -

View answer and explanation

Correct option: D

8 \overrightarrow{\mathrm{d}} \cdot(\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})

Option D (

8 \overrightarrow{{\mathrm{{d}}}} \cdot (\overrightarrow{{\mathrm{{a}}}} + \overrightarrow{{\mathrm{{b}}}} + \overrightarrow{{\mathrm{{c}}}})

) is correct because by taking the dot product of

\overrightarrow{{\mathrm{{d}}}}

with

(\overrightarrow{{\mathrm{{a}}}} + \overrightarrow{{\mathrm{{b}}}} + \overrightarrow{{\mathrm{{c}}}})

and using the given scalar triple product, we can find the value of

\lambda + \mu + \nu

30 Easy

30 of 296

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.

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