If $I$ is incentre of $\Delta \;ABC,$ then $I$ is

$\frac{{a{\bf{\vec a}} + b{\bf{\vec b}} + c\;{\bf{\vec c}}}}{{a + b + c}}$

$\frac{{a{\bf{\vec a}} + b{\bf{\vec b}} + c\;{\bf{\vec c}}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$

$\frac{1}{3}\left[ {{\bf{\vec a}} + {\bf{\vec b}} + \;{\bf{\vec c}}} \right]$

$\frac{{{\bf{\vec a}} + {\bf{\vec b}} + {\bf{\vec c}}}}{{a + b + c}}$

Let ${{\bf{\vec a}}}$ and ${{\bf{\vec b}}}$ are non-collinear vectors. If there exists scalars ${\rm{\alpha }},{\rm{\beta }}$ such that ${\rm{\alpha }}{\bf{\vec a}} + {\rm{\beta }}{\bf{\vec b}} = \vec 0$, then

${\rm{\alpha }} = {\rm{\beta }} = 0$

${\rm{\alpha }} = {\rm{\beta }} \ne 0$

${\rm{\alpha }} + {\rm{\beta }} = 0$

${\rm{\alpha }} \ne {\rm{\beta }}$

If the position vectors of the vertices of $\Delta \;ABC$ are $3\widehat {\bf{i}} + \widehat {\bf{j}} + 2{\bf{\hat k}},\;\;\widehat {\bf{i}} - 2\widehat {\bf{j}} + 7{\bf{\hat k}}$ and $ - 2\widehat {\bf{i}} + 3\widehat {\bf{j}} + 5{\bf{\hat k}},$ then the triangle $ABC$ is

Right angled, but not isosceles

If $A,B,C,D,E$ are five coplanar points, then $\overrightarrow {{\bf{DA}}} + \overrightarrow {{\bf{DB}}} + \overrightarrow {{\bf{DC}}} + \overrightarrow {{\bf{AE}}} + \overrightarrow {{\bf{BE}}} + \overrightarrow {{\bf{CE}}} $ is equal to

$3\;\overrightarrow {{\bf{DE}}} $

$2\;\overrightarrow {{\bf{DE}}} $

4$\overrightarrow {{\bf{ED}}} $

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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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211 Easy

211 of 318

Application +4 | -1

If the vectors represented by the sides

A B

and

B C

of the regular hexagon

A B C D E F

\mathbf{a}

and

\mathbf{b}

, then the vector represented by

\mathrm{AE}

will be

View answer and explanation

Correct option: A

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

Option A is correct because by using the properties of a regular hexagon,

\overrightarrow{AE} = 2\overrightarrow{BC} - \overrightarrow{AB} = 2\mathbf{b} - \mathbf{a}

212 Hard

212 of 318

Analyzing +4 | -1

\left|x_1\right|>\left|y_1\right|+\left|z_1\right|,\left|y_2\right|>\left|x_2\right|+\left|z_2\right|,\left|z_3\right|>\left|x_3\right|+\left|y_3\right|

,
then

x_1 \hat{\mathbf{i}}+y_1 \hat{\mathbf{j}}+z_1 \hat{\mathbf{k}}, x_2 \hat{\mathbf{i}}+y_2 \hat{\mathbf{j}}+z_2 \hat{\mathbf{k}}

and

x_3 \hat{\mathbf{i}}+y_3 \hat{\mathbf{j}}+z_3 \hat{\mathbf{k}}

are

View answer and explanation

Correct option: D

non-coplanar

The vectors are non-coplanar because the given inequalities contradict the condition for coplanarity, which requires the determinant to be zero.

213 Easy

213 of 318

Understanding +4 | -1

I

is incentre of

\Delta \;ABC,

then

I

View answer and explanation

Correct option: A

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

Option A is correct because the incenter of a triangle is a weighted average of the vertices' position vectors, with weights being the lengths of the opposite sides.

214 Medium

214 of 318

Application +4 | -1

The position vector of the point in which the line joining the points

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

and

3 \mathbf{k}-2 \mathbf{j}

cuts the plane through the origin and the points

4 \mathbf{j}

and

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

, is

View answer and explanation

Correct option: B

\frac{1}{5}(6 \mathbf{i}-10 \mathbf{j}+3 \mathbf{k})

Option B is correct because it represents the intersection point of the line joining the two given points and the plane passing through the origin and the other two points.

215 Easy

215 of 318

Analyzing +4 | -1

(2,3,9),(5,2,1),(1, \lambda, 8)

and

(\lambda, 2,3)

are coplanar, then the product of all possible values of

\lambda

is :

View answer and explanation

Correct option: D

\frac{95}{8}

Option D (

\frac{95}{8}

) is correct because the product of all possible values of

\lambda

is equal to the constant term divided by the coefficient of the highest power of

\lambda

in the determinant equation.

216 Easy

216 of 318

Application +4 | -1

If the vectors

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

and

\vec{b}=3 \hat{i}-4 \hat{j}

are the adjacent sides of a parallelogram. Then the vector along the bisector of the angle between

\vec{a}

and

\vec{b}

having magnitude

5 \sqrt{3}

View answer and explanation

Correct option: B

\pm(7 \hat{i}-\hat{j}+5 \hat{k})

Option B (

\pm(7 \hat{i}-\hat{j}+5 \hat{k})

) is correct because the vector along the angle bisector is proportional to the sum of the unit vectors along \vec{a} and \vec{b}, scaled to the given magnitude.

217 Easy

217 of 318

Application +4 | -1

(x, y, z) \neq(0,0,0)

and

(\mathbf{i}+\mathbf{j}+3 \mathbf{k}) x+(3 \mathbf{i}-3 \mathbf{j}+\mathbf{k}) \mathbf{y}+(-4 \mathbf{i}+5 \mathbf{j}) z=\lambda(x \mathbf{i}+\mathbf{y} \mathbf{j}+\mathbf{z k})

, then the value of

\lambda

will be

View answer and explanation

Correct option: D

0,-1

Option D (

0,-1

) is correct because substituting the given vector equation leads to the values of

\lambda

being 0 and -1.

218 Medium

218 of 318

Application +4 | -1

The line through

\mathbf{i}+3 \mathbf{j}+2 \mathbf{k}

and perpendicular to the lines

\mathbf{r}=(\mathbf{i}+2 \mathbf{j}-\mathbf{k})+\lambda(2 \mathbf{i}+\mathbf{j}+\mathbf{k})

View answer and explanation

Correct option: D

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

Option D is correct because the line passes through the given point and is perpendicular to the two given lines, hence parallel to their cross product.

219 Easy

219 of 318

Application +4 | -1

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

, where

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

are three non-coplanar vectors, then the value of

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

is given by

View answer and explanation

Correct option: A

Option A (3) is correct because by substituting the given expressions for p, q, and r, and using properties of the scalar triple product, the expression simplifies to 3.

220 Medium

220 of 318

Analyzing +4 | -1

The unit vectors orthogonal to the vector

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

and making equal angles with the

X

and

Y

axes is (are)

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it satisfies the orthogonality condition and equal angles with the X and Y axes.

221 Easy

221 of 318

Application +4 | -1

Let

ABC

be a triangle, the position vectors of whose vertices are respectively

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

and

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

Then, the

\Delta ABC

View answer and explanation

Correct option: C

Right angled isosceles

Option C (Right angled isosceles) is correct because the triangle has two sides of equal length and satisfies the Pythagorean theorem.

222 Easy

222 of 318

Understanding +4 | -1

Let

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

are non-collinear vectors. If there exists scalars

\mathrm{\alpha },\mathrm{\beta }

such that

\mathrm{\alpha }\mathbf{\vec a} + \mathrm{\beta }\mathbf{\vec b} = \vec 0

, then

View answer and explanation

Correct option: C

\mathrm{\alpha } = \mathrm{\beta } = 0

Option C (

\mathrm{\alpha } = \mathrm{\beta } = 0

) is correct because for non-collinear vectors, the only way for the linear combination to be zero is if both scalars are zero.

223 Medium

223 of 318

Application +4 | -1

Vector coplanar with vectors

\mathbf{i}+\mathbf{j}

and

\mathbf{j}+\mathbf{k}

and parallel to the vector

2 \mathbf{i}-2 \mathbf{j}-4 \mathbf{k}

, is

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it satisfies the conditions of being coplanar with

\mathbf{i}+\mathbf{j}

and

\mathbf{j}+\mathbf{k}

and parallel to

2 \mathbf{i}-2 \mathbf{j}-4 \mathbf{k}

224 Easy

224 of 318

Analyzing +4 | -1

Statement 1: If the points

(1,2,2),(2,1,2)

and

(2,2, z)

and

(1,1,1)

are coplanar, then

z=2

.

Statement 2: If the 4 points

P, Q, R

and

S

are coplanar, then the volume of the tetrahedron

P Q R S

is 0 .

View answer and explanation

Correct option: A

Statement 1 is false,, Statement 2 is true.

Statement 1 is false, and Statement 2 is true. Statement 1 is false because the correct value of z is 3, not 2, for the points to be coplanar, and Statement 2 is a fundamental property of coplanar points.

225 Easy

225 of 318

Analyzing +4 | -1

The vector equations of the two lines

L_1

and

L_2

are given by

L_1: \mathbf{r}=(2 \mathbf{i}+9 \mathbf{j}+13 \mathbf{k})+\lambda(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k})

and

\quad L_2: \mathbf{r}=(-3 \mathbf{i}+7 \mathbf{j}+p \mathbf{k})+\mu(-\mathbf{i}+2 \mathbf{j}-3 \mathbf{k})

.
Then, the lines

L_1

and

L_2

are

View answer and explanation

Correct option: C

intersecting lines for

p=-2

The lines

L_1

and

L_2

intersect when

p=-2

, which is found by setting up a determinant equation based on the direction vectors and position vectors of the lines.

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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

226 Easy

226 of 318

Application +4 | -1

If the position vectors of the vertices of

\Delta \;ABC

are

3\widehat \mathbf{i} + \widehat \mathbf{j} + 2\mathbf{\hat k},\;\;\widehat \mathbf{i} - 2\widehat \mathbf{j} + 7\mathbf{\hat k}

and

- 2\widehat \mathbf{i} + 3\widehat \mathbf{j} + 5\mathbf{\hat k},

then the triangle

ABC

View answer and explanation

Correct option: D

Equilateral

The correct answer is D (Equilateral) because all sides of the triangle have equal length, which is calculated using the position vectors of the vertices.

227 Easy

227 of 318

Analyzing +4 | -1

Let

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

three non-zero vectors such that no two of which are collinear and the vector

\vec a + \vec b

is collinear with

{\vec c}

and

\vec b + \vec c

is collinear with

{\vec a}

. Then,

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

View answer and explanation

Correct option: D

{\vec 0}

The correct answer is D (

{\vec 0}

) because the given conditions imply that the vectors are linearly dependent, leading to their sum being the zero vector.

228 Easy

228 of 318

Application +4 | -1

If a is a non – zero vector of modulus a and m is a non zero scalar, then ma is a unit vector, if

View answer and explanation

Correct option: C

m = \frac{1}{{\left| a \right|}}

Option C (

m = \frac{1}{{\left| a \right|}}

) is correct because for ma to be a unit vector, its magnitude must be 1, leading to the given condition.

229 Easy

229 of 318

Application +4 | -1

A, B, C, D and E are five coplanar points, then DA + DB + DC + AE + BE + CE is equal to

View answer and explanation

Correct option: B

3DE

Option B (3DE) is correct because by breaking down the vectors and using vector addition, the expression simplifies to 3DE.

230 Easy

230 of 318

Application +4 | -1

A,B,C,D,E

are five coplanar points, then

\overrightarrow {\mathbf{DA}} + \overrightarrow {\mathbf{DB}} + \overrightarrow {\mathbf{DC}} + \overrightarrow {\mathbf{AE}} + \overrightarrow {\mathbf{BE}} + \overrightarrow {\mathbf{CE}}

is equal to

View answer and explanation

Correct option: B

3\;\overrightarrow {\mathbf{DE}}

Option B (

3\;\overrightarrow {\mathbf{DE}}

) is correct because it is the simplified form of the given vector sum using vector addition properties.

231 Easy

231 of 318

Analyzing +4 | -1

The lines

\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}

and

\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}

are coplanar, if

View answer and explanation

Correct option: C

k=0

-3

The lines are coplanar if the determinant formed by the difference in position vectors and the direction vectors is zero, leading to k=0 or k=-3.

232 Easy

232 of 318

Analyzing +4 | -1

The points A(2 – x, 2, 2), B(2, 2 – y, 2), C(2, 2, 2 – z) and D(1, 1, 1) are coplanar, then locus of P(x, y, z) is

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because, given the coplanarity condition of the points A, B, C, and D, the locus of P(x, y, z) satisfies this equation.

233 Easy

233 of 318

Application +4 | -1

If the resultant of two forces of magnitudes P and Q acting at a point at an angle of

60^\circ

\sqrt 7 Q,

then

P/Q

View answer and explanation

Correct option: C

Option C (2) is correct because, using the law of cosines, we can relate the magnitudes of the forces P, Q, and their resultant, and then solve for the ratio P/Q.

234 Easy

234 of 318

Analyzing +4 | -1

Let a, b and c be distinct non – negative numbers and the vectors

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

lie in a plane then the quadratic equation

a{x^2} + 2cx + b = 0

has

View answer and explanation

Correct option: A

real and equal roots

Option A (real and equal roots) is correct because, given the coplanarity condition, the discriminant of the quadratic equation

a{x^2} + 2cx + b = 0

is zero, indicating real and equal roots.

235 Medium

235 of 318

Application +4 | -1

If the position vectors of the point

A, B, C

\mathbf{i}, \mathbf{j}, \mathbf{k}

respectively and

P

be a point such that

\overrightarrow{A B}=\overrightarrow{C P}

, then the position vector of

P

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it satisfies the condition

\overrightarrow{A B}=\overrightarrow{C P}

by finding the position vector of point P.

236 Easy

236 of 318

Application +4 | -1

\mathbf{p}=\mathbf{i}-2 \mathbf{j}+3 \mathbf{k}

and

\mathbf{q}=3 \mathbf{i}+\mathbf{j}+2 \mathbf{k}

, then a vector along

\mathbf{r}

which is linear combination of

\mathbf{p}

and

\mathbf{q}

and also perpendicular to

\mathbf{q}

View answer and explanation

Correct option: C

-\frac{1}{2}(\mathbf{i}+5 \mathbf{j}-4 \mathbf{k})

Option C (-

\frac{1}{2}(\mathbf{i}+5\mathbf{j}-4\mathbf{k})

) is correct because it represents a vector that is a linear combination of p and q and is also perpendicular to q.

237 Easy

237 of 318

Application +4 | -1

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

and

c = - \widehat i + 2\widehat j + 2\widehat k,

then

a + b + c

View answer and explanation

Correct option: C

4\widehat i - 4\widehat j

Option C (

4\widehat i - 4\widehat j

) is correct because it is the result of adding the three vectors component-wise.

238 Medium

238 of 318

Application +4 | -1

The position vectors of the points

A, B, C

and

D

are

3 {i}-2 {j}-{k}, 2 {i}+3 {j}-4 {k},-{i}+{j}+2 {k}

and

4 {i}+5 {j}+\lambda {k}

respectively. If the points

{A}, {B}, {C}

and

{D}

are on a plane, then the value of

\lambda

View answer and explanation

Correct option: B

\frac{-146}{17}

Option B (

\frac{{-146}}{{17}}

) is correct because the points are coplanar if the scalar triple product of the vectors formed by the points is zero, leading to the given value of

\lambda

239 Easy

239 of 318

Application +4 | -1

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

and

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

, then the vector in the direction of

{\vec a}

and having magnitude as

\left| {\vec b} \right|

, is

View answer and explanation

Correct option: C

\frac{7}{3}\left( {\hat i + 2\hat j + 2\hat k} \right)

Option C is correct. The required vector is found by multiplying the unit vector of

\vec a

by the magnitude of

\vec b

240 Easy

240 of 318

Application +4 | -1

A(-1,2,3), B(1,1,1)

and

C(2,-1,3)

are points on a plane. A unit normal vector to the plane

A B C

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it represents the unit normal vector to the plane ABC, derived from the cross product of two vectors in the plane.

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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