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 ${\bf{\vec a}} = \left( {2,1, - 1} \right),\;{\bf{\vec b}} = \left( {1, - 1,0} \right),\;{\bf{\vec c}} = \left( {5 - 1,1} \right),\;$ then unit vector parallel to ${\bf{\vec a}} + {\bf{\vec b}} - {\bf{\vec c}}$ but in opposite direction is

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

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

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

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JEE • MATHEMATICS • ADDITION OF VECTORS

Addition of vectors practice · Medium

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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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1 Medium

1 of 49

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 Medium

2 of 49

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.

3 Medium

3 of 49

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.

4 Medium

4 of 49

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

5 Medium

5 of 49

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.

6 Medium

6 of 49

Application +4 | -1

For a non-zero vector a, the set of real number satisfying the inequality

|(5-x) {a}| < |2 {a}|

consists of all

x

such that

View answer and explanation

Correct option: B

3 < x < 7

Option B (

3 < x < 7

) is correct because it represents the solution set for the inequality after simplification and solving for x.

7 Medium

7 of 49

Application +4 | -1

A straight line is given by

\mathbf{r}=(1+t) \mathbf{i}+3 t \mathbf{j}+(1-t) \mathrm{k}

, where

t \in R

. If this line lies in the plane

x+y+c z=d

, then the value of

(c+d)

View answer and explanation

Correct option: D

Option D (9) is correct because by substituting the line equation into the plane equation, we can find the values of c and d, and their sum is 9.

8 Medium

8 of 49

Analyzing +4 | -1

The lines

r = (\widehat i - \widehat j) + l(2\widehat i + \widehat k)

and

r = (2\widehat i - \widehat j) + m(\widehat i + \widehat j - \widehat k)

View answer and explanation

Correct option: A

do not intersect for any values of

l

and

m

Option A do not intersect for any values of

l

and

m

is correct because the system of equations derived from equating the two lines has no solution.

9 Medium

9 of 49

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.

10 Medium

10 of 49

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.

11 Medium

11 of 49

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.

12 Medium

12 of 49

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.

13 Medium

13 of 49

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

14 Medium

14 of 49

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.

15 Medium

15 of 49

Analyzing +4 | -1

Let

\vec{a}, \vec{b}

and

\vec{c}

be three non-zero vectors such that no two of these are collinear. If the vector

\vec{a}+2 \vec{b}

is collinear with

\vec{c}

and

\vec{b}+3 \vec{c}

is collinear with

\vec{a}

(

\lambda

being some non-zero scalar), then

\vec{a}+2 \vec{b}+6 \vec{c}

equals

View answer and explanation

Correct option: D

\overrightarrow{0}

Option D (

\overrightarrow{0}

) is correct because by manipulating the collinearity conditions, we find that

\vec{a}+2 \vec{b}+6 \vec{c}

equals the zero vector.

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16 Medium

16 of 49

Application +4 | -1

A unit vector a makes an angle

\frac{\pi }{4}

with Z – axis. If

a = \hat i + \hat j

is a unit vector, then a is equal to

View answer and explanation

Correct option: C

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

Option D (None of these) is incorrect. The correct answer is

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

, which is derived by solving the vector equations given the angle with the z-axis and the magnitude.

17 Medium

17 of 49

Analyzing +4 | -1

In a quadrilateral

P Q R S, \mathbf{P Q}=a, \mathbf{Q R}=b

\mathbf{S P}=a-b

. IfM is the mid-point of

Q R

and

X

is a point of

S M

such that,

S X=\frac{4}{5} S M

, then

View answer and explanation

Correct option: B

\mathbf{P X}=\frac{3}{5} \mathbf{P R}

Option B is correct because by expressing the vectors in terms of position vectors and using the given ratios, PX simplifies to (3/5)PR.

18 Medium

18 of 49

Application +4 | -1

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

are unit vectors such that

\bar{a}

is perpendicular to plane of

\bar{b}

and

\bar{c}

and the angle between

\bar{b} \& \bar{c}

\frac{\pi}{3}

the

|\bar{a}+\bar{b}+\bar{c}|

View answer and explanation

Correct option: A

Option A (2) is correct because the magnitude of the vector sum is calculated using the given conditions of unit vectors and perpendicularity.

19 Medium

19 of 49

Understanding +4 | -1

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

are three non-coplanar vector, then

\frac{\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}}{\mathbf{c} \times \mathbf{a} \cdot \mathbf{b}}+\frac{\mathbf{b} \cdot \mathbf{a} \times \mathbf{c}}{\mathbf{c} . \mathbf{a} \times \mathbf{b}}=

View answer and explanation

Correct option: A

Option A is correct because the expression simplifies to 1 - 1 = 0 due to the properties of scalar triple products.

20 Medium

20 of 49

Application +4 | -1

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

are non-coplanar vectors and

(x+y-3) \vec{a}_1+(2 x-y+2) \vec{a}_2+(2 x+y+\lambda) \vec{a}_3=\overrightarrow{0}

holds for some '

\mathrm{x}

' and '

\mathrm{y}

' then

\lambda

View answer and explanation

Correct option: C

-\frac{10}{3}

Option C (

\frac{{-10}}{{3}}

) is correct because since the vectors are non-coplanar, the coefficients must be zero, leading to a system of equations that can be solved for x, y, and

\lambda

21 Medium

21 of 49

Application +4 | -1

The position vectors of three points

{A}, {B}

and

{C}

are

{i}+{j}

{i}-{j}

and

l {i}+m {j}+n {k}

respectively. The vectors

{O A}

OB, OC

are coplanar if

View answer and explanation

Correct option: C

n=0, l, m

are scalar

Option C (

n=0, l, m

are scalar) is correct because for the vectors to be coplanar, their scalar triple product must be zero, which leads to the condition n=0.

22 Medium

22 of 49

Analyzing +4 | -1

\mathbf{p}=2 \mathbf{a}-3 \mathbf{b}, \mathbf{q}=\mathbf{a}-2 \mathbf{b}+\mathbf{c}

and

\mathbf{r}=-3 \mathbf{a}+\mathbf{b}+2 \mathbf{c}

, where

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

being non-zero non-coplanar vectors, then the vector

-2 \mathbf{a}+3 \mathbf{b}-\mathbf{c}

is equal to

View answer and explanation

Correct option: B

\frac{-7 \mathbf{q}+\mathbf{r}}{5}

Option B is correct because expressing the vector

-2\mathbf{a}+3\mathbf{b}-\mathbf{c}

as a linear combination of

\mathbf{p}, \mathbf{q}, \mathbf{r}

leads to the solution

\frac{-7 \mathbf{q}+\mathbf{r}}{5}

23 Medium

23 of 49

Application +4 | -1

A unit tangent vector at

t=2

on the curve

x=t^2+2, y=4 t^3-5, z=2 t^2-6 t

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the calculated unit tangent vector at t=2 does not match any of the given options.

24 Medium

24 of 49

Analyzing +4 | -1

|a + b| < |a - b|,

then the angle between a and b can lie in the interval.

View answer and explanation

Correct option: C

(\pi /2,\,\,3\pi /2)

Option C (

\pi /2,\,3\pi /2

\)) is correct because

|a + b| < |a - b|

implies that the angle between a and b is obtuse, lying between 90 and 270 degrees.

25 Medium

25 of 49

Application +4 | -1

The image of the point having the position vector

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

in the plane

\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0

View answer and explanation

Correct option: C

-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}

Option C (

-3 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}

) is correct. By finding the equation of the line passing through the given point and normal to the plane, we can find the image point.

26 Medium

26 of 49

Application +4 | -1

The value of

k

for which the points

A(1,0,3)

B(-1,3,4), C(1,2,1)

and

D(k, 2,5)

are coplanar, are

View answer and explanation

Correct option: D

-1

Option D (-1) is correct because the value of k is found by setting the scalar triple product of the vectors formed by the points to zero, indicating coplanarity.

27 Medium

27 of 49

Application +4 | -1

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

are the three non-coplanar vectors and

\mathbf{p}, \mathbf{q}, \mathbf{r}

are defined by the relations

p=\frac{b \times c}{[a b c]}, q=\frac{c \times a}{[a b c]}, r=\frac{a \times b}{[a b c]}

then

(a+b) \cdot p+(b+c)

\mathbf{q}+(\mathbf{c}+\mathbf{a}) \cdot \mathbf{r}=

View answer and explanation

Correct option: D

Option D (3) is correct because by using the properties of scalar triple product and reciprocal vectors, the expression simplifies to 3.

28 Medium

28 of 49

Application +4 | -1

\mathbf{\vec a} = \left( {2,1, - 1} \right),\;\mathbf{\vec b} = \left( {1, - 1,0} \right),\;\mathbf{\vec c} = \left( {5 - 1,1} \right),\;

then unit vector parallel to

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

but in opposite direction is

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it accurately calculates the unit vector parallel to

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

but in the opposite direction.

29 Medium

29 of 49

Application +4 | -1

Let

\overrightarrow{\mathrm{OA}}=\vec{a}, \overrightarrow{\mathrm{OB}}=2 \vec{a}+10 \vec{b}, \overrightarrow{O C}=\vec{b}

where

\mathrm{O}, \mathrm{A}, \mathrm{C}

are non collinear points. Let

\lambda

denote the area of the quadrilateral

\mathrm{OABC}

. Let

\mathrm{m}

denotes the area of parallelogram with

\overrightarrow{\mathrm{OA}}

and

\overrightarrow{\mathrm{OC}}

as adjacent sides. If

\lambda=2 \lambda \mathrm{m}

then find the value of

\lambda

View answer and explanation

Correct option: A

Option A (3) is correct because by calculating the areas of the quadrilateral and parallelogram and using the given relation, the value of lambda is found to be 3.

30 Medium

30 of 49

Application +4 | -1

A unit vector in the plane of the vectors

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

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

and orthogonal to

5 \mathbf{i}+2 \mathbf{j}+6 \mathbf{k}

View answer and explanation

Correct option: B

\frac{3 \mathrm{j}-\mathbf{k}}{\sqrt{10}}

Option B (

\frac{{3 \mathrm{j}-\mathbf{k}}}{{\sqrt{{10}}}}

) is correct because it satisfies the conditions of being in the plane of the given vectors and orthogonal to the specified vector.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 49 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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