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 · Mixed difficulty

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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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Showing 121-150 of 296 questions

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121 Easy

121 of 296

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}.

122 Easy

122 of 296

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.

123 Easy

123 of 296

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.

124 Easy

124 of 296

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.

125 Easy

125 of 296

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.

126 Easy

126 of 296

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.

127 Easy

127 of 296

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.

128 Easy

128 of 296

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.

129 Easy

129 of 296

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

130 Medium

130 of 296

Analyzing +4 | -1

a, b, c

and

A, B, C \in R-\{0\}

such that

a A+b B+c C+\sqrt{\left(a^2+b^2+c^2\right)\left(A^2+B^2+C^2\right)}=0

, then value of

\frac{a B}{b A}=\frac{b C}{c B}+\frac{c A}{a C}

View answer and explanation

Correct option: A

Option A (3) is correct because the given condition implies that the vectors are anti-parallel, leading to a constant ratio between corresponding components, and thus the expression evaluates to 3.

131 Easy

131 of 296

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|.

132 Easy

132 of 296

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.

133 Easy

133 of 296

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.

134 Easy

134 of 296

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.

135 Easy

135 of 296

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.

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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 296

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.

137 Easy

137 of 296

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}|

138 Easy

138 of 296

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.

139 Medium

139 of 296

Application +4 | -1

The position vectors of

\mathrm{P}, \mathrm{Q}, \mathrm{R}

with respect to

\mathrm{O}

are

3 \vec{i}-2 \vec{j}+\vec{k}, \vec{i}+3 \vec{j}+4 \vec{k}

and

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

respectively. Then the distance of

\mathrm{P}

from the plane

\mathrm{OQR}

View answer and explanation

Correct option: B

\frac{11}{3}

Option B (

\frac{{11}}{{3}}

) is correct because it represents the perpendicular distance from point P to the plane OQR, calculated using the projection of OP onto the normal vector of the plane.

140 Medium

140 of 296

Application +4 | -1

Let

\vec{a}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \vec{b}=\hat{i}+\hat{k}, \vec{c}=\hat{i}+2 \hat{j}-3 \hat{k}

be three given vectors. If

\vec{r}

is a vector such that

\vec{r} \times \vec{a}=\vec{c} \times \vec{a}

and

\vec{r} \cdot \vec{b}=0

, then

|\vec{r}|

is equal to

View answer and explanation

Correct option: D

\frac{11}{7} \sqrt{2}

Option D (

\frac{11}{7} \sqrt{2}

) is correct because it is the magnitude of vector

\vec{r}

found by solving the given vector equations.

141 Easy

141 of 296

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.

142 Easy

142 of 296

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

143 Medium

143 of 296

Application +4 | -1

The position vectors of the vertices

A, B

and

C

of triangle are

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

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

respectively. Let

I

denotes the length of the angle bisector

A D

\angle B A C

where

D

is on the line segment

B C,

then

2 I^2

equals:

View answer and explanation

Correct option: C

Option C (45) is correct because it is the result of calculating 2 times the square of the length of the angle bisector AD using vector methods.

144 Easy

144 of 296

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.

145 Easy

145 of 296

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

146 Easy

146 of 296

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.

147 Easy

147 of 296

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.

148 Easy

148 of 296

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.

149 Easy

149 of 296

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.

150 Easy

150 of 296

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.

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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