The equation of the plane perpendicular to the line $\frac{{x - 1}}{1} = \frac{{y - 2}}{{ - 1}} = \frac{{z + 1}}{2}{\rm{\;and\;passing\;through\;the\;point}}\left( {2,3,1} \right)\;{\rm{is\;}}$

${\bf{\vec r}}\cdot\left( {\widehat {\bf{i}} - \widehat {\bf{j}} + 2{\bf{\hat k}}} \right) = 1$

${\bf{\vec r}}\cdot\left( {\widehat {\bf{i}} + \widehat {\bf{j}} + 2{\bf{\hat k}}} \right) = 1$

${\bf{\vec r}}\cdot\left( {\widehat {\bf{i}} - \widehat {\bf{j}} + 2{\bf{\hat k}}} \right) = 7$

${\bf{\vec r}}\cdot\left( {\widehat {\bf{i}} + \widehat {\bf{j}} - 2{\bf{\hat k}}} \right) = 10$

The plane through the point $\left( { - 1, - 1, - 1} \right)$ and containing the line of intersection of the planes ${\bf{\vec r}}\,\, \cdot \,\,\left( {\widehat {{\bf{\dot i}}} + 3\,\widehat {{\bf{\dot j}}} - {\bf{\hat k}}} \right) = 0$ and ${\bf{\vec r}}\,\, \cdot \,\,\left( {\widehat {{\bf{\dot j}}} + 2\,{\bf{\hat k}}} \right) = 0$ is

${\bf{\vec r}}\,\, \cdot \,\,\left( {\widehat {{\bf{\dot i}}} + 2\widehat {{\bf{\dot j}}} - 3{\bf{\hat k}}} \right) = 0$

${\bf{\vec r}}\,\, \cdot \,\,\left( {\widehat {{\bf{\dot i}}} + 4\widehat {{\bf{\dot j}}} + {\bf{\hat k}}} \right) = 0$

${\bf{\vec r}}\,\, \cdot \,\,\left( {\widehat {{\bf{\dot i}}} + 5\widehat {{\bf{\dot j}}} - 5{\bf{\hat k}}} \right) = 0$

${\bf{\vec r}}\,\, \cdot \,\,\left( {\widehat {{\bf{\dot i}}} + \widehat {{\bf{\dot j}}} + 3{\bf{\hat k}}} \right) = 0$

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JEE • MATHEMATICS • DOT PRODUCT OF VECTORS

Dot product of vectors practice · Easy

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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 151-180 of 241 questions

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151 Easy

151 of 241

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.

152 Easy

152 of 241

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}

153 Easy

153 of 241

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.

154 Easy

154 of 241

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.

155 Easy

155 of 241

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.

156 Easy

156 of 241

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.

157 Easy

157 of 241

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.

158 Easy

158 of 241

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.

159 Easy

159 of 241

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.

160 Easy

160 of 241

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}

161 Easy

161 of 241

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.

162 Easy

162 of 241

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}

163 Easy

163 of 241

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.

164 Easy

164 of 241

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.

165 Easy

165 of 241

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.

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166 Easy

166 of 241

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.

167 Easy

167 of 241

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.

168 Easy

168 of 241

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

169 Easy

169 of 241

Application +4 | -1

A particle is acted on by a force of 6 units in the direction

9\widehat \mathbf{i} + 6\widehat \mathbf{j} + 2\mathbf{\hat k}

and is displaced from the point

3\widehat \mathbf{i} + 4\widehat \mathbf{j} - 15\mathbf{\hat k}

to the point

7\widehat \mathbf{i} - 6\widehat \mathbf{j} + 8\mathbf{\hat k}

. The work done is

View answer and explanation

Correct option: C

Option C (12) is correct because it is the result of the dot product of the force vector and the displacement vector.

170 Easy

170 of 241

Application +4 | -1

(\mathbf{a} \times \mathbf{b})^2+(\mathbf{a} \cdot \mathbf{b})^2=144

and

|\mathbf{a}|=4

, then

|\mathbf{b}|=

View answer and explanation

Correct option: C

Option C (3) is correct because it uses the identity relating the dot product and cross product to the magnitudes of the vectors to solve for |b|.

171 Easy

171 of 241

Analyzing +4 | -1

The vectors

\mathbf{a}

and

\mathbf{b}

are not perpendicular and

\mathbf{c}

and

\mathbf{d}

are two vectors satisfying

\mathbf{b} \times \mathbf{c}=\mathbf{b} \times \mathbf{d}

and

\mathbf{a} \cdot \mathbf{b}=0

. Then, the vector

\mathbf{d}

is equal to

View answer and explanation

Correct option: C

\mathbf{c}-\left(\frac{\mathrm{a} \cdot \mathrm{c}}{\mathbf{a} \cdot \mathbf{b}}\right) \mathbf{b}

Option C (

\mathbf{c}-\left(\frac{{\mathrm{a} \cdot \mathrm{c}}}{{\mathbf{a} \cdot \mathbf{b}}}\right) \mathbf{b}

) is correct because it satisfies the given conditions

\mathbf{b} \times \mathbf{c}=\mathbf{b} \times \mathbf{d}

and

\mathbf{a} \cdot \mathbf{d}=0

172 Easy

172 of 241

Application +4 | -1

Let

a = 2\widehat i - 3\widehat j + 4\widehat k

and

\;b = 7\hat i\mathrm{ } + \mathrm{ }\hat j\mathrm{ } - 6\hat k.

r \times a = r \times b,r.(\widehat i + 2\widehat j + \widehat k) = - 3

then.

r(2\hat i - \mathrm{ }3\hat j + \widehat k)

is equal to

View answer and explanation

Correct option: A

Option A (12) is correct. The solution involves using the properties of the cross product and dot product of vectors to find the value of r.(2i - 3j + k).

173 Easy

173 of 241

Application +4 | -1

\vec a,\;\vec b,\;\vec c

are vectors such that

\vec a.\vec b = 0

and

\vec a + \vec b = \vec c

, then

View answer and explanation

Correct option: A

{\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} = {\left| {\vec c} \right|^2}

Option A (

{\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} = {\left| {\vec c} \right|^2}

) is correct because it represents the Pythagorean theorem in vector form, given that

\vec a

and

\vec b

are orthogonal.

174 Easy

174 of 241

Application +4 | -1

The equation of the plane perpendicular to the line

\frac{{x - 1}}{1} = \frac{{y - 2}}{{ - 1}} = \frac{{z + 1}}{2}\mathrm{\;and\;passing\;through\;the\;point}\left( {2,3,1} \right)\;\mathrm{is\;}

View answer and explanation

Correct option: B

\mathbf{\vec r}\cdot\left( {\widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k}} \right) = 1

The correct answer is C (

\mathbf{\vec r}\cdot\left( {\widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k}} \right) = 7

) because it is the equation of the plane that is perpendicular to the given line and passes through the given point.

175 Easy

175 of 241

Application +4 | -1

Let

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

and

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

be three vectors. A vector

\vec{v}

in the plane of

\vec{a}

and

\vec{b}

, whose projection on

\vec{c}

\frac{1}{\sqrt{3}}

, is given by

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it represents a vector in the plane of

\vec{a}

and

\vec{b}

with the specified projection on

\vec{c}

176 Easy

176 of 241

Analyzing +4 | -1

Let the cosine of angle between the vectors

\mathrm{p}

and

\mathbf{q}

\lambda

such that

2 \mathbf{p}+\mathbf{q}=\hat{\mathbf{i}}+\hat{\mathbf{j}}

and

\mathbf{p}+2 \mathbf{q}=\hat{\mathbf{i}}-\hat{\mathbf{j}}

, then

\lambda

is equal to

View answer and explanation

Correct option: B

-\frac{4}{5}

The correct answer is B (-

\frac{4}{5}

) because the cosine of the angle between the vectors p and q is calculated using the given equations and the dot product formula.

177 Easy

177 of 241

Application +4 | -1

The work done by the force

\vec F = 2\hat i - \hat j - \hat k

in moving an object along the vector

3\hat i + 2\hat j - 5\hat k

View answer and explanation

Correct option: C

9 units

Option C (9 units) is correct because it calculates the work done by taking the dot product of the force and displacement vectors.

178 Easy

178 of 241

Analyzing +4 | -1

The angle

\theta

between two non-zero vectors

\mathbf{a}

and

\mathbf{b}

satisfies the relation

\cos \theta=(\mathbf{a} \times \hat{\mathbf{i}}) \cdot(\mathbf{b} \times \hat{\mathbf{i}})+(\mathbf{a} \times \hat{\mathbf{j}}) \cdot(\mathbf{b} \times \hat{\mathbf{j}})+(\mathbf{a} \times \hat{\mathbf{k}}) \cdot(\mathbf{b} \times \hat{\mathbf{k}})

then the least value of

|\mathbf{a}|+|\mathbf{b}|

is equal to (where

\theta \neq 90^{\circ}

)

View answer and explanation

Correct option: C

\sqrt{2}

Option C is correct because by simplifying the given expression using vector identities and applying AM >= GM, the least value of |a| + |b| is found to be

\sqrt{2}

179 Easy

179 of 241

Application +4 | -1

If three vectors

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

satisfy

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

and

|\mathbf{a}|=3

|\mathbf{b}|=5,|\mathbf{c}|=7

, then the angle between

\mathbf{a}

and

\mathbf{b}

View answer and explanation

Correct option: C

60^{\circ}

Option C (

60^{\circ}

) is correct because by using the given condition and squaring the equation, we can find the cosine of the angle between vectors a and b.

180 Easy

180 of 241

Application +4 | -1

The plane through the point

\left( { - 1, - 1, - 1} \right)

and containing the line of intersection of the planes

\mathbf{\vec r}\,\, \cdot \,\,\left( {\widehat {\mathbf{\dot i}} + 3\,\widehat {\mathbf{\dot j}} - \mathbf{\hat k}} \right) = 0

and

\mathbf{\vec r}\,\, \cdot \,\,\left( {\widehat {\mathbf{\dot j}} + 2\,\mathbf{\hat k}} \right) = 0

View answer and explanation

Correct option: A

\mathbf{\vec r}\,\, \cdot \,\,\left( {\widehat {\mathbf{\dot i}} + 2\widehat {\mathbf{\dot j}} - 3\mathbf{\hat k}} \right) = 0

Option A is correct because the equation of the plane is found by considering the linear combination of the two given planes and using the given point to solve for the constant.

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