If ${\bf{\hat u}}\;{\rm{and\;}}{\bf{\hat v}}$ unit vectors and ${\rm{\theta }}$ is the acute angle between them, then $2{\bf{\hat u}} \times 3{\bf{\hat v}}$ is a unit vector for

Exactly one value of ${\rm{\theta }}$

Exactly two values of ${\rm{\theta }}$

More than two values of ${\rm{\theta }}$

The unit vector perpendicular to ${\bf{\hat i}} - {\bf{\hat j}}$ and coplanar with ${\bf{\hat i}} + 2{\bf{\hat j}}{\rm{\;and\;}}2{\bf{\hat i}} + 3{\bf{\hat j}}$ is

$\frac{1}{{\sqrt 2 }}\left( {{\bf{\hat i}} + {\bf{\hat j}}} \right)$

$\frac{{2{\bf{\hat i}} - 5{\bf{\hat j}}}}{{\sqrt {2a} }}$

$2{\bf{\hat i}} + 5{\bf{\hat j}}$

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

A unit vector in the plane of $\widehat {\bf{i}} + 2\widehat {\bf{j}} + {\bf{\hat k}}$ and $\widehat {\bf{i}} + \widehat {\bf{j}} + 2{\bf{\hat k}}$ and perpendicular to 2$\widehat {\bf{i}} + \widehat {\bf{j}} + {\bf{\hat k}}$ is

$\frac{{\widehat {\bf{j}} - {\bf{\hat k}}}}{{\sqrt 2 }}$

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

$\frac{{\widehat {\bf{i}} + \widehat {\bf{j}}}}{{\sqrt 2 }}$

$\frac{{\widehat {\bf{j}} + {\bf{\hat k}}}}{{\sqrt 2 }}$

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JEE • MATHEMATICS • DOT PRODUCT OF VECTORS

Dot product of vectors practice · Easy

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Showing 31-60 of 241 questions

31 Easy

31 of 241

Application +4 | -1

\mathbf{\hat u}\;\mathrm{and\;}\mathbf{\hat v}

unit vectors and

\mathrm{\theta }

is the acute angle between them, then

2\mathbf{\hat u} \times 3\mathbf{\hat v}

is a unit vector for

View answer and explanation

Correct option: D

Exactly one value of

\mathrm{\theta }

Option D (Exactly one value of

\mathrm{\theta }

) is correct because the sine function has a unique value between 0 and 90 degrees for a given value between 0 and 1.

32 Easy

32 of 241

Analyzing +4 | -1

Let

a = \hat i - 2\hat j + \hat k\,\,\mathrm{and}\,\,b = \hat i - \hat j + \hat k

be two vectors. If

c

is a vector such that

b \times c = \,b \times a\,\,\mathrm{and}\,\,c.\,a = 0

then

c \cdot b

is equal to

View answer and explanation

Correct option: C

- \frac{1}{2}

Option C (

- \frac{1}{2}

) is correct because it is the result of calculating the dot product of vector c and vector b after finding vector c using the given conditions.

33 Easy

33 of 241

Application +4 | -1

{a}, {b}, {c}

are unit vectors such that

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

then

{c} \cdot {a}

is equal to

View answer and explanation

Correct option: B

-\frac{1}{2}

Option B (

-\frac{1}{2}

) is correct because given the condition

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

, the dot product

{c} \cdot {a}

is equal to -1/2.

34 Easy

34 of 241

Application +4 | -1

The distance between the line

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

and the plane

\mathbf{\vec r}\cdot\left( {\widehat \mathbf{i} + 5\widehat \mathbf{j} + \mathbf{\hat k}} \right) = 5

View answer and explanation

Correct option: C

\frac{{10}}{{3\sqrt 3 }}

Option C (

\frac{{10}}{{3\sqrt 3 }}

) is correct because it represents the perpendicular distance between the given line and plane, calculated using the formula for the distance between a point on the line and the plane.

35 Easy

35 of 241

Application +4 | -1

The vector equation of the line passing through the points

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

and

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

View answer and explanation

Correct option: A

\mathbf{\vec r} = 3\widehat \mathbf{i} + 2\widehat \mathbf{j} + \mathbf{\hat k} + \mathrm{\;\lambda }\left( { - 5\widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k}} \right)

The correct answer is A (

\mathbf{\vec r} = 3\widehat \mathbf{i} + 2\widehat \mathbf{j} + \mathbf{\hat k} + \mathrm{\;\lambda }\left( { - 5\widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k}} \right)

) because it correctly represents the vector equation of a line passing through the given points.

36 Easy

36 of 241

Application +4 | -1

\vec{x}=3 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}-\hat{\mathrm{k}}, \overrightarrow{\mathrm{y}}=\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}

and

\vec{z}=3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-12 \hat{\mathrm{k}}

,
then the magnitude of the projection of

\vec{x} \times \vec{y}

\vec{z}

is:

View answer and explanation

Correct option: C

Option B (5) is correct because it is the maximum possible value of the scalar triple product

|\vec{u} \times \vec{v} \cdot \vec{w}|

37 Easy

37 of 241

Application +4 | -1

The angle between a diagonal of a cube and one of its edges is

View answer and explanation

Correct option: A

\cos ^{-1}(1 / \sqrt{3})

Option A (

\cos ^{-1}(1 / \sqrt{3})

) is correct because the angle between a diagonal of a cube and one of its edges is

\cos ^{-1}(1 / \sqrt{3})

38 Easy

38 of 241

Application +4 | -1

If the position vectors of

A,B\;\mathrm{and\;}C

are respectively

2\widehat \mathbf{i} - \widehat \mathbf{j} + \mathbf{\hat k},\mathrm{\;}\widehat \mathbf{i} - 3\widehat \mathbf{j} - 5\mathbf{\hat k}\mathrm{\;and\;}3\widehat \mathbf{i} - 4\widehat {\mathbf{\dot J}} - 4\mathbf{\hat k}\mathrm{\;}

then

\mathrm{co}{\mathrm{s}^2}\;A

is equal to

View answer and explanation

Correct option: C

\frac{{35}}{{41}}

Option C (

\frac{{35}}{{41}}

) is correct because it is the correct calculation of

\cos^2 A

using the cosine rule and the given position vectors.

39 Easy

39 of 241

Application +4 | -1

\left| {\vec a} \right| = \left| {\vec b} \right| = \left| {\vec a + \vec b} \right| = 1

, then

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

is equal to

View answer and explanation

Correct option: C

\sqrt 3

Option C (

\sqrt 3

) is correct because it is calculated using the properties of vector magnitudes and the given conditions.

40 Easy

40 of 241

Application +4 | -1

Let

a

and

b

be two vectors such that

\left| {\mathrm{2a + 3b}} \right|\mathrm{ = }\,\left| {\mathrm{3a + b}} \right|

and the angle between a and b is

60^\circ .

\frac{1}{8}a

is a unit vector, then

\left| \mathrm{b} \right|

is equal to

View answer and explanation

Correct option: C

Option C (5) is correct because it represents the magnitude of vector b, calculated using the given conditions and the properties of dot products.

41 Easy

41 of 241

Application +4 | -1

The total work done by two forces

{{\mathbf{\vec F}}_1} = 2\widehat \mathbf{i} - \widehat \mathbf{j}

and

{{\mathbf{\vec F}}_2} = 3\widehat \mathbf{i} + 2\widehat \mathbf{j} - \mathbf{\hat k}

acting on a particle when it is displaced from the point

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

5\widehat \mathbf{i} + 5\widehat \mathbf{j} + 3\mathbf{\hat k}

View answer and explanation

Correct option: D

11 units

Option D (11 units) is correct because it is the dot product of the total force and the displacement vector.

42 Easy

42 of 241

Application +4 | -1

If the vectors

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

and

2 \mathbf{i}-\mathbf{j}+8 \mathbf{k}

are perpendicular, then

\lambda

View answer and explanation

Correct option: C

Option C (14) is correct because the dot product of perpendicular vectors is zero, allowing us to solve for lambda.

43 Easy

43 of 241

Application +4 | -1

A parallelogram is constructed on

3\mathbf{\vec a} + \mathbf{\vec b}\mathrm{\;and\;}\mathbf{\vec a} - 4\mathbf{\vec b}

, where

\left| {\mathbf{\vec a}\left| { = 6\mathrm{\;and\;}} \right|\mathbf{\vec b}} \right| = 8\mathrm{\;and\;}\mathbf{\vec a}\mathrm{\;and\;}\mathbf{\vec b}

are anti-parallel, then the length of the longer diagonal is

View answer and explanation

Correct option: D

The correct answer is D. The length of the longer diagonal is calculated by finding the magnitude of the sum of the two vectors representing the sides of the parallelogram, considering they are anti-parallel.

44 Easy

44 of 241

Application +4 | -1

The direction cosines of the resultant of the vectors

(\mathbf{i}+\mathbf{j}+\mathbf{k}),(-\mathbf{i}+\mathbf{j}+\mathbf{k}),(\mathbf{i}-\mathbf{j}+\mathbf{k})

and

(\mathbf{i}+\mathbf{j}-\mathbf{k})

, are

View answer and explanation

Correct option: D

\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)

Option D (

\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)

) is correct because the direction cosines are found by normalizing the resultant vector.

45 Easy

45 of 241

Analyzing +4 | -1

If the angle between the vectors

\mathbf{a}=\hat{\mathbf{i}}+(\cos x) \hat{\mathbf{j}}+\hat{\mathbf{k}}

and

\mathbf{b}=\left(\sin ^2 x-\sin x\right) \hat{\mathbf{i}}-(\cos x) \hat{\mathbf{j}}+(3-4 \sin x) \hat{\mathbf{k}}

is obtuse and

x \in\left(0, \frac{\pi}{2}\right)

, then the exhaustive set of values of '

x

' is equal to

View answer and explanation

Correct option: B

x \in\left(\frac{\pi}{6}, \frac{\pi}{2}\right)

Option B (

x \in (\frac{\pi}{6}, \frac{\pi}{2})

) is correct because the angle between the vectors is obtuse, implying their dot product is negative, which leads to the solution

x \in (\frac{\pi}{6}, \frac{\pi}{2})

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46 Easy

46 of 241

Application +4 | -1

The angle between

{a}+2 {b}

and

{a}-3 {b}

|{a}|

=1,|{b}|=2

and angle between

{a}

and

{b}

60^{\circ}

View answer and explanation

Correct option: B

\cos ^{-1} \frac{-24}{\sqrt{21} \sqrt{31}}

Option B (

\cos ^{-1} \frac{-24}{\sqrt{21} \sqrt{31}}

) is correct because the angle between the vectors

{a}+2 {b}

and

{a}-3 {b}

is calculated using the dot product formula and the given magnitudes and angle between {a} and {b}.

47 Easy

47 of 241

Application +4 | -1

The summation of two unit vectors is a third unit vector, then the modulus of the difference of the unit vectors is

View answer and explanation

Correct option: A

\sqrt 3

Option A (

\sqrt 3

) is correct because the magnitude of the difference of two unit vectors, given that their sum is also a unit vector, can be derived using vector algebra.

48 Easy

48 of 241

Analyzing +4 | -1

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

and

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

then angle between

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

View answer and explanation

Correct option: B

\frac{\pi }{3}

Option B (

\frac{\pi }{3}

) is correct because it is the angle between the vectors that satisfies the given vector equation and magnitudes.

49 Easy

49 of 241

Understanding +4 | -1

If a is any vector in space, then

View answer and explanation

Correct option: A

\mathbf{a}=(\mathbf{a} \cdot \mathbf{i}) \mathbf{i}+(\mathbf{a} . \mathbf{j}) \mathbf{j}+(\mathbf{a} \cdot \mathbf{k}) \mathbf{k}

Option A (

\mathbf{a}=(\mathbf{a} \cdot \mathbf{i}) \mathbf{i}+(\mathbf{a} . \mathbf{j}) \mathbf{j}+(\mathbf{a} \cdot \mathbf{k}) \mathbf{k}

) is correct because it represents the decomposition of vector

\mathbf{a}

into its components along the coordinate axes.

50 Easy

50 of 241

Application +4 | -1

The distance of the point

B

with position vector

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

from the line passing through the point

A

with position vector

4 {i}+2 {j}+2 {k}

and parallel to the vector

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

View answer and explanation

Correct option: A

\sqrt{10}

Option A (

\sqrt{10}

) is correct because the distance is found by recognizing that AB is perpendicular to the line, making the distance simply |AB|.

51 Easy

51 of 241

Application +4 | -1

The unit vector perpendicular to

\mathbf{\hat i} - \mathbf{\hat j}

and coplanar with

\mathbf{\hat i} + 2\mathbf{\hat j}\mathrm{\;and\;}2\mathbf{\hat i} + 3\mathbf{\hat j}

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is a unit vector perpendicular to

\mathbf{\hat i} - \mathbf{\hat j}

and coplanar with the other given vectors.

52 Easy

52 of 241

Application +4 | -1

Let

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

be two vectors perpendicular to each other in the

xy

-plane. Then, a vector in the same plane having projections 1 and 2 along

{\vec b}

and

{\vec c}

, respectively, is

View answer and explanation

Correct option: B

2\hat i - \hat j

Option B (

2\hat i - \hat j

) is correct because it satisfies the conditions of being in the xy-plane and having projections 1 and 2 along the given vectors.

53 Easy

53 of 241

Application +4 | -1

The distance between the line

\mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}})

and the plane

\mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5

View answer and explanation

Correct option: C

\frac{10}{3 \sqrt{3}}

Option C (

\frac{10}{3 \sqrt{3}}

) is correct because the line is parallel to the plane, and the distance is calculated using the formula for the distance between a point on the line and the plane.

54 Easy

54 of 241

Application +4 | -1

A particle acted on by constant forces

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

and

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

is displaced from the point

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

to the point

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

. The total work done by the forces is

View answer and explanation

Correct option: D

40 units.

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

55 Easy

55 of 241

Application +4 | -1

Let

{a}={i}+{j}+{k}, {b}={i}-{j}+{k}

and

{c}={i}-{j}-{k}

be three vectors. A vector

{v}

in the plane of

{a}

and

{b}

, whose projection on

{c}

1 / \sqrt{3}

, is given by

View answer and explanation

Correct option: C

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

Option C (3i - j + k) is correct because it lies in the plane of a and b, and its projection on c is 1/

\sqrt{3}

56 Easy

56 of 241

Analyzing +4 | -1

Let a, b,c, be vectors of length

3,4,5

respectively and

\mathbf{a}

be perpendicular to

(\mathbf{b}+\mathbf{c}), \mathbf{b}

(\mathbf{c}+\mathbf{a})

and

\mathbf{c}

(\mathbf{a}+\mathbf{b})

, then the value of

(\mathbf{a}+\mathbf{b}+\mathbf{c})

View answer and explanation

Correct option: D

5 \sqrt{2}

The correct answer is D (

5 \sqrt{2}

) because the magnitude of the sum of the vectors is calculated using the given perpendicularity conditions and vector lengths.

57 Easy

57 of 241

Application +4 | -1

Let

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

and

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

be three vectors. A vector of the type

\vec{b}+\lambda \vec{c}

for some scalar

\lambda

, whose projection on

\vec{a}

is of magnitude

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

is :

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because by finding the vector

\vec{d} = \vec{b} + \lambda \vec{c}

and then using the projection formula, we can solve for

\lambda

and find the vector.

58 Easy

58 of 241

Application +4 | -1

\theta

be the angle between the vectors

a=2 i+2 j-k

and

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

, then

View answer and explanation

Correct option: A

\cos \theta=\frac{4}{21}

Option A (

\cos \theta=\frac{4}{21}

) is correct because it is the result of applying the dot product formula to find the cosine of the angle between the two vectors.

59 Easy

59 of 241

Analyzing +4 | -1

A unit vector in the plane of

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

and

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

and perpendicular to 2

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

View answer and explanation

Correct option: D

\frac{{\widehat \mathbf{j} - \mathbf{\hat k}}}{{\sqrt 2 }}

Option D (

\frac{{\widehat \mathbf{j} - \mathbf{\hat k}}}{{\sqrt 2 }}

) is correct because it represents a unit vector in the plane of the given vectors and is perpendicular to the specified vector.

60 Easy

60 of 241

Application +4 | -1

\hat{x}, \hat{y}

and

\hat{z}

are three unit vectors in three-dimensional space, then the minimum value of

|\hat{x}+\hat{y}|^2+|\hat{y}+\hat{z}|^2+|\hat{z}+\hat{x}|^2

View answer and explanation

Correct option: B

Option B (3) is correct because the minimum value is found by using the inequality

(\hat{x}+\hat{y}+\hat{z})^2 \geq 0

and manipulating the given expression.

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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