Let $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{k}}-\hat{\mathbf{i}}$. If $\mathbf{d}$ is $a$ unit vector such that $\mathbf{a} \cdot \mathbf{d}=0=(b \mathbf{c}$ ) , then $\mathbf{d}$ is equal to

$\pm \frac{(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})}{\sqrt{3}}$

$\pm \frac{(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{3}}$

If ${\bf{\vec p}} = \widehat {\bf{i}} + \widehat {\bf{j}},{\bf{\vec q}} = 4{\bf{\hat k}} - \widehat {\bf{j}}\;{\rm{and}}\;{\bf{\vec r}} = \widehat {\bf{i}} + {\bf{\hat k}},$ then the unit vector in the direction of $3{\bf{\vec p}} + {\bf{\vec q}} - 2{\bf{\vec r}}$ is

$\frac{1}{3}\left( {\widehat {\bf{i}} + 2\widehat {\bf{j}} + 2{\bf{\hat k}}} \right)$

$\frac{1}{3}\left( {\widehat {\bf{i}} - 2\widehat {\bf{j}} - 2{\bf{\hat k}}} \right)$

$\frac{1}{3}\left( {\widehat {\bf{i}} - 2\widehat {\bf{j}} + 2{\bf{\hat k}}} \right)$

$\widehat {\bf{i}} + 2\widehat {\bf{j}} + 2{\bf{\hat k}}$

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JEE • MATHEMATICS • ADDITION OF VECTORS

Addition of vectors practice · Mixed difficulty

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1 Medium

1 of 318

Application +4 | -1

Let

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

and

\mathbf{c}=\hat{\mathbf{k}}-\hat{\mathbf{i}}

. If

\mathbf{d}

a

unit vector such that

\mathbf{a} \cdot \mathbf{d}=0=(b \mathbf{c}

) , then

\mathbf{d}

is equal to

View answer and explanation

Correct option: A

\pm \frac{(\hat{i}+\hat{j}-2 \hat{k})}{\sqrt{6}}

Option A is correct because the unit vector

\mathbf{d}

is perpendicular to

\mathbf{a}

and lies in the plane of

\mathbf{b}

and

\mathbf{c}

, which leads to

\mathbf{d} = \pm \frac{(\hat{i}+\hat{j}-2 \hat{k})}{\sqrt{6}}

2 Easy

2 of 318

Application +4 | -1

A vector

{\mathbf{\vec a}}

has components

2\;p

and 1 with respect to a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense, if this respect to new system

{\mathbf{\vec a}}

has components

p + 1

and 1, then

View answer and explanation

Correct option: B

p = 1\;\mathrm{or\;}p = \frac{{ - 1}}{2}

Option B (

p = 1\mathrm{\;or\;}p = \frac{{ - 1}}{2}

) is correct because the magnitude of the vector remains unchanged under rotation, leading to a quadratic equation with these solutions.

3 Easy

3 of 318

Application +4 | -1

\vec{a}, \vec{b}

and

\vec{c}

are three mutually perpendicular unit vetors and

\vec{d}

is a unit vector which makes equal angles with

\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}

ānd

\overrightarrow{\mathrm{c}}

then the value of

|\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{d}}|^2

View answer and explanation

Correct option: B

4+2 \sqrt{3}

Option B (

4+2 \sqrt{3}

) is correct because it involves using the properties of mutually perpendicular unit vectors and the given condition about the unit vector d to find the magnitude squared of the vector sum.

4 Easy

4 of 318

Application +4 | -1

The non-zero vectors are

\vec{a}, \vec{b}

and

\vec{c}

are related by

\vec{a}=8 \vec{b}

and

\vec{c}=-7 \vec{b}

. Then the angle between

\vec{a}

and

\vec{c}

View answer and explanation

Correct option: D

\pi

Option D (

\pi

) is correct because the vectors

\vec{{a}}

and

\vec{{c}}

are scalar multiples of

\vec{{b}}

with opposite signs, indicating they are anti-parallel.

5 Easy

5 of 318

Application +4 | -1

If the vectors

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

and

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

ar coplanar, then the value of

\lambda

is equal

View answer and explanation

Correct option: C

-13 / 9

Option C (-13/9) is correct because the scalar triple product of coplanar vectors is zero, leading to the equation that solves for lambda.

6 Easy

6 of 318

Analyzing +4 | -1

The medium

AD

of the triangle

ABC

is bisected at

E,\;BE

meets

AC

F

, then

AF:AC =

View answer and explanation

Correct option: B

1/3

Option B (1/3) is correct because using vector methods and the section formula, the ratio AF:AC can be determined to be 1/3.

7 Easy

7 of 318

Application +4 | -1

If a vector

{\vec r}

of magnitude

3\sqrt 6

is directed along the bisector of the angle between the vectors

\vec a = 7\hat i - 4\hat j - 4\hat k

and

\vec b = - 2\hat i - \hat j + 2\hat k

, then

\vec r =

View answer and explanation

Correct option: A

\hat i - 7\hat j + 2\hat k

The correct answer is A (

\hat i - 7\hat j + 2\hat k

) because it is the vector along the angle bisector with the given magnitude.

8 Medium

8 of 318

Application +4 | -1

The value of

b

such that scalar product of the vectors

(\mathbf{i}+\mathbf{j}+\mathbf{k})

with the unit vector parallel to the sum of the vectors

(2 \mathbf{i}+4 \mathbf{j}-5 \mathbf{k})

and

(\mathbf{b i}+2 \mathbf{j}+3 \mathbf{k})

is 1 , is

View answer and explanation

Correct option: D

Option D (1) is correct because the value of b that satisfies the given condition regarding the scalar product of vectors is found to be 1.

9 Easy

9 of 318

Application +4 | -1

The value of

\alpha

for which the vectors

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

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

and

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

are coplanar is

View answer and explanation

Correct option: B

-3

Option B (-3) is correct because it makes the scalar triple product of the three vectors equal to zero, indicating that they are coplanar.

10 Easy

10 of 318

Application +4 | -1

\vec{a}

and

\vec{b}

are non-collinear vectors, then the value of

\alpha

for which the vectors

\vec{u}=(\alpha-2) \vec{a}+\vec{b}

and

\vec{v}=(2+3 a) \vec{a}-3 \vec{b}

are collinear is:

View answer and explanation

Correct option: B

\frac{2}{3}

Option B (

\frac{{2}}{{3}}

) is correct because for

\vec{{u}}

and

\vec{{v}}

to be collinear, one must be a scalar multiple of the other. Solving the resulting equations gives

\alpha = \frac{{2}}{{3}}

11 Easy

11 of 318

Application +4 | -1

The point having position vectors

2\widehat i + 3\widehat j + 4\widehat k,\,\,3\widehat i + 4\widehat j + 2\widehat k

and

4\widehat i + 2\widehat j + 3\widehat k

are the vertices of

View answer and explanation

Correct option: C

equilateral triangle

Option C (equilateral triangle) is correct because the distances between all three points are equal, indicating that the points form an equilateral triangle.

12 Easy

12 of 318

Application +4 | -1

If the position vectors of the vertices

A, B, C

of a triangle

A B C

are

7 \mathbf{j}+10 \mathbf{k},-\mathbf{i}+6 \mathbf{j}+6 \mathbf{k}

and

-4 \mathbf{i}+9 \mathbf{j}+6 \mathbf{k}

respectively, the triangle is

View answer and explanation

Correct option: D

Right angled and isosceles also

Option D (Right angled and isosceles also) is correct because the triangle's sides satisfy both the conditions for a right-angled triangle (Pythagorean theorem) and an isosceles triangle (two sides equal).

13 Easy

13 of 318

Application +4 | -1

\mathbf{\vec p} = \widehat \mathbf{i} + \widehat \mathbf{j},\mathbf{\vec q} = 4\mathbf{\hat k} - \widehat \mathbf{j}\;\mathrm{and}\;\mathbf{\vec r} = \widehat \mathbf{i} + \mathbf{\hat k},

then the unit vector in the direction of

3\mathbf{\vec p} + \mathbf{\vec q} - 2\mathbf{\vec r}

View answer and explanation

Correct option: A

\frac{1}{3}\left( {\widehat \mathbf{i} + 2\widehat \mathbf{j} + 2\mathbf{\hat k}} \right)

Option A (1/3(i + 2j + 2k)) is correct because it calculates the unit vector in the direction of the given linear combination of vectors.

14 Easy

14 of 318

Application +4 | -1

Let

\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=10 \vec{a}+2 \vec{b}

and

\overrightarrow{O C}=\vec{b}

where

\mathrm{A}

and

\mathrm{C}

are non collinear points. Let 'p' denote the area of quadrilateral

O A B C

and let '

q

' denote the area of parallelogram with

\overrightarrow{O A}

and

\overrightarrow{O C}

as adjacent sides. If

\mathrm{p}=\mathrm{kq}

then

\mathrm{k}=

View answer and explanation

Correct option: B

Option B (6) is correct because the area of the quadrilateral OABC is 6 times the area of the parallelogram formed by vectors

\vec{a}

and

\vec{b}

15 Easy

15 of 318

Analyzing +4 | -1

\vec a,\vec b,\vec c

are the position vectors of the vertices of an equilateral triangle whose orthocenter is at the origin, then

View answer and explanation

Correct option: A

\vec a + \vec b + \vec c = \vec 0

The correct answer is A (

\vec a + \vec b + \vec c = \vec 0

) because in an equilateral triangle with the orthocenter at the origin, the centroid also lies at the origin, implying the sum of the position vectors is zero.

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16 Easy

16 of 318

Analyzing +4 | -1

Let the vectors

\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}

\vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}

and

\vec{c}=t \hat{i}-t \hat{j}+\hat{k}, t \in \mathbf{R}

be such that for

\alpha, \beta, \gamma \in \mathbf{R}, \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0}

\Rightarrow \alpha=\beta=\gamma=0

. Then, the set of all values of

t

View answer and explanation

Correct option: C

Equal to

\mathbf{R}-\{0\}

Option D (Equal to

\mathbf{R}-\{0\}

) is correct because the determinant of the matrix formed by the vectors is non-zero for all

t

except

t=0

, ensuring the vectors are linearly independent.

17 Easy

17 of 318

Application +4 | -1

If a, b and c are position vector of A, B and C respectively of

\Delta ABC

and if

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

then the distance between the centroid and incentre of

\Delta ABC

View answer and explanation

Correct option: C

\frac{1}{3}

Option C (

\frac{1}{3}

) is correct because it is calculated by finding the magnitude of the difference between the incenter and centroid position vectors.

18 Easy

18 of 318

Application +4 | -1

Let u be a vector coplanar with the vectors

a = 2\hat i + 3\hat j - \hat k

and

b = \hat j + \hat k

If u is perpendicular to a and u. b = 24, then

{\left| \mathrm{u} \right|^2}

is equal to

View answer and explanation

Correct option: A

336

Option A (336) is correct because it is the result of solving for the magnitude squared of vector u using the given conditions of coplanarity, perpendicularity, and dot product.

19 Easy

19 of 318

Application +4 | -1

A vector a has the compounds 2p and 1 w.r.t a rectangular Cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If with respect to a new system, a has components (p+1) and 1, then

View answer and explanation

Correct option: B

p = 1\,\,{\text{or}}\,\,p = - \frac{1}{3}

Option B (

p = 1\,\,{\text{or}}\,\,p = - \frac{1}{3}

) is correct because the magnitude of the vector remains unchanged after rotation, allowing us to equate the magnitudes and solve for p.

20 Easy

20 of 318

Application +4 | -1

If the position vector of the vertices,

A,B,C

of a

\Delta ABC

are

7\mathbf{\hat j} + 10\mathbf{\hat k},\mathrm{\;} - \mathbf{\hat i} + 6\mathbf{\hat j} + 6\mathbf{\hat k}

and

- 4\mathbf{\hat i} + 9\mathbf{\hat j} + 6\mathbf{\hat k}

respectively, then triangle is

View answer and explanation

Correct option: D

Right angled and isosceles also

The correct answer is D. The triangle is right angled and isosceles because two sides have equal length and satisfy the Pythagorean theorem.

21 Easy

21 of 318

Understanding +4 | -1

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

are non-coplanar vectors and

x\vec a + y\vec b + z\vec c = 0

, then

View answer and explanation

Correct option: B

x,y,z

are necessarily zero

Option B (

x,y,z

are necessarily zero) is correct because non-coplanar vectors are linearly independent.

22 Easy

22 of 318

Application +4 | -1

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

and

{c}=3 {i}+{j}

, then

t

such that

{a}+t{b}

is at right angle to

{c}

will be equal to

View answer and explanation

Correct option: A

Option A (5) is correct because the value of t is found by setting the dot product of (a+tb) and c to zero, since they are at right angles.

23 Easy

23 of 318

Understanding +4 | -1

G

and

G^{\prime}

be the centroids of the triangles

A B C

and

A^{\prime} B^{\prime} C^{\prime}

respectively, then

\overrightarrow{A A^{\prime}}+\overrightarrow{B B^{\prime}}+\overrightarrow{C C^{\prime}}=

View answer and explanation

Correct option: D

3 \overrightarrow{\mathrm{GG}}

Option D (

3 \overrightarrow{\mathrm{GG^{\prime}} }

) is correct because the sum of the vectors connecting corresponding vertices of the triangles is three times the vector connecting their centroids.

24 Easy

24 of 318

Application +4 | -1

A unit vector which is perpendicular to the vector

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

and is coplanar with the vectors

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

and

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

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it satisfies the conditions of being perpendicular to the given vector and coplanar with the other two vectors.

25 Medium

25 of 318

Analyzing +4 | -1

'I'

is the incentre of

\Delta ABC

whose corresponding sides are a, b, c respectively.

a\,IA + b\,IB + c\,IC

is always equal to

View answer and explanation

Correct option: A

Option A (0) is correct because the weighted sum of the vectors from the incenter to the vertices is zero.

26 Easy

26 of 318

Application +4 | -1

If the vectors

4\hat i + 11\hat j + m\hat k,\,\,\,7\hat i + 2\hat j + 6\hat k

and

\hat i + 5\hat j + 4\hat k

are coplanar, then

m

is equal to

View answer and explanation

Correct option: C

Option C (10) is correct because it satisfies the condition for coplanar vectors, which can be determined using the determinant of the matrix formed by the vector components.

27 Medium

27 of 318

Analyzing +4 | -1

Let

a, b, c

be three distinct real numbers, none equal to one. If the vectors

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

and

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

are coplanar, then

\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}

is equal to

View answer and explanation

Correct option: D

Option D is correct because the coplanarity condition of the vectors leads to an equation that simplifies to

\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1

28 Easy

28 of 318

Application +4 | -1

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

and

\mathbf{\vec b} = \left( { - 2,m} \right)

are two collinear vectors, then

m

is equal to

View answer and explanation

Correct option: A

Option A (2) is correct because for two vectors to be collinear, one must be a scalar multiple of the other, leading to m = 2.

29 Medium

29 of 318

Understanding +4 | -1

\bar{a}, \bar{b}, \bar{c}, \bar{d}

are the sides of a parallelogram taken in an order, then

View answer and explanation

Correct option: C

\bar{a}+\bar{c}=\bar{b}+\bar{d}=\overline{0}

Option D (

\bar{a}+\bar{c}=\bar{b}+\bar{d}=\overline{0}

) is correct because in a parallelogram, opposite sides are equal and opposite in direction, thus their sum is zero.

30 Easy

30 of 318

Application +4 | -1

\overrightarrow a ,\overrightarrow b ,\overrightarrow c

are three non-coplanar vector such that

\left[ {\begin{array}{cccccccccccccccccccc} {\overrightarrow a + \overrightarrow b + \overrightarrow c }&{\overrightarrow a - \overrightarrow b + \overrightarrow c }&{2\overrightarrow a + \overrightarrow b - \overrightarrow c } \end{array}\,} \right] = k\,\left[ {\begin{array}{cccccccccccccccccccc} {\overrightarrow a }&{\overrightarrow b }&{\overrightarrow c } \end{array}} \right]

then k is

View answer and explanation

Correct option: D

Option D (6) is correct because the scalar triple product can be expressed as a determinant, and the determinant can be simplified to find the value of k.

Revision notes and key takeaways

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

Key Takeaways

✓ This set contains 318 single-correct-answer MCQs on ADDITION OF VECTORS.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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