The unit vector in $ZOX$ plane and making angle $45^\circ $ and $60^\circ $ respectively with${\bf{\vec a}} = 2\widehat {\bf{i}} + 2\widehat {\bf{j}} - {\bf{\hat k}}\;{\rm{and}}\;{\bf{\vec b}} = 0\widehat {\bf{i}} + \widehat {\bf{j}} - {\bf{\hat k}},$ is

$\frac{1}{{3\sqrt 2 }}\widehat {\bf{i}} + \frac{4}{{3\sqrt 2 }}\widehat {\bf{j}} + \frac{1}{{3\sqrt 2 }}\;{\bf{\hat k}}\;$

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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 181-210 of 296 questions

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181 Easy

181 of 296

Application +4 | -1

If the scalar projection of the vector

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

on the vector

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

\frac{1}{{\sqrt {30} }}\mathrm{\;then\;the\;value\;of\;}x\;\mathrm{is\;}

View answer and explanation

Correct option: A

- 3/2

Option A (

- 3/2

) is correct because the scalar projection formula is used, and solving for x gives -3/2.

182 Easy

182 of 296

Application +4 | -1

The value of

c

so that for all real

x

, then vectors o

cx\;\widehat \mathbf{i} - 6\widehat \mathbf{j} + 3\mathbf{\hat k},\;x\widehat \mathbf{i} + 2\widehat \mathbf{j} + 2cx\mathbf{\hat k}

make an obtuse angle are

View answer and explanation

Correct option: C

- \frac{4}{3} < c < 0

Option C (

- \frac{4}{3} < c < 0

) is correct because for the angle to be obtuse, the dot product must be negative, leading to a quadratic inequality whose solution gives the range of c.

183 Easy

183 of 296

Application +4 | -1

If the vectors

\mathbf{\vec a} = \widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k},\;\mathbf{\vec b} = 2\widehat \mathbf{i} + 4\widehat \mathbf{j} + \mathbf{\hat k}

and

\mathbf{\vec c} = \mathrm{\lambda }\widehat \mathbf{i} + \widehat \mathbf{j} + \mathrm{\mu }\mathbf{\hat k}

are mutually orthogonal, then

\left( {\lambda ,\mathrm{\mu }} \right)

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the values of

\lambda

and

\mu

are found by using the orthogonality condition (dot product equals zero) between the vectors.

184 Medium

184 of 296

Application +4 | -1

The values of

c

so that for all real

x

, the vectors

c x {i}-6 {j}+3 {k}

and

x {i}+2 {j}+2 c x {k}

make an obtuse angle with each other, are

View answer and explanation

Correct option: B

\frac{-4}{3} < c < 0

Option B (

\frac{{-4}}{{3}} < c < 0

) is correct because for the vectors to make an obtuse angle, their dot product must be negative, leading to the given inequality for

c

185 Easy

185 of 296

Application +4 | -1

\mathrm{\theta }

is the angle between the lines

AB

and

AC

where

A,B

and

C

are the three points with coordinates

\left( {1,2, - 1} \right),\left( {2,0,3} \right),\left( {3, - 1,2} \right)

respectively, then

\sqrt {462} \cos \mathrm{\theta }

is equal to

View answer and explanation

Correct option: A

Option A (20) is correct because using the dot product formula to find the cosine of the angle between the vectors AB and AC, and then multiplying by

\sqrt{462}

, yields 20.

186 Easy

186 of 296

Application +4 | -1

The position vector of a point

P

r = x\widehat i + y\hat j + z\widehat k

, where

x,y, z \in N

and

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

. If

\mathbf{r} \cdot \mathbf{a}=20

, then the number of possible position of

P

View answer and explanation

Correct option: A

Option A (81) is correct because it represents the number of possible positions of point P, calculated by finding the number of positive integer solutions to the given equation.

187 Easy

187 of 296

Application +4 | -1

\theta

be the angle between the unit vectors a and

\mathbf{b}

, then

\cos \frac{\theta}{2}=

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it directly follows from the formula relating the magnitude of the sum of two unit vectors to the cosine of half the angle between them.

188 Easy

188 of 296

Application +4 | -1

\vec{A}=3 \mathbf{i}+\mathbf{j}+2 \mathbf{k}

and

\vec{B}=2 \mathbf{i}-2 \mathbf{j}+4 \mathbf{k}

and

\theta

is the angle between

\vec{A}

and

\vec{B}

, then the value of

\sin \theta

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it calculates sin(theta) using the dot product to find cos(theta) and then using the trigonometric identity sin²(theta) + cos²(theta) = 1.

189 Easy

189 of 296

Application +4 | -1

Let

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

and

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

. A vector in the plane of

{a}

and

{b}

whose projection on c is

1 / \sqrt{3}

View answer and explanation

Correct option: C

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

Option C (2i + j + 2k) is correct because it lies in the plane of a and b, and its projection on c is 1/

\sqrt{3}

190 Easy

190 of 296

Application +4 | -1

The vectors

2\hat i + 3\hat j - 4\hat k

and

a\hat i + b\hat j + c\hat k

are perpendicular when

View answer and explanation

Correct option: B

a = 4,\;b = 4,\;c = 5

Option B (

a = 4, b = 4, c = 5

) is correct because the dot product of the two vectors is zero, satisfying the condition for perpendicularity.

191 Easy

191 of 296

Analyzing +4 | -1

The resultant of forces

\vec{P}

and

\vec{Q}

\vec{R}

. If

\vec{Q}

is doubled then

\vec{R}

is doubled. If the direction of

\vec{Q}

is reversed, then

\vec{R}

is again doubled. Then

P^2: Q^2: R^2

View answer and explanation

Correct option: C

2: 3: 2

The correct answer is C (

2: 3: 2

) because it is derived by setting up and solving equations based on the given conditions about the resultant of two forces.

192 Medium

192 of 296

Application +4 | -1

If the equation of the plane passing through the point

(-1,2,0)

and parallel to the lines

\frac{x}{3}=\frac{y+1}{0}=\frac{z-2}{-1}

and

\frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-1}

a x+b y+c z=1

, then the value of

(a+b+c)

, is

View answer and explanation

Correct option: C

Option C (5) is correct because after finding the normal vector and the equation of the plane, the sum of the coefficients a, b, and c is 5.

193 Easy

193 of 296

Analyzing +4 | -1

Let

\mathbf{u}, \mathbf{v}

and

\mathbf{w}

be such that

|\mathbf{u}|=1,|\mathbf{v}|=2

|\mathbf{w}|=3

. If the projection

\mathbf{v}

along

\mathbf{u}

is equal to that of

\mathbf{w}

along

\mathbf{u}

and

\mathbf{v}, \mathbf{w}

are perpendicular to each other, then

\mathbf{u}-\mathbf{v}+\mathbf{w} \mid

equals.

View answer and explanation

Correct option: A

\sqrt{14}

The correct answer is A (

\sqrt{14}

) because the magnitude of the vector sum is found using the projection condition and perpendicularity of v and w.

194 Easy

194 of 296

Application +4 | -1

\theta

be the angle between the unit vectors

\mathbf{a}

and

\mathbf{b}

, then

\mathbf{a}-\sqrt{2} \mathbf{b}

will be a unit vector if

\theta=

View answer and explanation

Correct option: B

\frac{\pi}{4}

Option B (

\frac{\pi}{4}

) is correct because it satisfies the condition for a-sqrt(2)b to be a unit vector, derived from the dot product and magnitudes.

195 Easy

195 of 296

Application +4 | -1

\vec a = - 2\hat i + \hat j + \hat k,\;\vec b = \hat i + 5\hat j

and

\vec c = 4\hat i + 4\hat j - 2\hat k

, then the projection of

3\vec a - 2\vec b

on the axis of the vector

{\vec c}

View answer and explanation

Correct option: B

- 11

Option B (

-11

) is correct because the projection of

3\vec a - 2\vec b

on the axis of

{\vec c}

is calculated using the dot product and equals -11.

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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

196 Easy

196 of 296

Analyzing +4 | -1

A vector whose modulus is

\sqrt{51}

and makes the same angle with

\mathbf{a}=\frac{\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{3}, \mathbf{b}=\frac{-4 \hat{\mathbf{i}}-3 \hat{\mathbf{k}}}{5}

and

\mathbf{c}=\hat{\mathbf{j}}

, will be

View answer and explanation

Correct option: D

\pm(5 \hat{\mathbf{i}}-\hat{\mathbf{j}}-5 \hat{\mathbf{k}})

Option D (

\pm(5 \hat{\mathbf{i}}-\hat{\mathbf{j}}-5 \hat{\mathbf{k}})

) is correct because it satisfies the given conditions of having a modulus of

\sqrt{51}

and making the same angle with the given vectors.

197 Easy

197 of 296

Understanding +4 | -1

The vector

p=({a} \cdot {c}) {b}-({a} \cdot {b}) {c}

is perpendicular to

View answer and explanation

Correct option: C

{a}

Option A (

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

) is correct because it represents

\sin (\theta / 2)

when

a

and

b

are unit vectors and

\theta

is the angle between them.

198 Medium

198 of 296

Application +4 | -1

The unit vector in

ZOX

plane and making angle

45^\circ

and

60^\circ

respectively with

\mathbf{\vec a} = 2\widehat \mathbf{i} + 2\widehat \mathbf{j} - \mathbf{\hat k}\;\mathrm{and}\;\mathbf{\vec b} = 0\widehat \mathbf{i} + \widehat \mathbf{j} - \mathbf{\hat k},

View answer and explanation

Correct option: B

\frac{1}{{\sqrt 2 }}\widehat \mathbf{i} - \frac{1}{{\sqrt 2 }}\;\mathbf{\hat k}

Option B (

\frac{{1}}{{\sqrt 2 }}\widehat {\bf{{i}}} - \frac{{1}}{{\sqrt 2 }}\;{\bf{{\hat k}}}

) is correct because it satisfies the conditions of being a unit vector in the ZOX plane and making the specified angles with the given vectors.

199 Easy

199 of 296

Analyzing +4 | -1

Magnitude of vectors

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

are 3,4,5 respectively. If

{\mathbf{\vec a}}

and

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

and

\mathbf{\vec c} + \mathbf{\vec a},\;\mathbf{\vec c}

and

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

are mutually perpendicular, then magnitude of

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

View answer and explanation

Correct option: C

5\sqrt 2

Option C (

5\sqrt 2

) is correct because using the given perpendicularity conditions, we find that

|\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c}| = 5\sqrt{2}

200 Easy

200 of 296

Application +4 | -1

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

and

\left| {\mathbf{\vec a}} \right| = \sqrt {37} ,\;\left| {\mathbf{\vec b}} \right| = 3,\;\left| {\mathbf{\vec c}} \right| = 4

, then the angle between

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

View answer and explanation

Correct option: C

60^\circ

Option C (

60^\circ

) is correct because it is the angle between vectors

\mathbf{\vec b}

and

\mathbf{\vec c}

that satisfies the given vector equation and magnitudes.

201 Easy

201 of 296

Application +4 | -1

\vec a = 2\hat i + 2\hat j + 3\hat k,\;\hat b = - \hat i + 2\hat j + \hat k

and

\vec c = 3\hat i + \hat j

, then

\vec a + t\;\vec b

is perpendicular to

{\vec c}

, if

t

is equal to

View answer and explanation

Correct option: A

Option A (8) is correct because it satisfies the condition for perpendicularity between the vector sum and vector

{\vec c}

202 Easy

202 of 296

Application +4 | -1

\left| a \right| = 2,\,\,\left| b \right| = 3,\,\,\left| c \right| = 4

and

a + b +c = 0

then the value of

b \cdot c + c \cdot a + a \cdot b

is equal to

View answer and explanation

Correct option: D

-29/2

Option D (-29/2) is correct because it is derived from the given magnitudes and the condition a + b + c = 0 using the dot product.

203 Medium

203 of 296

Application +4 | -1

The equation of the plane passing through the line of intersection of the planes

r.(\hat i + \hat j + \hat k) = 1

and

r.(2\hat i + 3\hat j - \hat k) + 4 = 0

and parallel to the

X

-axis is

View answer and explanation

Correct option: A

r.(\hat j - 3\hat k) + 6 = 0

Option A (

r.(\hat j - 3\hat k) + 6 = 0

) is correct because it represents the equation of a plane passing through the intersection of two given planes and parallel to the X-axis.

204 Easy

204 of 296

Understanding +4 | -1

\theta

be the angle between two vectors

\mathbf{a}

and

\mathbf{b}

, then a.b

\geq 0

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the dot product a.b is non-negative when the angle between the vectors is between 0 and 90 degrees.

205 Easy

205 of 296

Analyzing +4 | -1

\mathbf{p}, \mathbf{q}

are two non-collinear vectors such that

(b-c) p \times q+(c-a) \mathbf{p}+(a-b) q=0

where

a, b, c

are lengths of sides of a triangle, then the triangle is

View answer and explanation

Correct option: C

equilateral

Option C (equilateral) is correct because the given equation implies a = b = c, which defines an equilateral triangle.

206 Medium

206 of 296

Application +4 | -1

The distance of a point

(7,-3,-4)

from the plane passing through the points

(2,-3,1),(-1,1,-2)

and

(3,-4,2)

View answer and explanation

Correct option: C

5 \sqrt{2}

The correct answer is C (

5 \sqrt{{2}}

) because it represents the distance of the point from the plane passing through the given points, calculated using the formula for the distance between a point and a plane.

207 Easy

207 of 296

Application +4 | -1

The vectors

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

and

(1+m) {i}-2 m {j}+{k}

include an acute angle for

View answer and explanation

Correct option: B

m < -2

m>\frac{-1}{2}

Option B (

m < -2

m>\frac{{-1}}{{2}}

) is correct because it represents the range of values for m that make the dot product of the two vectors positive, indicating an acute angle between them.

208 Easy

208 of 296

Application +4 | -1

Let

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

and

\mathbf{c}=x \hat{\mathbf{i}}+(x-2) \hat{\mathbf{j}}-\hat{\mathbf{k}}

. If the vector

\mathbf{c\, l i e s}

in the plane of

\mathbf{a}

and

\mathbf{b}

, then

x

equal to

View answer and explanation

Correct option: D

-2

Option D (-2) is correct because if

\mathbf{c}

lies in the plane of

\mathbf{a}

and

\mathbf{b}

, then

\mathbf{a}

\mathbf{b}

, and

\mathbf{c}

are coplanar, implying their scalar triple product is zero.

209 Easy

209 of 296

Application +4 | -1

\mathbf{a}

and

\mathbf{b}

are mutually perpendicular vectors, then

(a+b)^2=

View answer and explanation

Correct option: B

a-b

Option D (

(a-b)^2

) is correct because when a and b are mutually perpendicular, their dot product is zero, leading to

(a+b)^2 = a^2 + b^2 = (a-b)^2

210 Medium

210 of 296

Application +4 | -1

The projection of

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

on the line whose equation is

\vec{r}=(3+\lambda) \hat{i}+(3-2 \lambda) \hat{j}+(5+6 \lambda) \hat{k}

where

\lambda

is a scalar parameter, is

View answer and explanation

Correct option: B

\frac{32}{\sqrt{41}}

Option B (

\frac{32}{\sqrt{41}}

) is correct because it is the projection of the given vector onto the line's direction vector.

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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Dot product of vectors practice · Mixed difficulty

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