Find the position vector of a point P which divides the line joining two points A and B whose position vectors are i ^ + 2 j ^ - k ^ and - i ^ + j ^ + k ^ respectively in the ratio 2 : 1 i internally i i externally

Given, A = i ^ + 2 j ^ - k ^ , B = - i ^ + j ^ + k ^ We know that, the position vector of a point P dividing a line segment joining the points A and B whose position vectors are a → and b → respectively, in the ratio m : n externally and internally are given by m b → - n a → m - n and m b → + n a → m + n respectively. So, position vector of a point P which divides the line A B in the ratio of 2 : 1 externally is O P → = 2 - i ^ + j ^ + k ^ - 1 i ^ + 2 j ^ - k ^ 2 - 1 ⇒ O P → = - 2 i ^ + 2 j ^ + 2 k ^ - ( i ^ + 2 j ^ - k ^ ) 2 - 1 = - 2 i ^ + 2 j ^ + 2 k ^ - i ^ + 2 j ^ - k ^ = - 3 i ^ + 3 k ^ . And position vector of a point P which divides the line A B in the ratio 2 : 1 internally is O P → = 2 - i ^ + j ^ + k ^ + 1 · i ^ + 2 j ^ - k ^ 2 + 1 ⇒ O P → = - 1 3 i ^ + 4 3 j ^ + 1 3 k ^ .

Find the position vector of a point P which divides the line joining two points A and B whose position vectors are i ^ + 2 j ^ - k ^ and - i ^ + j ^ + k ^ respectively in the ratio 2 : 1 i internally i i externally

Given, A = i ^ + 2 j ^ - k ^ , B = - i ^ + j ^ + k ^ We know that, the position vector of a point P dividing a line segment joining the points A and B whose position vectors are a → and b → respectively, in the ratio m : n externally and internally are given by m b → - n a → m - n and m b → + n a → m + n respectively. So, position vector of a point P which divides the line A B in the ratio of 2 : 1 externally is O P → = 2 - i ^ + j ^ + k ^ - 1 i ^ + 2 j ^ - k ^ 2 - 1 ⇒ O P → = - 2 i ^ + 2 j ^ + 2 k ^ - ( i ^ + 2 j ^ - k ^ ) 2 - 1 = - 2 i ^ + 2 j ^ + 2 k ^ - i ^ + 2 j ^ - k ^ = - 3 i ^ + 3 k ^ . And position vector of a point P which divides the line A B in the ratio 2 : 1 internally is O P → = 2 - i ^ + j ^ + k ^ + 1 · i ^ + 2 j ^ - k ^ 2 + 1 ⇒ O P → = - 1 3 i ^ + 4 3 j ^ + 1 3 k ^ .

If the position vector of one end of a line segment A B → is 2 i ^ + 3 j ^ - k ^ and the position vector of its middle point is 3 i ^ + j ^ + k ^ , then find the position vector of the other end.

Let A → = 2 i ^ + 3 j ^ - k ^ & B → = r → Middle point of A B → = 3 i ^ + j ^ + k ^ Now, 3 i ^ + j ^ + k ^ = 2 i ^ + 3 j ^ - k ^ + r → 2 ⇒ 6 i ^ + j ^ + k ^ = 2 i ^ + 3 j ^ - k ^ + r → ⇒ r → = 4 i ^ + 3 j ^ + 7 k ^ So, the point B → is 4 i ^ + 3 j ^ + 7 k ^ .

If the position vectors of the points A → , B → and C → are a → , b → and 3 a → – 2 b → respectively, then prove that the points A → , B → and C → are collinear.

Given: a → , b → and 3 a → – 2 b → are the position vectors of A → , B → and C → respectively. To prove: The given points are collinear. Proof: A B → = b → - a → , B C → = 3 a → - 3 b → , A C → = 2 a → - 2 b → A B → × B C → . A C → = b → - a → × 3 a → - 3 b → . ( 2 a → - 2 b → ) = - 3 a → - b → × a → - b → . 2 a → - 2 b → = 0 (As a → × a → = 0 → ) Hence, A → , B → , C → are collinear.

The position vectors of four points P , Q , R and S are 2 a → + 4 c → , 5 a → + 3 3 b → + 4 c → , - 2 3 b → + c → and 2 a → + c → respectively. Prove that P Q → is parallel to R S → .

Given: Position vectors of P , Q , R & S are 2 a → + 4 c → , 5 a → + 3 3 b → + 4 c → , - 2 3 b → + c → and 2 a → + c → respectively. To prove: P Q → is parallel to R S → P Q → = 3 a → + 3 3 b → R S → = 2 a → + 2 3 b → P Q → × R S → = 3 a → + 3 3 b → × ( 2 a → + 2 3 b → ) ⇒ P Q → × R S → = 6 a → + 3 b → × ( a → + 3 b → ) = 0 → Hence, P Q → ∥ R S → .

Find the condition that the three points whose position vectors are a → = a i ^ + b j ^ + c k ^ , b → = i ^ + c j ^ and c → = - i ^ - j ^ are collinear.

Given, a → = a i ^ + b j ^ + c k ^ , b → = i ^ + c j ^ , c → = - i ^ - j ^ To find, The condition for the vectors to be collinear A B → = 1 - a i ^ + c - b j ^ - c k ^ B C → = - 2 i ^ + - 1 - c j ^ A C → = - 1 - a i ^ + - 1 - b j ^ - c k ^ If A , B , C are collinear then A B → = λ B C → ⇒ 1 - a i ^ + c - b j ^ - c k ^ = λ - 2 i ^ + - 1 - c j ^ ⇒ 1 - a = - 2 λ , c - b = λ - 1 - c , - c = 0 c = 0 and - b = λ - 1 ⇒ λ = b ⇒ 1 - a = - 2 λ ⇒ a - 2 b = 1 .

Vectors a → and b → are non-collinear. Find for what values of x , vectors c → = x - 2 a → + b → and d → = 2 x + 1 a → - b → are collinear?

Given: a → and b → are non-collinear c → = x - 2 a → + b → d → = 2 x + 1 a → - b → To find: x such that they are collinear. For collinear, c → + λ d → = 0 x - 2 a → + b → + λ 2 x + 1 a → - b → = 0 ⇒ x - 2 + λ 2 x + 1 a → + 1 - λ b → = 0 Now using a → & b → are non-collinear so their coefficient must be zero, So, x - 2 + λ 2 x + 1 = 0 . . . ( i ) And, 1 - λ = 0 . . . ( ii ) ⇒ λ = 1 Now, put λ = 1 in equation ( i ) , we get x = 1 3 .

E M B I B E

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Exercise for Session 2>Q 16

MEDIUM

JEE Advanced

IMPORTANT

Earn 100

Find the position vector of a point $P$ which divides the line joining two points $A$ and $B$ whose position vectors are $\hat{i} + 2 \hat{j} - \hat{k}$ and $- \hat{i} + \hat{j} + \hat{k}$ respectively in the ratio $2 : 1$

$(i)$ internally
$(i i)$ externally

Important Questions on Vector Algebra

MEDIUM

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Exercise for Session 2>Q 16

Find the position vector of a point $P$ which divides the line joining two points $A & B$ , whose position vectors are $\hat{i} + 2 \hat{j} - \hat{k} and - \hat{i} + \hat{j} + \hat{k}$ , respectively in the ratio of $2 : 1$ externally.