Find the magnitude and direction of r → 1 - r → 2 when r → 1 = r → 2 = 5 and r 1 → points North –East while points r 2 → North –West.

To find the required magnitude and direction of r 1 → - r 2 → . Here, r 1 → = O A → pointing North-East and r 2 → = O B → pointing North-West. Where O A → = O B → = 5 . As shown in figure, ∠ B O A = 90 ° r 1 → - r 2 → = B A → (Using triangle law) Clearly, ∠ B O A is right-angled at O . B A → 2 = O A → 2 + O B → 2 =5 2 +5 2 =50 B A → = 5 5 Or r 1 → - r 2 → = 5 5 i.e., r 1 → – r 2 → has magnitude 5 5 and points from West to East.

Let O A C B be a parallelogram with O → at the origin and O C → as 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.

To find out the ratio, it is given that O A C B be a parallelogram shown as Here, O D → = 1 2 B C → ⇒ O P → + P D → = 1 2 B P → + P C → ; [using triangle law of vector addition] ⇒ 2 O P → + 2 P D → = B P → + P C → ⇒ – 2 P O → + 2 P D → = – P B → + P C → ⇒ P B → + 2 P D → = P C → + 2 P O → ⇒ P B → +2 P D → 1 +2 = P C → +2 P O → 1 +2 The common point P → of B D → and C O → divides each in the ratio 2 : 1 .

∆ A B C is a triangle with the point P on side B C → such that 3 B P → = 2 P C → , the point Q is on the line C A → such that 4 C Q → = Q A → . Find the ratio in which the line joining the common point R of A P → and B Q → and the point S divides A B → .

It is given that A B C is a triangle with the point P on side B C → such that 3 B P → = 2 P C → , the point Q is on the line C A → such that 4 C Q → = Q A → . Let S be point of intersection of A B → and C R → . Let A be the origin 0 → and the position vectors of the points B , C , P , Q , R and S be b → , c → , p → , q → , r → and s → , respectively. ⇒ p → = 3 b → +2 c → 5 . . . i And q → = 4 c → 5 . . . ii Equating c → from equation i & ii ⇒ 5 p → -3 b → 2 = 5 q → 4 ⇒ 10 p → – 6 b → =5 q → 10 p → = 5 q → + 6 b → ⇒ 10 p → 11 = 5 q → +6 b → 11 = r → . . . iii Equating p → from equation i & iii ⇒ 11 r → 10 = p → = 3 b → +2 c → 5 11 r → = 6 b → + 4 c → ⇒ 11 r → - 4 c → = 6 b → 11 r → –4 c → 7 = 6 7 b → = s → , Thus S divides A B → in ratio 6 : 1

Let r → 1 , r → 2 , r → 3 , . . . r n → be the position vectors of points P 1 , P 2 , P 3 , . . . P n relative to an origin O . Show that if the vector's equation a 1 r → 1 + a 2 r → 2 + a 3 r → 3 . . . + a n r → n = 0 holds, then a similar equation will also hold good with respect to any other origin O ' if a 1 + a 2 + a 3 . . . + a n = 0

Given, a 1 r → 1 + a 2 r → 2 + a 3 r → 3 . . . + a n r → n = 0 Let the position vector of O ' with reference to O as the origin be α . Then, Now, O ' P i → = Position vector ⇒ O ' P i → = P i - Position vector of O ' ⇒ O ' P i → = r → i - α . . . i where, i = 1, 2 , … , n Let the position vectors of P 1 , P 2 , P 3 , … , P n with respect to O ' as the origin be R → 1 , R → 2 , R → 3 , … , R → n , respectively. Then, R → i = O ' P i = r i - α , i = 1,2 , … , n , [using equation i ] Now, a 1 R → 1 + a 2 R → 2 + … + a n R → n = 0 ⇒ ∑ i = 1 n a i R → i = 0 ⇒ ∑ i = 1 n a i r → i - α = 0 ⇒ ∑ i = 1 n a i r → i - ∑ i = 1 n a i α = 0 ⇒ 0 - α ∑ i = 1 n a i = 0 ; ∵ ∑ i = 1 n a i r → i = 0 given ⇒ ∑ i = 1 n a i = 0 Thus, ∑ i = 1 n a i R → i = 0 will hold good, if ∑ i = 1 n a i = 0 .

Let O A B C D be a pentagon in which the sides O A → and C B → are parallel and the sides O D → and A B → are parallel as shown in figure. Also, O A → : C B → = 2 : 1 and O D → : A B → = 1 : 3 . If the diagonals O C → and A D → meet at X , find O X → : X C → .

We have, Let the position vectors of A , B , C and D be a → , b → , c → and d → , respectively. Then O A → : C B → = 2 : 1 ⇒ O A → = 2 C B → ⇒ a → = 2 b → – c → ⋯ i And O D → : A B → = 1 : 3 ⇒ 3 O D → = A B → ⇒ 3 d → = b → – a → = b → – 2 b → – c → [using equation i ] ⇒ 3 d → = – b → + 2 c → … ii Let O X → : X C → = λ :1 and A X → : X D → = μ :1 . Now, X divides O C → in the ratio λ :1 . Therefore, P.V. of x → = λ c → λ +1 . . . iii X also divides A D → in the ratio μ :1 P.V. of x → = μ d → + a → μ +1 . . . iv From equations iii and iv , we get λ c → λ +1 = μ d → + a → μ +1 ⇒ λ λ +1 c → = μ μ +1 d → + 1 μ +1 a → ⇒ λ λ +1 c → = μ μ +1 – b → +2 c → 3 + 1 μ +1 2 b → – c → [using equations i and ii ] ⇒ λ λ +1 c → = 6 – μ 3 μ +1 b → + 2 μ 3 μ +1 – 2 μ +1 c → ⇒ λ λ +1 c → = 6 – μ 3 μ +1 b → + 2 μ –6 3 μ +1 c → ⇒ 6 – μ 3 μ +1 b → + 2 μ –6 3 μ +1 – λ λ +1 c → = 0 ⇒ 6 – μ 3 μ + 1 = 0 and 2 μ –6 3 μ +1 – λ λ +1 = 0 (as b → and c → are non –collinear) ⇒ λ = 2 5 Hence, O X → : X C → = 2 : 5 .

If u → , v → & w → is a linearly independent system of vectors, examine the system p → , q → & r → where p → = cos a u → + cos b v → + cos c w → q → = sin a u → + sin b v → + sin c w → r → = sin x + a u → + sin x + b v → + sin x + c w → for linearly dependent.

To examine the given system, let l , m & n be scalars such that l p → + m q → + n r → = 0 l cos a u → + cos b v → + cos c w → + m sin a u → + sin b v → + sin c w → + n sin x + a u → + sin x + b v → + sin x + c w → = 0 l cos a + m sin a + n sin x + a = 0 … i l cos b + m sin b + n sin x + b = 0 … ii l cos c + m sin c + n sin x + c = 0 … iii This is a homogeneous system of linear equations in l , m & n The determinant of the coefficient matrix is Δ = cos a sin a sin x + a cos b sin b sin x + b cos c sin c sin x + c = cos a sin a 0 cos b sin b 0 cos c sin c 0 = 0 (using c 3 → c 3 - sin x c 1 - cos x c 2 ) So, the above system of equations has non-trivial solutions also. This means that l , m & n may attain non-zero values also. Hence, the given system of vectors is a linearly dependent system of vectors.

Amit M Agarwal Solutions for Chapter: Vector Algebra, Exercise 10: Vector Algebra Exercise 7

Author:Amit M Agarwal

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Vector Algebra, Exercise 10: Vector Algebra Exercise 7

Attempt the practice questions on Chapter 1: Vector Algebra, Exercise 10: Vector Algebra Exercise 7 with hints and solutions to strengthen your understanding. Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY solutions are prepared by Experienced Embibe Experts.

Questions from Amit M Agarwal Solutions for Chapter: Vector Algebra, Exercise 10: Vector Algebra Exercise 7 with Hints & Solutions

EASY

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Vector Algebra Exercise 7>Q 112

A vector $\vec{a}$ has components ${\vec{a}}_{1}, {\vec{a}}_{2} & {\vec{a}}_{3}$ in a right-handed rectangular Cartesian system $O X Y Z$ . The coordinate system is rotated about $Z$ -axis through the angle $\frac{π}{2}$ . Find the components of $\vec{a}$ in new system.

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Vector Algebra Exercise 7>Q 113

Find the magnitude and direction of ${\vec{r}}_{1} - {\vec{r}}_{2}$ when $|{\vec{r}}_{1}| = |{\vec{r}}_{2}| = 5$ and $\vec{r_{1}}$ points North –East while points $\vec{r_{2}}$ North –West.

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Vector Algebra Exercise 7>Q 114

Let $O A C B$ be a parallelogram with $\vec{O}$ at the origin and $\vec{O C}$ as a diagonal. Let $\vec{D}$ be the midpoint of $\vec{O A}$ . Using vector methods, prove that $\vec{B D}$ and $\vec{C O}$ intersect in the same ratio, determine this ratio.

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Vector Algebra Exercise 7>Q 115

$∆ A B C$ is a triangle with the point $P$ on side $\vec{B C}$ such that $3 \vec{B P} = 2 \vec{P C}$ , the point $Q$ is on the line $\vec{C A}$ such that $4 \vec{C Q} = \vec{Q A}$ . Find the ratio in which the line joining the common point $R$ of $\vec{A P}$ and $\vec{B Q}$ and the point $S$ divides $\vec{A B}$ .

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Vector Algebra Exercise 7>Q 116

In a $∆ A B C$ internal angle bisectors $A I, B I$ and $C I$ are produced to meet opposite sides in $A^{'}, B^{'}$ and $C^{'}$ , respectively. Prove that the maximum value of $\frac{A I \cdot B I \cdot C I}{A A^{'} \cdot B B^{'} \cdot C C^{'}}$ is $\frac{8}{27}$

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Vector Algebra Exercise 7>Q 117

Let ${\vec{r}}_{1}, {\vec{r}}_{2}, {\vec{r}}_{3}, . . . \vec{r_{n}}$ be the position vectors of points $P_{1}, P_{2}, P_{3}, . . . P_{n}$ relative to an origin $O$ . Show that if the vector's equation $a_{1} {\vec{r}}_{1} + a_{2} {\vec{r}}_{2} + a_{3} {\vec{r}}_{3} . . . + a_{n} {\vec{r}}_{n} = 0$ holds, then a similar equation will also hold good with respect to any other origin $O^{'}$ if $a_{1} + a_{2} + a_{3} . . . + a_{n} = 0$

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Vector Algebra Exercise 7>Q 118

Let $O A B C D$ be a pentagon in which the sides $\vec{O A}$ and $\vec{C B}$ are parallel and the sides $\vec{O D}$ and $\vec{A B}$ are parallel as shown in figure. Also, $|\vec{O A}| : |\vec{C B}| = 2 : 1$ and $|\vec{O D}| : |\vec{A B}| = 1 : 3$ . If the diagonals $\vec{O C}$ and $\vec{A D}$ meet at $X$ , find $|\vec{O X}| : |\vec{X C}|$ .

Question Image

HARD

JEE Advanced

IMPORTANT

Skills in Mathematics for JEE MAIN & ADVANCED VECTORS & 3D GEOMETRY>Chapter 1 - Vector Algebra>Vector Algebra Exercise 7>Q 119

If $\vec{u}, \vec{v} & \vec{w}$ is a linearly independent system of vectors, examine the system $\vec{p}, \vec{q} & \vec{r}$ where

$\vec{p} = (\cos a) \vec{u} + (\cos b) \vec{v} + (\cos c) \vec{w}$

$\vec{q} = (\sin a) \vec{u} + (\sin b) \vec{v} + (\sin c) \vec{w}$

$\vec{r} = \sin (x + a) \vec{u} + \sin (x + b) \vec{v} + \sin (x + c) \vec{w}$ for linearly dependent.

Amit M Agarwal Solutions for Chapter: Vector Algebra, Exercise 10: Vector Algebra Exercise 7

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Vector Algebra, Exercise 10: Vector Algebra Exercise 7

Questions from Amit M Agarwal Solutions for Chapter: Vector Algebra, Exercise 10: Vector Algebra Exercise 7 with Hints & Solutions

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