Unit vector which is perpendicular to both the vectors $3\widehat {\bf{i}} + \widehat {\bf{j}} + 2{\bf{\hat k}}$ and $2\widehat {\bf{i}} - 2\widehat {\bf{j}} + 4{\bf{\hat k}}$ is

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

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

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

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

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JEE • MATHEMATICS • DOT PRODUCT OF VECTORS

Dot product of vectors practice · Easy

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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 91-120 of 241 questions

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91 Easy

91 of 241

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.

92 Easy

92 of 241

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.

93 Easy

93 of 241

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.

94 Easy

94 of 241

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.

95 Easy

95 of 241

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.

96 Easy

96 of 241

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.

97 Easy

97 of 241

Application +4 | -1

A unit vector

{c}

perpendicular to

{a}={i}-{j}

and coplanar with

{a}

and

{b}={i}+{k}

View answer and explanation

Correct option: A

\frac{1}{\sqrt{6}}({i}+{j}+2 {k})

Option A (

\frac{1}{\sqrt{6}}({i}+{j}+2 {k})

) is correct because it satisfies the conditions of being perpendicular to {a} and coplanar with {a} and {b}.

98 Easy

98 of 241

Application +4 | -1

The cartesian equation of the plane

\mathbf{r}=(1+\lambda-\mu) \hat{\mathbf{i}}+(2-\lambda) \hat{\mathbf{j}}+(3-2 \lambda+2 \mu) \hat{\mathbf{k}}

, is

View answer and explanation

Correct option: C

2 x+z=5

Option C is correct because the cartesian equation is derived from the given vector equation by finding the normal vector and using the point on the plane.

99 Easy

99 of 241

Analyzing +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

\hat{\mathbf{n}}

is a unit vector such that

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

and

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

, then

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

is equal to

View answer and explanation

Correct option: C

Option C (3) is correct because the unit vector

\hat{\mathbf{n}}

is found to be

\hat{\mathbf{k}}

-\hat{\mathbf{k}}

, and the magnitude of the dot product of

\mathbf{w}

and

\hat{\mathbf{n}}

is 3.

100 Easy

100 of 241

Application +4 | -1

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

are the unit vectors such that

{\mathbf{\vec a}}

is perpendicular to the plane

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

and the angle between

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

\frac{\pi }{3}

then

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

is equal to

View answer and explanation

Correct option: C

\mathrm{\;} \pm 2

Option C ( ± 2) is correct because it calculates the magnitude of the vector sum of three unit vectors, where one is perpendicular to the plane of the other two.

101 Easy

101 of 241

Understanding +4 | -1

a

and

{b}

are unit vectors and

\theta

is the angle between

{a}

and

{b}

then

\sin (\theta / 2)

is equal to

View answer and explanation

Correct option: A

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

Option D (

\mathbf{O M}=\frac{1}{2}(\mathbf{O B}+\mathbf{O A})

) is correct because the position vector of the midpoint is the average of the position vectors of the endpoints.

102 Easy

102 of 241

Analyzing +4 | -1

If the vectors

3 p+q, 5 p-3 q

and

2 p+q, 4 p-2 q

are pairs of mutually perpendicular vectors, then

\sin (\mathbf{p q})

View answer and explanation

Correct option: B

\sqrt{55} / 8

Option D (

\sqrt{247} / 16

) is correct because it is calculated by using the perpendicularity conditions to find the cosine of the angle and then calculating the sine.

103 Easy

103 of 241

Analyzing +4 | -1

A non-zero vector

{\mathbf{\vec a}}

is parallel to the line of intersection of the plane determined by vectors

\widehat \mathbf{i},\widehat \mathbf{i} - \widehat \mathbf{j}

and the plane determined by the vectors

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

\widehat \mathbf{i} - \mathbf{\hat k}

. The angle between

{\mathbf{\vec a}}

and

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

View answer and explanation

Correct option: C

\frac{\pi }{4}

Option C (

\frac{\pi }{4}

) is correct because the angle between the vector parallel to the line of intersection and the given vector is calculated using the dot product formula.

104 Easy

104 of 241

Application +4 | -1

The angle between the vectors

{i}-{j}+{k}

and

-{i}+{j}+2 {k}

View answer and explanation

Correct option: C

90^{\circ}

Option C (

90^{\circ}

) is correct because the dot product of the two vectors is 0, indicating they are orthogonal.

105 Easy

105 of 241

Analyzing +4 | -1

The vectors

\mathbf{a}=2 \lambda^2 \hat{\mathbf{i}}+4 \lambda \hat{\mathbf{j}}+\hat{\mathbf{k}}

and

\mathbf{b}=7 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}

make an obtuse angle whereas the angle between

\mathbf{b}

and

\hat{\mathbf{k}}

is acute and less than

\pi / 6

View answer and explanation

Correct option: D

null set

The correct answer is D (null set) because the conditions for the angle between a and b being obtuse and the angle between b and k being acute and less than

\pi/6

lead to contradictory constraints on

\lambda

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106 Easy

106 of 241

Understanding +4 | -1

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

, then

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

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

View answer and explanation

Correct option: C

Zero

Option C (Zero) is correct because the dot product of (a+b) and (a-b) simplifies to |a|^2 - |b|^2, which is zero when |a| = |b|.

107 Easy

107 of 241

Application +4 | -1

Let

{\mathbf{\vec a}}

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

The correct answer is D (1) because it is found by applying the law of cosines to the magnitude of the difference of two unit vectors.

108 Easy

108 of 241

Analyzing +4 | -1

Let

\vec{a}

be a vector which is perpendicular to the vector

3 \hat{i}+\frac{1}{2} \hat{j}+2 \hat{k}

. If

\vec{a} \times(2 \hat{i}+\hat{k})=2 \hat{i}-13 \hat{j}-4 \hat{k}

, then the projection of the vector

\vec{a}

on the vector

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

is :

View answer and explanation

Correct option: C

\frac{5}{3}

Option C (

\frac{5}{3}

) is correct because it is the projection of vector

\vec{a}

on the given vector, calculated using the dot product and magnitudes.

109 Easy

109 of 241

Application +4 | -1

A unit vector in the

x y

-plane which is perpendicular to

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

View answer and explanation

Correct option: B

\frac{1}{5}(3 \mathbf{i}+4 \mathbf{j})

Option B (

\frac{1}{5}(3 \mathbf{i}+4 \mathbf{j})

) is correct because it's a unit vector in the xy-plane that is perpendicular to the given vector.

110 Easy

110 of 241

Application +4 | -1

Unit vector which is perpendicular to both the vectors

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

and

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

View answer and explanation

Correct option: D

\frac{{\widehat \mathbf{i} - \widehat \mathbf{j} - \mathbf{\hat k}}}{{\sqrt 3 }}

The correct answer is D. The unit vector perpendicular to two vectors is found by taking the cross product of the vectors and then normalizing the result.

111 Easy

111 of 241

Application +4 | -1

The angle between the vectors

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

and

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

when

\mathbf{\vec a} = \left( {1,1,4} \right)

and

\mathrm{\;}\mathbf{\vec b} = \left( {1, - 1,4} \right)

View answer and explanation

Correct option: B

90^\circ

Option B (

90^\circ

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

112 Easy

112 of 241

Application +4 | -1

|\vec{a}|=2,|\vec{b}|=3

and

|2 \vec{a}-\vec{b}|=5

, then

|2 \vec{a}+\vec{b}|

equals:

View answer and explanation

Correct option: C

Option C (5) is correct because by using the given information and the properties of dot products, we can find the angle between the vectors and then calculate

|2 \vec{a}+\vec{b}|

113 Easy

113 of 241

Application +4 | -1

A force of magnitude 5 unit acting along the vector

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

displaces the point of applications from (1,2,3) to (5,3,7) then the work done is

View answer and explanation

Correct option: B

50/3 unit

Option B (50/3 unit) is correct because it is the result of the dot product of the force vector and the displacement vector.

114 Easy

114 of 241

Application +4 | -1

\mathbf{a}=4 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}

and

\mathbf{b}=3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}

, then the component of

\mathbf{a}

along

\mathbf{b}

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents the projection of vector

\mathbf{a}

onto vector

\mathbf{b}

, calculated using the formula for vector projection.

115 Easy

115 of 241

Application +4 | -1

Let

\vec{a}

and

\vec{b}

be two vectors, Let

|\vec{a}|=1,|\vec{b}|=4

and

\vec{a} \cdot \vec{b}=2

. If

\vec{c}=(2 \vec{a} \times \vec{b})-3 \vec{b}

, then the value of

\vec{b} \cdot \vec{c}

View answer and explanation

Correct option: B

-48

Option B (-48) is correct because

\vec{{b}} \cdot (2 \vec{{a}} \times \vec{{b}}) = 0

and

\vec{{b}} \cdot \vec{{b}} = |\vec{{b}}|^2 = 16

, so

\vec{{b}} \cdot \vec{{c}} = -3(16) = -48

116 Easy

116 of 241

Application +4 | -1

\mathrm{\vec \alpha } = 2\,\widehat {\mathbf{\dot i}} + 3\widehat {\,\mathbf{\dot j}} - \mathbf{\hat k},\;\,\,\;\mathrm{\vec \beta } = - \widehat {\mathbf{\dot i}} + 2\,\widehat {\mathbf{\dot j}} - 4\,\mathbf{\hat k},\;\,\,\;\mathrm{\vec \gamma } = \widehat {\mathbf{\dot i}} + \widehat {\mathbf{\dot j}} + \mathbf{\hat k}

, then what is the value of

\left( {\mathbf{\vec a} \times \mathbf{\vec b}} \right)\,\, \cdot \,\,\left( {\mathrm{\vec \alpha } \times \mathrm{\vec \gamma }} \right)\,\,?

View answer and explanation

Correct option: C

- 74

Option C (-74) is correct because it is the result of calculating the scalar triple product using the given vectors.

117 Easy

117 of 241

Application +4 | -1

A vector

\mathbf{b}

, which is collinear with vector

\mathbf{a}=

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

and satisfies

\mathbf{a} \cdot \mathbf{b}=2

is given by

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it is collinear with

\mathbf{a}

and satisfies the dot product condition

\mathbf{a} \\. \mathbf{b} = 2

118 Easy

118 of 241

Understanding +4 | -1

\theta

is the angle between vectors

{\vec a}

and

{\vec b}

such that

\vec a\,.\,\vec b \ge 0

, then

View answer and explanation

Correct option: C

0 \le \theta \le \frac{\pi }{2}

Option C (

0 \le \theta \le \frac{\pi }{2}

) is correct because the dot product of two vectors is non-negative when the angle between them is between 0 and 90 degrees.

119 Easy

119 of 241

Application +4 | -1

Area of rectangle having vertices A, B, C and D with position vector

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

and

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

View answer and explanation

Correct option: A

2 sq unit

Option A (2 sq unit) is correct because the area of the rectangle is calculated by finding the lengths of the sides using the position vectors and then multiplying them.

120 Easy

120 of 241

Application +4 | -1

\mathbf{x}

and

\mathbf{y}

are two unit vectors and

\pi

is the angle between them, then

\frac{1}{2}|x-y|

is equal to

View answer and explanation

Correct option: C

Option C (1) is correct because the magnitude of the difference between two unit vectors is related to the angle between them, and in this case, simplifies to 1.

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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