Find the position vector of a point R which divides the line joining the two points P and Q with position vectors O P → = 2 a → + b → and O Q → = a → - 2 b → respectively in the ratio 1 : 2 internally and externally.

It is given that P and Q are two points with position vectors O P → = 2 a → + b → and O Q → = a → - 2 b → respectively. When R divides P Q internally in the ratio 1 : 2 , then Position vector of R = 1 × ( a → - 2 b → ) + 2 × ( 2 a → + b → ) 1 + 2 = 5 a → 3 When R divides PQ externally in the ratio 1 : 2 , then Position vector of R = 1 × ( a → - 2 b → ) - 2 × ( 2 a → + b → ) 1 - 2 = - 3 a → - 4 b → - 1 = 3 a → + 4 b → .

Let a → , b → , c → , d → be the position vectors of the four distinct points A , B , C , D . If b → - a → = c → - d → then show that A B C D is a parallelogram.

Consider the quadrilateral A B C D with a → , b → , c → and d → as the position vectors of the four distinct points A , B , C and D . Given that b → - a → = c → - d → ⇒ A B → = D C → ⇒ A B ‖ D C Recall that b → - a → = c → - d → ⇒ b → - c → = a → - d → ⇒ C B → = D A → ⇒ C B ‖ D A Thus, the opposite sides pairs of A B C D are equal and parallel. Hence, A B C D is a parallelogram.

Show that the four points A , B , C , D with position vectors a , → b , → c , → d → respectively such that 3 a → - 2 b → + 5 c → - 6 d → = 0 , are coplanar. Also, find the position vector of the point of intersection of the line segments A C and B D .

Let A C and B D intersects at a point P . We have, 3 a → - 2 b → + 5 c → - 6 d → = 0 → . ⇒ 3 a → + 5 c → = 2 b → + 6 d → Since sum of coefficients on both sides of the above equation is 8 , so we divide the equation on both sides by 8 . i.e., 3 a → + 5 c → 8 = 2 b → + 6 d → 8 ⇒ 3 a → + 5 c → 3 + 5 = 2 b → + 6 d → 2 + 6 Therefore, P divides A C in the ratio of 3 : 5 and P divides B D in the ratio of 2 : 6 Therefore, position vector of the point of intersection of A C and B D will be 3 a → + 5 c → 8 or 2 b → + 6 d → 8 . Since A C and B D intersect at point P , therefore A , B , C and D are coplanar.

Show that the four points P , Q , R , S with position vectors p → , q → , r → , s → respectively such that 5 p → - 2 q → + 6 r → - 9 s → = 0 → ,are coplanar. Also, find the position vector of the point of intersection of the line segments P R and Q S .

Let the point of intersection of the line segments P R and Q S is A . Then 5 p → - 2 q → + 6 r → - 9 s → = 0 → ⇒ 5 p → + 6 r → = 2 q → + 9 s → The sum of the coefficients on both the sides of the above equation is 11 .So, we divide the given equation with 11 . ⇒ 5 p → + 6 r → 11 = 2 q → + 9 s → 11 ⇒ 5 p → + 6 r → 5 + 6 = 2 q → + 9 s → 2 + 9 Therefore, A divides P R in the ratio of 5 : 6 and Q S in the ratio of 2 : 9 The position vector of the point of intersection of the line segment is 5 p + 6 r ' 11 , 2 q + 9 s 11 .

E M B I B E

MATHEMATICS CLASS XII VOLUME-2>Chapter 4 - Algebra of Vectors>EXERCISE

R D Sharma Solutions for Chapter: Algebra of Vectors, Exercise 3: EXERCISE

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Algebra of Vectors, Exercise 3: EXERCISE

Attempt the free practice questions on Chapter 4: Algebra of Vectors, Exercise 3: EXERCISE with hints and solutions to strengthen your understanding. MATHEMATICS CLASS XII VOLUME-2 solutions are prepared by Experienced Embibe Experts.

Questions from R D Sharma Solutions for Chapter: Algebra of Vectors, Exercise 3: EXERCISE with Hints & Solutions

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 4 - Algebra of Vectors>EXERCISE>Q 1

Find the position vector of a point $R$ which divides the line joining the two points $P$ and $Q$ with position vectors $\vec{O P} = 2 \vec{a} + \vec{b} and \vec{O Q} = \vec{a} - 2 \vec{b}$ respectively in the ratio $1 : 2$ internally and externally.

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 4 - Algebra of Vectors>EXERCISE>Q 2

Let $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ be the position vectors of the four distinct points $A, B, C, D$ . If $\vec{b} - \vec{a} = \vec{c} - \vec{d}$ then show that $A B C D$ is a parallelogram.

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 4 - Algebra of Vectors>EXERCISE>Q 3

If $\vec{a}, \vec{b}$ are the position vectors of $A, B$ respectively, find the position vector of a point $C$ in $A B$ produced such that $A C = 3 A B$ and that a point $D$ in $B A$ produced such that $B D = 2 B A$ .

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 4 - Algebra of Vectors>EXERCISE>Q 4

Show that the four points $A, B, C, D$ with position vectors $\vec{a,} \vec{b,} \vec{c,} \vec{d}$ respectively such that $3 \vec{a} - 2 \vec{b} + 5 \vec{c} - 6 \vec{d} = 0,$ are coplanar. Also, find the position vector of the point of intersection of the line segments $A C$ and $B D .$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 4 - Algebra of Vectors>EXERCISE>Q 5

Show that the four points $P, Q, R, S$ with position vectors $\vec{p}, \vec{q}, \vec{r}, \vec{s}$ respectively such that $5 \vec{p} - 2 \vec{q} + 6 \vec{r} - 9 \vec{s} = \vec{0}$ ,are coplanar. Also, find the position vector of the point of intersection of the line segments $P R$ and $Q S$ .

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 4 - Algebra of Vectors>EXERCISE>Q 6

The vertices $A, B, C$ of triangle $A B C$ have respectively position vectors $\vec{a}, \vec{b}, \vec{c}$ with respect to a given origin $O$ . Show that the point $D$ where the bisector of $∠ A$ meets $B C$ has position vector $\vec{d} = \frac{β \vec{b} + γ \vec{c}}{β + γ}, where β = |\vec{c} - \vec{a}| and, γ = |\vec{a} - \vec{b}|$ .Hence, deduce that the incentre $I$ has position vector $\frac{α \vec{a} + β \vec{b} + γ \vec{c}}{α + β + γ}, where α = |\vec{b} - \vec{c}|$ .

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