Addition of Vectors

Interactive Feed

JEE Main 2013

If the vectors $A B = 3 \hat{i} + 4 \hat{k}$ and $A C = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle $A B C$ , then the length of the median through $A$ is

18

(B)

72

(C)

33

(D)

45

JEE Main 2008

LEVELJEE Main

The vector $a = α \hat{i} + 2 \hat{j} + β \hat{k}$ lies in the plane of the vectors $b = \hat{i} + \hat{j}$ and $c = \hat{j} + \hat{k}$ and bisects the angle between $b$ and $c$ . Then which one of the following gives possible values of $α$ and $β$ ?

(A)

α = 2, β = 2

(B)

α = 1, β = 2

(C)

α = 2, β = 1

(D)

α = 1, β = 1

JEE Main 2007

LEVELJEE Main

The resultant of two forces Pn and 3n is a force of 7n. If the direction of 3n force were reversed, the resultant would be $19 n$ . The value of P is

A particle has two velocities of equal magnitude inclined to each other at an angle $θ$ . If one of them is halved, the angle between the other and the original resultant velocity is bisected by the new resultant. Then $θ$ is

(A)

9 0^{\circ}

(B)

12 0^{\circ}

(C)

4 5^{\circ}

(D)

6 0^{\circ}

JEE Main 2005

LEVELBoard

If $C$ is the mid point of $A B$ and $P$ is any point outside $A B$ , then

(A)

P A + P B = 2 P C

(B)

P A + P B = P C

(C)

P A + P B + 2 P C = 0

(D)

P A + P B + P C = 0

JEE Main 2005

LEVELJEE Advanced

ABC is a triangle. Forces $P, Q, R$ acting along IA, IB, and IC respectively are in equilibrium, where I is the incentre of $△ A B C$ . Then $P : Q : R$ is

(A)

sin A : sin B : sin C

(B)

sin \frac{A}{2} : sin \frac{B}{2} : sin \frac{C}{2}

(C)

cos \frac{A}{2} : cos \frac{B}{2} : cos \frac{C}{2}

(D)

cos A : cos B : cos C

JEE Main 2005

LEVELJEE Main

A and B are two like parallel forces. A couple of moment H lies in the plane of A and B and is contained with them. The resultant of A and B after combining is displaced through a distance

(A)

\frac{2 H}{A - B}

(B)

\frac{H}{A + B}

(C)

\frac{H}{2 ( A + B )}

(D)

\frac{H}{A - B}

JEE Main 2005

LEVELJEE Main

The resultant R of two forces acting on a particle is at right angles to one of them and its magnitude is one third of the other force. The ratio of larger force to the smaller one is

2

22

JEE Main 2004

LEVELJEE Main

Let $a, b$ and $c$ be three non-zero vectors such that no two of these are collinear. If the vector $a + 2 b$ is collinear with $c$ and $b + 3 c$ is collinear with $a$ ( $λ$ being some non-zero scalar) then $a + 2 b + 6 c$ equals

(A)

0

(B)

λ b

(C)

λ c

(D)

λ a

JEE Main 2004

LEVELJEE Main

With two forces acting at point, the maximum affect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the forces are

(A)

(2 + \frac{1}{2} 3) N

and

(2 - \frac{1}{2} 3) N

(B)

(2 + 3) N

and

(2 - 3) N

(C)

(2 + \frac{1}{2} 2) N

and

(2 - \frac{1}{2} 2) N

(D)

(2 + 2) N

and

(2 - 2) N

JEE Main 2003

LEVELJEE Main

The vectors $A B = 3 \hat{i} + 4 \hat{k}$ and $A C = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle ABC. The length of the median through A is

(A)

288

(B)

18

(C)

72

(D)

33

JEE Main 2003

LEVELJEE Main

The resultant of forces $P$ and $Q$ is $R$ . If $Q$ is doubled then $R$ is doubled. If the direction of $Q$ is reversed, then $R$ is again doubled. Then $P^{2} : Q^{2} : R^{2}$ is

The sum of two forces is 18 N and resultant whose direction is at right angles to the smaller force is 12 N. The magnitude of the two forces are

Show, by vector methods, that the angular bisectors of a triangle are concurrent and find an expression for the position vector of the point of concurrency in terms of the position vectors of the vertices.

JEE Advanced 1988

LEVELJEE Main

Let $O A C B$ be a parallelogram with $O$ at the origin and $O C$ a diagonal. Let $D$ be the midpoint of $O A$ . Using vector methods prove that $B D$ and $C O$ intersect in the same ratio. Determine this ratio.

JEE Advanced 1998

LEVELJEE Main

Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezium lies on the line passing through the mid-points of the parallel sides. (You may assume that the trapezium is not a parallelogram.)

JEE Main 2018 (15 April Evening)

LEVELJEE Main

If the position vectors of the vertices A, B and C of a $Δ A B C$ are respectively $4 \hat{i} + 7 \hat{j} + 8 \hat{k}$ , $2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $2 \hat{i} + 5 \hat{j} + 7 \hat{k}$ , then the position vector of the point, where the bisector of $∠ A$ meets BC is :-

(A)

\frac{1}{2} (4 \hat{i} + 8 \hat{j} + 11 \hat{k})

(B)

\frac{1}{3} (6 \hat{i} + 13 \hat{j} + 18 \hat{k})

(C)

\frac{1}{4} (8 \hat{i} + 14 \hat{j} + 19 \hat{k})

(D)

\frac{1}{3} (6 \hat{i} + 11 \hat{j} + 15 \hat{k})

JEE Main 2020 - 7 Jan (Morning)

LEVELJEE Main

A vector $a = α \hat{i} + 2 \hat{j} + β \hat{k} (α, β \in R)$ lies in the plane of the vectors, $b = \hat{i} + \hat{j}$ and $c = \hat{i} - \hat{j} + 4 \hat{k}$ . If $a$ bisects the angle between $b$ and $c$ , then:

(A)

a \cdot \hat{k} + 2 = 0

(B)

a \cdot \hat{k} + 4 = 0

(C)

a \cdot \hat{k} - 2 = 0

(D)

a \cdot \hat{k} + 5 = 0

JEE Main 2020 - 6 Sep (Morning)

LEVELJEE Main

If $a$ and $b$ are unit vectors, then the greatest value of $3 ∣ a + b ∣ + ∣ a - b ∣$ is

JEE Main 2025 April

LEVELJEE Main

Let the three sides of a triangle $A B C$ be given by the vectors $2 \hat{i} - \hat{j} + \hat{k}, \hat{i} - 3 \hat{j} - 5 \hat{k}$ and $3 \hat{i} - 4 \hat{j} - 4 \hat{k}$ . Let $G$ be the centroid of the triangle $A B C$ . Then $6 (∣ A G ∣^{2} + ∣ B G ∣^{2} + ∣ C G ∣^{2})$ is equal to ________

JEE Main 2025 (January)

LEVELJEE Main

Let A, B, C be three points in xy-plane, whose position vector are given by $3 \hat{i} + \hat{j}$ , $\hat{i} + 3 \hat{j}$ and $a \hat{i} + (1 - a) \hat{j}$ respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors $O A$ and $O B$ is $\frac{9}{2}$ then the sum of all the possible values of a is :

(A)

(B)

9/2

Let the position vectors of the points $P, Q, R$ and $S$ be $a = \hat{i} + 2 \hat{j} - 5 \hat{k}$ , $b = 3 \hat{i} + 6 \hat{j} + 3 \hat{k}$ , $c = \frac{17}{5} \hat{i} + \frac{16}{5} \hat{j} + 7 \hat{k}$ and $d = 2 \hat{i} + \hat{j} + \hat{k}$ , respectively. Then which of the following statements is true?

(A)

The points

P, Q, R

and

S

are NOT coplanar

(B)

\frac{b + 2 d}{3}

is the position vector of a point which divides

P R

internally in the ratio

5 : 4

(C)

\frac{b + 2 d}{3}

is the position vector of a point which divides

P R

externally in the ratio

5 : 4

(D)

The square of the magnitude of the vector

b \times d

is 95

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the vectors AB=3i^+4k^ and AC=5i^−2j^​+4k^ are the sides of a triangle ABC, then the length of the median through A is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The resultant of two forces Pn and 3n is a force of 7n. If the direction of 3n force were reversed, the resultant would be 19​n. The value of P is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If C is the mid point of AB and P is any point outside AB, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABC is a triangle. Forces P,Q​,R acting along IA, IB, and IC respectively are in equilibrium, where I is the incentre of △ABC. Then P:Q:R is

A and B are two like parallel forces. A couple of moment H lies in the plane of A and B and is contained with them. The resultant of A and B after combining is displaced through a distance

The resultant R of two forces acting on a particle is at right angles to one of them and its magnitude is one third of the other force. The ratio of larger force to the smaller one is

With two forces acting at point, the maximum affect is obtained when their resultant is 4N. If they act at right angles, then their resultant is 3N. Then the forces are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The vectors AB=3i^+4k^ and AC=5i^−2j^​+4k^ are the sides of a triangle ABC. The length of the median through A is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The resultant of forces P and Q​ is R. If Q​ is doubled then R is doubled. If the direction of Q​ is reversed, then R is again doubled. Then P2:Q2:R2 is

The sum of two forces is 18 N and resultant whose direction is at right angles to the smaller force is 12 N. The magnitude of the two forces are

Show, by vector methods, that the angular bisectors of a triangle are concurrent and find an expression for the position vector of the point of concurrency in terms of the position vectors of the vertices.

Prove, by vector methods or otherwise, that the point of intersection of the diagonals of a trapezium lies on the line passing through the mid-points of the parallel sides. (You may assume that the trapezium is not a parallelogram.)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A vector a=αi^+2j^​+βk^(α,β∈R) lies in the plane of the vectors, b=i^+j^​ and c=i^−j^​+4k^. If a bisects the angle between b and c, then:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a and b are unit vectors, then the greatest value of 3​∣a+b∣+∣a−b∣ is

If the vectors $A B = 3 \hat{i} + 4 \hat{k}$ and $A C = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle $A B C$ , then the length of the median through $A$ is

The resultant of two forces Pn and 3n is a force of 7n. If the direction of 3n force were reversed, the resultant would be $19 n$ . The value of P is

If $C$ is the mid point of $A B$ and $P$ is any point outside $A B$ , then

ABC is a triangle. Forces $P, Q, R$ acting along IA, IB, and IC respectively are in equilibrium, where I is the incentre of $△ A B C$ . Then $P : Q : R$ is

The vectors $A B = 3 \hat{i} + 4 \hat{k}$ and $A C = 5 \hat{i} - 2 \hat{j} + 4 \hat{k}$ are the sides of a triangle ABC. The length of the median through A is

The resultant of forces $P$ and $Q$ is $R$ . If $Q$ is doubled then $R$ is doubled. If the direction of $Q$ is reversed, then $R$ is again doubled. Then $P^{2} : Q^{2} : R^{2}$ is

A vector $a = α \hat{i} + 2 \hat{j} + β \hat{k} (α, β \in R)$ lies in the plane of the vectors, $b = \hat{i} + \hat{j}$ and $c = \hat{i} - \hat{j} + 4 \hat{k}$ . If $a$ bisects the angle between $b$ and $c$ , then:

If $a$ and $b$ are unit vectors, then the greatest value of $3 ∣ a + b ∣ + ∣ a - b ∣$ is