If ${\rm{\vec \alpha }} + {\rm{\vec \beta }} + {\rm{\vec \gamma }} = a{\rm{\vec \delta }}\;{\rm{and}}\;{\rm{\vec \beta }} + {\rm{\vec \gamma }} + {\rm{\vec \delta }} = b{\rm{\vec \alpha }}\;{\rm{and}}\;{\rm{\vec \alpha }},{\rm{\;\vec \beta }},{\rm{\;\vec \gamma }}$ are non-coplanar and ${{\rm{\vec \alpha }}}$ is not parallel to ${{\rm{\vec \delta }}}$, then ${\rm{\vec \alpha }} + {\rm{\vec \beta }} + {\rm{\vec \gamma }}$ $ + {\rm{\vec \delta }}$ equals

$\left( {a + b} \right){\rm{\vec \gamma }}$

In $\Delta ABC,\;{\rm{if\;}}2{\rm{\;}}\overrightarrow {{\bf{AC}}} = 3\overrightarrow {{\bf{CB}}} ,\;$ then $2\overrightarrow {{\bf{OA}}} + 3\overrightarrow {{\bf{OB}}} $ equals

$5\;\overrightarrow {{\bf{OC}}} $

$ - \overrightarrow {{\bf{OC}}} $

$4\;\overrightarrow {{\bf{OC}}} $

If $ABCD$ be a parallelogram and $M$ be the point of intersection of the diagonals. If $O$ is any point, then $\overrightarrow {{\bf{OA}}} + \overrightarrow {{\bf{OB}}} + \overrightarrow {{\bf{OC}}} + \overrightarrow {{\bf{OD}}} $ is

$4\overrightarrow {{\bf{OM}}} $

$3\overrightarrow {{\bf{OM}}} $

$2\overrightarrow {{\bf{OM}}} $

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JEE • MATHEMATICS • ADDITION OF VECTORS

Addition 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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61 Easy

61 of 318

Application +4 | -1

If the position vectors of A, B, C and D are

2\hat i + \hat j,\,\,\hat i - 3\hat j,\,\,3\hat i - 2\hat j

and

\hat i + \lambda \hat j,

respectively and

AB\parallel CD,

then

\lambda

will be

View answer and explanation

Correct option: B

-6

Option B (-6) is correct because it satisfies the condition for parallel vectors AB and CD, where one is a scalar multiple of the other.

62 Easy

62 of 318

Analyzing +4 | -1

\mathrm{\vec \alpha } + \mathrm{\vec \beta } + \mathrm{\vec \gamma } = a\mathrm{\vec \delta }\;\mathrm{and}\;\mathrm{\vec \beta } + \mathrm{\vec \gamma } + \mathrm{\vec \delta } = b\mathrm{\vec \alpha }\;\mathrm{and}\;\mathrm{\vec \alpha },\mathrm{\;\vec \beta },\mathrm{\;\vec \gamma }

are non-coplanar and

{\mathrm{\vec \alpha }}

is not parallel to

{\mathrm{\vec \delta }}

, then

\mathrm{\vec \alpha } + \mathrm{\vec \beta } + \mathrm{\vec \gamma }

+ \mathrm{\vec \delta }

equals

View answer and explanation

Correct option: C

The correct answer is C (0) because, given the conditions and the non-coplanar nature of the vectors, the sum of all vectors must equal zero.

63 Medium

63 of 318

Application +4 | -1

The lines whose vector equations are

r=2 \hat{i}-3 \hat{j}+7 \hat{k}+\lambda(2 \hat{i}+p \hat{j}+5 \hat{k})

and

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

are perpendicular for all values of

\lambda

and

\mu

p=

View answer and explanation

Correct option: A

Option A (6) is correct because the dot product of the direction vectors of the two lines must be zero for them to be perpendicular, leading to p = 6.

64 Easy

64 of 318

Application +4 | -1

\Delta ABC,\;\mathrm{if\;}2\mathrm{\;}\overrightarrow {\mathbf{AC}} = 3\overrightarrow {\mathbf{CB}} ,\;

then

2\overrightarrow {\mathbf{OA}} + 3\overrightarrow {\mathbf{OB}}

equals

View answer and explanation

Correct option: A

5\;\overrightarrow {\mathbf{OC}}

Option A (

5\;\overrightarrow {\mathbf{OC}}

) is correct because using the given relation and vector addition, we can express

2\overrightarrow {\mathbf{OA}} + 3\overrightarrow {\mathbf{OB}}

5\overrightarrow {\mathbf{OC}}

65 Easy

65 of 318

Application +4 | -1

If the vectors

\mathbf{\vec a} = 2\widehat \mathbf{i} + \widehat \mathbf{j} + 4\mathbf{\hat k},\;\mathbf{\vec b} = 4\widehat \mathbf{i} - 2\widehat \mathbf{j} + 3\mathbf{\hat k}

and

\mathbf{\vec c} = 2\widehat \mathbf{i} - 3\widehat \mathbf{j} - \mathrm{\lambda }\mathbf{\hat k}

are coplanar, then the value of

\mathrm{\lambda }

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because when the vectors are coplanar, the determinant of their components equals zero, leading to lambda = 1.

66 Easy

66 of 318

Analyzing +4 | -1

Orthocentre of an equilateral triangle ABC is the origin O. If OA = a, OB = b, OC = c, then AB+2BC+3CA is equal to

View answer and explanation

Correct option: B

Option B (3a) is correct because, in an equilateral triangle, the centroid and orthocenter coincide, and using the given condition OA + OB + OC = 0, the expression AB + 2BC + 3CA simplifies to 3a.

67 Easy

67 of 318

Application +4 | -1

The point collinear with

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

among the following is

View answer and explanation

Correct option: C

\left( {0, - 4, - 6} \right)

Option C ((0, -4, -6)) is correct because the scalar triple product of the vectors formed by the points is zero, indicating collinearity.

68 Easy

68 of 318

Application +4 | -1

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

and

\mathbf{c}=\mathbf{i}+\alpha \mathbf{j}

are coplanar vectors, the value of

\alpha

View answer and explanation

Correct option: C

\frac{4}{3}

Option C (

\frac{4}{3}

) is correct because the scalar triple product of coplanar vectors is zero, leading to the equation that solves for

\alpha = \frac{4}{3}

69 Easy

69 of 318

Application +4 | -1

Let a unit vector

\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}

make angles

\frac{\pi}{2}, \frac{\pi}{3}

and

\frac{2 \pi}{3}

with the vectors

\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}

and

\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}

respectively, If

\vec{v}=\frac{1}{2} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}

, then

|\hat{u}-\vec{v}|^2

is equal to

View answer and explanation

Correct option: D

\frac{5}{2}

The correct answer is D (5/2) because it is the squared magnitude of the difference between the unit vector

\hat{u}

and the vector

\vec{v}

, calculated using the given angle conditions.

70 Medium

70 of 318

Analyzing +4 | -1

The points

(5,-4,2)(4,-3,1),(7,-6,4)

and

(8,-7,5)

are the vertices of

View answer and explanation

Correct option: C

a parallelogram

The correct answer is C (a parallelogram) because the opposite sides are parallel, but adjacent sides are not perpendicular.

71 Medium

71 of 318

Analyzing +4 | -1

The line, that is coplanar to the line

\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}

, is

View answer and explanation

Correct option: B

\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}

Option B (

\frac{{x+1}}{{-1}}=\frac{{y-2}}{{2}}=\frac{{z-5}}{{5}}

) is correct because the lines are coplanar, meaning the distance between them is zero.

72 Easy

72 of 318

Application +4 | -1

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

and

\vec{c}=\hat{i}+\hat{j}+(2 r-1) \hat{k}

are three vectors such that

\vec{c}

is parallel to the plane of

\vec{a}

and

\vec{b}

, then

r

is equal to

View answer and explanation

Correct option: C

Option C (0) is correct because if

\vec{{c}}

is parallel to the plane of

\vec{{a}}

and

\vec{{b}}

, then

\vec{{a}}

\vec{{b}}

, and

\vec{{c}}

are coplanar, meaning their scalar triple product is zero. Solving the determinant equation gives

r=0

73 Easy

73 of 318

Analyzing +4 | -1

Let

a, b

and

c

be distinct non-negative numbers and the vectors

a \hat{\mathbf{i}}+a \hat{\mathbf{j}}+c \hat{\mathbf{k}}, \hat{\mathbf{i}}+\hat{\mathbf{k}}, c \hat{\mathbf{i}}+c \hat{\mathbf{j}}+b \hat{\mathbf{k}}

lie in a plane, then the quadratic equation

a x^2+2 c x+b=0

has

View answer and explanation

Correct option: A

real and equal roots

Option A is correct because the condition for coplanarity leads to

c^2 = ab

, which implies the discriminant of the quadratic equation is zero, resulting in real and equal roots.

74 Medium

74 of 318

Analyzing +4 | -1

Let the vectors

\vec{u}_{1}=\hat{i}+\hat{j}+a \hat{k}, \, \vec{u}_{2}=\hat{i}+b \hat{j}+\hat{k}

and

\vec{u}_{3}=c \hat{i}+\hat{j}+\hat{k}

be coplanar. If the vectors

\vec{v}_{1}=(a+b) \hat{i}+c \hat{j}+c \hat{k}

\vec{v}_{2}=a \hat{i}+(b+c) \hat{j}+a \hat{k}

and

\vec{v}_{3}=b \hat{i}+b \hat{j}+(c+a) \hat{k}

are also coplanar, then

6(a+b+c)

is equal to

View answer and explanation

Correct option: C

Option C, 12, is the correct value of

6(a+b+c)

when the given vectors

\vec{u}_1, \vec{u}_2, \vec{u}_3

and

\vec{v}_1, \vec{v}_2, \vec{v}_3

are coplanar. This is found by setting the determinants of the matrices formed by the vectors to zero and solving the resulting equations.

75 Easy

75 of 318

Analyzing +4 | -1

If a, b and c are any three vectors forming a linearly independent system, then

\theta \in R,

[a\cos \theta + b\sin \theta + \cos 2\theta ,\,\,\,a\cos \left( {\frac{{2\pi }}{3} + \theta } \right) + b\sin \left( {\frac{{2\pi }}{3} + \theta } \right) + c\,\cos 2\left( {\frac{{2\pi }}{3} + \theta } \right)

a\cos \left( {\theta - \frac{{2\pi }}{3}} \right) + b\sin \left( {\theta - \frac{{2\pi }}{3}} \right) + c\,\cos 2\left( {\theta - \frac{{2\pi }}{3}} \right)

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the determinant simplifies to 0, making the scalar triple product zero.

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76 Easy

76 of 318

Analyzing +4 | -1

Let

a, b, c

denote the lengths of the sides of

a

triangle such that

(a-b) \mathbf{u}+(b-c) \mathbf{v}+(c-a)(\mathbf{u} \times \mathbf{v})=0

for any two non-collinear vectors

u

and

\mathbf{v}

, then the triongle is

View answer and explanation

Correct option: B

equilateral

Option B (equilateral) is correct because for the given vector equation to hold for any non-collinear vectors u and v, the coefficients must be zero, implying a=b=c.

77 Easy

77 of 318

Analyzing +4 | -1

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

are three non-coplanar vectors such that

\vec a + \;\vec b + \;\vec c = \alpha \vec d

and

\vec b + \vec c + \vec d = \beta \vec a

, then

\vec a + \vec b + \vec c + \vec d

is equal to

View answer and explanation

Correct option: A

{\vec 0}

Option A (

\vec 0

) is correct because the given conditions lead to a contradiction unless the sum of the vectors is zero.

78 Easy

78 of 318

Understanding +4 | -1

For two particular vectors

\mathbf{A}

and

\mathbf{B}

, it is known that

A \times B=B \times A

. What must be true about the two vectors?

View answer and explanation

Correct option: C

One of the two vectors is a scalar multiple of the other vector

The correct answer is C (One of the two vectors is a scalar multiple of the other vector) because if

\mathbf{A} \times \mathbf{B} = \mathbf{B} \times \mathbf{A}

, then

\mathbf{A}

and

\mathbf{B}

must be parallel or one of them is a zero vector, meaning one is a scalar multiple of the other.

79 Easy

79 of 318

Application +4 | -1

a = 2\widehat i + 5\widehat j

and

b = 2\widehat i - \widehat j,

then the unit vector along

a + b

will be

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it represents the unit vector in the direction of the sum of the given vectors.

80 Easy

80 of 318

Analyzing +4 | -1

Let a, b and c be three non-zero vectors which are pairwise non-collinear. If

a + 3b

collinear with c and

b + 2c

is collinear with a, then

a + 3b + 6c

View answer and explanation

Correct option: D

Option D (0) is correct because the collinearity conditions lead to a linear combination of a, b, and c equaling the zero vector.

81 Medium

81 of 318

Application +4 | -1

\vec{a}

and

\vec{c}

are unit vectors and

|\vec{b}|=4

. If angle between

\vec{b}

and

\vec{c}

\cos ^{-1}\left(\frac{1}{4}\right)

and

\vec{a} \times \vec{b}=2 \vec{a} \times \vec{c}

, then

\vec{b}=\lambda \vec{a}+2 \vec{c}

, where

\lambda

is equal to

View answer and explanation

Correct option: D

\pm 4

Option D (

\pm 4

) is correct because by using the given conditions and vector properties, we can derive that

\lambda = \pm 4

82 Medium

82 of 318

Application +4 | -1

The position vector of a point at a distance of

3 \sqrt{11}

units from

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

on a line passing through the points

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

and

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

View answer and explanation

Correct option: B

-8 \mathbf{i}-4 \mathbf{j}-\mathbf{k}

Option B (-8i - 4j - k) is the correct position vector, found by considering both directions along the line at the given distance.

83 Easy

83 of 318

Application +4 | -1

Let

\vec a,\overrightarrow {\;b} ,\;\vec c

are three non-coplanar vectors such that

\overrightarrow {{r_1}} = \vec a - \vec b + \vec c,\;\overrightarrow {{r_2}} = \vec b + \vec c - \vec a,\;\overrightarrow {{r_3}} = \vec c + \vec a + \vec b,\;\vec r = 2\vec a - 3\vec b + 4\vec c

\vec r = {\lambda _1}\overrightarrow {{r_1}} + {\lambda _2}\overrightarrow {{r_2}} + {\lambda _3}\overrightarrow {{r_3}}

, then

View answer and explanation

Correct option: B

{\lambda _1} + {\lambda _3} = 3

The correct answer is B. By expressing

\vec r

as a linear combination of

\overrightarrow {{r_1}}

\overrightarrow {{r_2}}

, and

\overrightarrow {{r_3}}

and solving the resulting system of equations, we find that

{\lambda _1} + {\lambda _3} = 3

84 Easy

84 of 318

Application +4 | -1

If the vectors

\vec a = 2\hat i + 3\hat j + 6\hat k

are collinear and

\left| {\vec b} \right| = 21

, then

\vec b =

View answer and explanation

Correct option: A

\pm 3\left( {2\hat i + 3\hat j + 6\hat k} \right)

Option A (

\pm 3\left( {2\hat i + 3\hat j + 6\hat k} \right)

) is correct because it represents a vector collinear with

\vec a

and having a magnitude of 21.

85 Easy

85 of 318

Application +4 | -1

If the vector b is collinear with the vector

a = (2\sqrt 2 , - 1,4)

and

\left| b \right| = 0

,then

View answer and explanation

Correct option: C

2a \pm b = 0

Option C (

2a \pm b = 0

) is correct because it represents the relationship between vectors a and b when b is collinear with a and |b| = 10.

86 Easy

86 of 318

Analyzing +4 | -1

The lines

x=a y-1=z-2

and

x=3 y-2=b z-2,(a b \neq 0)

are coplanar, if:

View answer and explanation

Correct option: B

b=1, a \in R-\{0\}

Option B (b=1, a ∈ R-{0}) is correct because it satisfies the coplanarity condition derived from setting the scalar triple product of the direction vectors and the vector connecting the lines to zero.

87 Easy

87 of 318

Understanding +4 | -1

ABCD

be a parallelogram and

M

be the point of intersection of the diagonals. If

O

is any point, then

\overrightarrow {\mathbf{OA}} + \overrightarrow {\mathbf{OB}} + \overrightarrow {\mathbf{OC}} + \overrightarrow {\mathbf{OD}}

View answer and explanation

Correct option: B

4\overrightarrow {\mathbf{OM}}

Option B (

4\overrightarrow {\mathbf{OM}}

) is correct because the sum of the position vectors of the vertices of a parallelogram with respect to any point O is four times the position vector of the intersection point M of the diagonals.

88 Easy

88 of 318

Application +4 | -1

If three points A, B and C are collinear, whose position vectors are

\hat i - 2\hat j - 8\hat k,\,\,\,5\hat i - 2\hat k{\text{ }}\,\,{\text{and}}\,\,\,{\text{ }}11\hat i + 3\hat j + 7\hat k

respectively, then the ratio in which B divides AC is

View answer and explanation

Correct option: B

02:03

Option B (02:03) is correct because it represents the ratio in which point B divides the line segment AC, calculated using the section formula.

89 Easy

89 of 318

Application +4 | -1

If the position vectors of the points

A, B, C, D

2 \mathbf{i}+3 \mathbf{j}+5 \mathbf{k}, \quad \mathbf{i}+2 \mathbf{j}+3 \mathbf{k},-5 \mathbf{i}+4 \mathbf{j}-2 \mathbf{k}

and

\mathbf{i}+10 \mathbf{j}+10 \mathbf{k}

respectively, then

View answer and explanation

Correct option: B

\overrightarrow{A B} \| \overrightarrow{C D}

Option B (

\overrightarrow{A B} || \overrightarrow{C D}

) is correct because the vectors

\overrightarrow{A B}

and

\overrightarrow{C D}

are scalar multiples of each other, indicating they are parallel.

90 Easy

90 of 318

Application +4 | -1

The value of λ, for which the four points

2\widehat \mathbf{i} + 3\widehat \mathbf{j} - \mathbf{\hat k},\mathrm{\;\;}\widehat \mathbf{i} - 2\widehat \mathbf{j} + 3\mathbf{\hat k},3\widehat \mathbf{i} + 4\widehat \mathbf{j} - 2\mathbf{\hat k}\mathrm{\;},\widehat \mathbf{i} - 6\widehat \mathbf{j} + \mathrm{\lambda }\mathbf{\hat k}\mathrm{\;}

are coplanar, is

View answer and explanation

Correct option: C

Option C (6) is correct because it is the value of λ that makes the four points coplanar, which is determined by setting the scalar triple product of the vectors formed by these points equal to zero.

Revision notes and key takeaways

Review ADDITION OF VECTORS in your textbook or class notes, then use this set to check your understanding.

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