If 4$\widehat {\bf{i}} + 7\widehat {\bf{j}} + 8{\bf{\hat k}},\;2\widehat {\bf{i}} + 3\widehat {\bf{j}} + 4{\bf{\hat k}}$ and $2\widehat {\bf{i}} + 5\widehat {\bf{j}} + 7{\bf{\hat k}}$ are the position vectors of the vertices $A,B\;{\rm{and\;}}C$ respectively of triangle $ABC$. The position vector of the point where the bisector of angle $A$ meets $BC$ is

$\frac{1}{2}\left( {6\widehat {\bf{i}} + 13\widehat {\bf{j}} + 18{\bf{\hat k}}} \right)$

$\frac{2}{3}\left( {6\widehat {\bf{i}} + 12\widehat {\bf{j}} - 8{\bf{\hat k}}} \right)$

$\frac{1}{3}\left( { - 6\widehat {\bf{i}} - 8\widehat {\bf{j}} - 9{\bf{\hat k}}} \right)$

$\frac{2}{3}\left( { - 6\widehat {\bf{i}} - 12\widehat {\bf{j}} + 8{\bf{\hat k}}} \right)$

If $P,\;Q,\;R$ are the mid-points of the sides $AB,\;BC$ and $CA$ of ${\rm{\Delta }}ABC$ are $O$ is a point within the triangle, then $\vec OA + \vec OB + \vec OC = $

$2\left( {\vec OP + \vec OQ + \vec OR} \right)$

$4\left( {\vec OP + \vec OQ + \vec OR} \right)$

$6\left( {\vec OP + \vec OQ + \vec OR} \right)$

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JEE • MATHEMATICS • ADDITION OF VECTORS

Addition of vectors practice · Mixed difficulty

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151 Medium

151 of 318

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.

152 Easy

152 of 318

Analyzing +4 | -1

Vectors

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

and

c = 3\hat i + \hat j + 4\hat k

are so placed that the end point of one vector is the starting point of the next vector. Then the vectors are

View answer and explanation

Correct option: B

coplanar but cannot form a triangle

Option B (coplanar but cannot form a triangle) is correct because the vectors satisfy a + b = c, indicating they are coplanar, and since they form a closed figure, they cannot form a triangle.

153 Easy

153 of 318

Application +4 | -1

If the vectors

\mathrm{\alpha }\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}

\widehat \mathbf{i} + \mathrm{\beta }\widehat \mathbf{j} + \mathbf{\hat k}

\widehat \mathbf{i} + \widehat \mathbf{j} + \mathrm{\gamma }\mathbf{\hat k}\left( {\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma } \ne 1} \right)

are coplanar, then the value of

\frac{1}{{1 - \mathrm{\alpha }}} + \frac{1}{{1 - \mathrm{\beta }}} - \frac{1}{{1 - \mathrm{\gamma }}}\;\mathrm{is}

View answer and explanation

Correct option: C

Option C (1) is correct because the coplanarity condition leads to a determinant equation, which simplifies to the given expression equaling 1.

154 Easy

154 of 318

Application +4 | -1

The unit vector parallel to the resultant vector of

2\widehat i + 4\widehat j - 5\widehat k

and

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

View answer and explanation

Correct option: A

\frac{1}{7}(3\widehat i + 6\widehat j - 2\widehat k)

Option A (

\frac{1}{7}(3\widehat i + 6\widehat j - 2\widehat k)

) is correct because it is the unit vector in the direction of the resultant vector.

155 Easy

155 of 318

Application +4 | -1

If a = (1, -1) and b = (-2, m) are two collinear vectors, then m is equal to

View answer and explanation

Correct option: C

Option C (2) is correct because it represents the value of m when vector b is collinear with vector a, calculated using the condition for collinearity.

156 Medium

156 of 318

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

157 Easy

157 of 318

Application +4 | -1

If ABCD is parallelogram,

AB = 2\hat i + 4\hat j - 5\hat k

and

AD = \hat i + 2\hat j + 3\hat k,

then the unit vectors in the direction of BD is

View answer and explanation

Correct option: C

\frac{1}{{\sqrt {69} }}( - \hat i - 2\hat j + 8\hat k)

Option C (

\frac{1}{{\sqrt {69} }}( - \hat i - 2\hat j + 8\hat k)

) is correct because it represents the unit vector in the direction of BD, calculated by subtracting the vectors AB and AD.

158 Easy

158 of 318

Application +4 | -1

For what value of

\mathrm{m}

, are the points with position vector

10\hat i + 3\hat j,12\hat i - 5\hat j\,\,{\text{and}}\,m\hat i + 11\hat j

collinear?

View answer and explanation

Correct option: A

Option A (8) is correct because for the points to be collinear, the area of the triangle formed by them must be zero, leading to m = 8.

159 Easy

159 of 318

Application +4 | -1

If the position vectors of A and B are

\hat i + 3\hat j - 7\hat k\,\,\,{\text{and}}\,\,\,5\hat i - 2\hat j + 4\hat k,

then the direction cosine of AB along Y – axis is

View answer and explanation

Correct option: B

- \frac{5}{{\sqrt {162} }}

Option B (

- \frac{{5}}{{\sqrt {162} }}

) is correct because it represents the direction cosine of the vector AB along the Y-axis, calculated as the Y-component divided by the magnitude of AB.

160 Easy

160 of 318

Understanding +4 | -1

Let

ABCD

be a parallelogram and

M

be the point of intersection of the diagonals. If

O

is any point, then

\vec OA + \vec OB + \vec OC + \vec OD =

View answer and explanation

Correct option: B

4\;\vec OM

Option B (

4\;\vec OM

) is correct because M is the midpoint of both diagonals, and the sum of the position vectors of the vertices is 4 times the position vector of the intersection point.

161 Easy

161 of 318

Understanding +4 | -1

In the

\Delta ABC

, AB = a, AC = c and BC = b, then

View answer and explanation

Correct option: B

a + b – c = 0

Option B (a + b – c = 0) is correct because it represents the vector sum of the sides of the triangle ABC.

162 Medium

162 of 318

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.

163 Easy

163 of 318

Application +4 | -1

If O is the origin and the position vector of A is

4\widehat i + 5\widehat j,

then a unit vector parallel to OA is

View answer and explanation

Correct option: C

\frac{1}{{\sqrt {41} }}(4\widehat i + 5\widehat j)

Option C (

\frac{1}{{\sqrt {41} }}(4\widehat i + 5\widehat j)

) is correct because it represents the unit vector in the direction of OA.

164 Easy

164 of 318

Application +4 | -1

OP = 8

and

OP

makes an angle

45^\circ

and

60^\circ

with OX – axis and OY – axis respectively, then OP is equal to

View answer and explanation

Correct option: B

4(\sqrt 2 \widehat i + \widehat j \pm \widehat k)

Option B (

4(\sqrt 2 \widehat i + \widehat j \pm \widehat k)

) is correct because it represents the vector OP with magnitude 8 and the given angles with the OX and OY axes.

165 Easy

165 of 318

Application +4 | -1

2\widehat \mathbf{i} + 4\widehat \mathbf{j} - 5\mathbf{\hat k}

and

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

are adjacent side of a parallelogram, then the lengths of its diagonals are

View answer and explanation

Correct option: A

7,\sqrt {69}

Option A (

7,\sqrt {69}

) is correct because it represents the magnitudes of the sum and difference of the given vectors, which are the diagonals of the parallelogram.

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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

166 Easy

166 of 318

Application +4 | -1

Let there be two points

A, B

on the curve

y=x^2

in the plane

O X Y

satisfying

{O A} \cdot {i}=1

and

{O B} \cdot {i}=-2

then the length of the vector

2 O A-3 O B

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the calculated length of the vector 2OA - 3OB is 2

\sqrt{41}

, which is not among the options.

167 Easy

167 of 318

Analyzing +4 | -1

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

be the position vectors of the points

A, B, C

and

D

respectively referred to same origin

O

such that no three of these points are collinear and

\vec{a}+\vec{c}=\vec{b}+\vec{d}

, then quadrilateral

A B C D

is a

View answer and explanation

Correct option: D

Parallelogram

Option D (Parallelogram) is correct because the given condition implies that the midpoints of the diagonals AC and BD coincide, which is a property of parallelograms.

168 Easy

168 of 318

Application +4 | -1

If 4

\widehat \mathbf{i} + 7\widehat \mathbf{j} + 8\mathbf{\hat k},\;2\widehat \mathbf{i} + 3\widehat \mathbf{j} + 4\mathbf{\hat k}

and

2\widehat \mathbf{i} + 5\widehat \mathbf{j} + 7\mathbf{\hat k}

are the position vectors of the vertices

A,B\;\mathrm{and\;}C

respectively of triangle

ABC

. The position vector of the point where the bisector of angle

A

meets

BC

View answer and explanation

Correct option: A

\frac{1}{2}\left( {6\widehat \mathbf{i} + 13\widehat \mathbf{j} + 18\mathbf{\hat k}} \right)

Option A (

\frac{1}{2}\left( {6\widehat \mathbf{i} + 13\widehat \mathbf{j} + 18\mathbf{\hat k}} \right)

) is correct because it represents the position vector of the point where the angle bisector of angle A intersects BC, calculated using the section formula and the angle bisector theorem.

169 Easy

169 of 318

Analyzing +4 | -1

The number of the distinct real values of λ, for which the vectors

- {\mathrm{\lambda }^2}\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}

\widehat \mathbf{i} - {\mathrm{\lambda }^2}\widehat \mathbf{j} + \mathbf{\hat k}\mathrm{\;}

and

\widehat \mathbf{i} + \widehat \mathbf{j} - {\mathrm{\lambda }^2}\mathbf{\hat k}\mathrm{\;}

are coplanar, is

View answer and explanation

Correct option: C

Two

Option C (Two) is correct because solving the determinant equation for coplanarity yields two distinct real values for λ, which are

\pm \sqrt{2}

170 Easy

170 of 318

Understanding +4 | -1

\mathbf{M}

is the mid point of

\mathbf{A B}

and

\mathbf{O}

is any point, then

View answer and explanation

Correct option: D

\mathbf{O M}=\frac{1}{2}(\mathbf{O B}+\mathbf{O A})

Option C (no solution) is correct because the equation

a \sin x+b \cos x=c

has no solution when

|c| > \sqrt{a^2+b^2}

171 Easy

171 of 318

Application +4 | -1

The vectors

{a}={i}+{j}+m {k}, {b}={i}+{j}+(m+1) {k}

and

{c}={i}-{j}+m {k}

are coplanar if

m

is equal to

View answer and explanation

Correct option: D

None of these

Option D (

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

) is correct because it represents the magnitude of the projection of u x v on w, calculated using the scalar projection formula.

172 Easy

172 of 318

Application +4 | -1

Let

a = 2\hat i + {\lambda _1}\hat j + 3\hat k,\,\,b = 4\hat i + (3 - {\lambda _2})\hat j + 6\hat k\mathrm{ }

and

c = 3\hat i + 6\hat j + ({\lambda _3} - 1)\hat k

be three vectors such that b = 2a and a is perpendicular to c. Then a possible value of

({\lambda _1},{\lambda _2},{\lambda _3})

View answer and explanation

Correct option: C

\left( { - \frac{1}{2},4,0} \right)

Option A (with slope 2/3) is correct because the locus of the centroid is a line with the same slope as the original line.

173 Medium

173 of 318

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}

174 Medium

174 of 318

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.

175 Easy

175 of 318

Analyzing +4 | -1

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

are linearly independent vectors and

\Delta = \left| {\begin{array}{cccccccccccccccccccc} {\vec a}&{\vec b}&{\vec c}\\ {\vec a.\vec a}&{\vec a.\vec b}&{\vec a.\vec c}\\ {\vec a.\vec c}&{\vec b.\vec c}&{\vec c.\vec c} \end{array}} \right|

, then

View answer and explanation

Correct option: C

\Delta =

any non-zero value

Option C (

\Delta =

any non-zero value) is correct because if the determinant is zero, the vectors would be linearly dependent, contradicting the given information.

176 Easy

176 of 318

Application +4 | -1

A vector

{b}

collinear with

{a}=2 \sqrt{2} {i}-{j}+4{k}

of length 10 is given by

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because it is collinear with the given vector and has a length of 10.

177 Easy

177 of 318

Application +4 | -1

If the points whose position vectors are

2\hat i + \hat j + \hat k,\;6\hat i - \hat j + 2\hat k

and

14\hat i - 5\hat j + p\hat k

are collinear, then

p =

View answer and explanation

Correct option: B

Option D (4) is correct. If the points are collinear, the vectors formed by them are parallel, allowing us to find the value of p.

178 Easy

178 of 318

Understanding +4 | -1

P,\;Q,\;R

are the mid-points of the sides

AB,\;BC

and

CA

\mathrm{\Delta }ABC

are

O

is a point within the triangle, then

\vec OA + \vec OB + \vec OC =

View answer and explanation

Correct option: B

\vec OP + \vec OQ + \vec OR

Option B (

\vec OP + \vec OQ + \vec OR

) is correct because it directly follows from expressing

\vec OA + \vec OB + \vec OC

in terms of the midpoints P, Q, and R.

179 Easy

179 of 318

Application +4 | -1

The position vectors of three consecutive vertices of a parallelogram are

\mathbf{i}+\mathbf{j}+\mathbf{k}, \mathbf{i}+3 \mathbf{j}+5 \mathbf{k}

and

7 \mathbf{i}+9 \mathbf{j}+11 \mathbf{k}

. The position vector of the fourth vertex is

View answer and explanation

Correct option: B

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

The correct answer is D (

6 \mathbf{i}+8 \mathbf{j}+10 \mathbf{k}

) because in a parallelogram, the midpoints of the diagonals coincide, allowing us to solve for the fourth vertex.

180 Easy

180 of 318

Application +4 | -1

If vectors

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

and

a\widehat \mathbf{i} + b\widehat \mathbf{j} - 15\mathbf{\hat k}

are collinear, then

View answer and explanation

Correct option: D

a = 9,\;b = 3

Option D (

a = 9, b = 3

) is correct because collinear vectors are scalar multiples of each other, leading to the ratios of their components being equal.

Revision notes and key takeaways

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