Find the vector equation of the line joining the points 2 i ^ + j ^ + 3 k ^ and - 4 i ^ + 3 j ^ - k ^ .

Let, a → = 2 i ^ + j ^ + 3 k ^ and b → = - 4 i ^ + 3 j ^ - k ^ The vector equation of the line passing through a → , b → is r → = ( 1 - t ) a → + t b → , t ∈ R = 2 i ^ + j ^ + 3 k ^ + t [ - 4 i ^ + 3 j ^ - k ^ - 2 i ^ - j ^ - 3 k ^ ] ∴ The vector equation is r ¯ = 2 i ^ + j ^ + 3 k ^ + t [ - 6 i ^ + 2 j ^ - 4 k ^ ] .

If a → , b → , c → are non-coplanar find the point of intersection of the line passing through the points 2 a → + 3 b → - c → , 3 a → + 4 b → - 2 c → with the line joining the points a → - 2 b → + 3 c → , a → - 6 b → + 6 c → .

Let, O A → = 2 a → + 3 b → - c → , O B → = 3 a → + 4 b → - 2 c → , O C → = a → - 2 b → + 3 c → and O A → = a → - 6 b → + 6 c → The vector equation of the line joining the points O A → , O B → is r → = O A → + t ( O B → - O A → ) , t ∈ R = 2 a → + 3 b → - c → + t 3 a → + 4 b → - 2 c → - 2 a → - 3 b → + c → r → = 2 a → + 3 b → - c → + t a → + b → - c → ...(i) The vector equation of the line joining the points O C → , O D → is r → = O C → + s ( O D → - O C → ) , s ∈ R = a → - 2 b → + 3 c → + s a → - 6 b → + 6 c → - a → + 2 b → - 3 c → r → = a → - 2 b → + 3 c → + s - 4 b → + 3 c → = a → + ( - 2 - 4 s ) b → + ( 3 + 3 s ) c → ...(ii) Comparing a → , b → , c → coefficients of (i) and (ii), we get 2 + t = 1 ⇒ t = - 1 And 3 + t = - 2 - 4 s ⇒ 2 = - 2 + 4 s ⇒ s = - 1 Substitute t in (i), r → = 2 a → + 3 b → - c → + - 1 a → + b → - c → r → = a → + 2 b → ∴ The vector equation is r ¯ = a → + 2 b → .

Find vector equation of the plane which passes through the points 2 i ^ + 4 j ^ + 2 k ^ , 2 i ^ + 3 j ^ + 5 k ^ and parallel to the vector 3 i ^ - 2 j ^ + k ^ . Also find point where this plane meets line joining points 2 i ^ + j ^ + 3 k ^ and 4 i ^ - 2 j ^ + 3 k ^ .

Vector equation of plane passing through points A a → , B b → and parallel to C c → is r → = ( 1 - s ) a → + s b → + t c → Let a → = 2 i ^ + 4 j ^ + 2 k ^ , b → = 2 i ^ + 3 j ^ + 5 k ^ , c → = 3 i - 2 j ^ + k ^ , d → = 2 i ^ + j ^ + 3 k ^ and e → = 4 i ^ - 2 j ^ + 3 k ^ Equation of the plane is r → = ( 1 - s ) ( 2 i ^ + 4 j ^ + 2 k ^ ) + s ( 2 i ^ + 3 j ^ + 5 k ^ ) + t ( 3 i ^ - 2 j ^ + k ^ ) = ( 2 - 2 s + 2 s + 3 t ) i ^ + ( 4 - 4 s + 3 s - 2 t ) j ^ + ( 2 - 2 s + 5 s + t ) k ^ Equation of the line is r → = ( 1 - p ) d → + p e → , where p is a scalar r → = ( 1 - p ) ( 2 i ^ + j ^ + 3 k ^ ) + p ( 4 i ^ - 2 j ^ + 3 k ^ ) = ( 2 + 2 p ) i ^ + ( 1 - 3 p ) j ^ + 3 k ^ By comparing the like terms, we get 3 t + 2 = 2 p + 2 ⇒ 3 t - 2 p = 0 . . . . ( i ) 4 - s - 2 t = 1 - 3 p ⇒ 2 t + s - 3 p = 3 . . . . ( i i ) 2 + 3 s + t = 3 ⇒ 3 s + t = 1 . . . . . ( i i i ) Multiply i i by 3 , we have 6 t + 3 s - 9 p = 9 . . . i v Subtracting ( i i ) & ( i v ) , we get 5 t - 9 p = 8 . . . . ( v ) Solving i & v we get p = - 24 17 To find point of intersection put p in ( 2 ) r → = 2 - 48 17 i ^ + 1 + 72 17 j ^ + 3 k ^ = - 14 17 i ^ + 89 17 j ^ + 3 k ^ ∴ Point of intersection is - 14 17 , 89 17 , 3 .

Find the vector equation of the plane passing through the points 4 i ^ - 3 j ^ - k ^ , 3 i ^ + 7 j ^ - 10 k ^ and 2 i ^ + 5 j ^ - 7 k ^ and show that the point i ^ + 2 j ^ - 3 k ^ lies in the plane.

Let a → = 4 i ^ - 3 j ^ - k ^ , b → = 3 i ^ + 7 j ^ - 10 k ^ and c → = 2 i ^ + 5 j ^ - 7 k ^ The vector equation of the plane passing through a → , b → , c → is r → = ( 1 - s - t ) a → + s b → + t c → , s , t ∈ R = a → + s ( b → - a → ) + t ( c → - a → ) = 4 i ^ - 3 j ^ - k ^ + s ( 3 i ^ + 7 j ^ - 10 k ^ - 4 i ^ + 3 i ^ + k ^ ) + t ( 2 i ^ + 5 j ^ - 7 k ^ - 4 i ^ + 3 j ^ + k ^ ) ⇒ r → = 4 i ^ - 3 i ^ - k ^ + s ( - i ^ + 10 j ^ - 9 k ^ ) + t ( - 2 i ^ + 8 j ^ - 6 k ^ ) Suppose r → = i ^ + 2 j ^ - 3 k ^ lies on plane ⇒ i ^ + 2 j ^ - 3 k ^ = 4 i ^ - 3 j ^ - k ^ + s ( - i ^ + 10 j ^ - 9 k ^ ) + t ( - 2 i ^ + 8 j ^ - 6 k ^ ) Comparing i ^ , j ^ , k ^ on both sides 1 = 4 - s - 2 t ⇒ 2 t + s = 3 . . . . . ( 1 ) 2 = - 3 + 10 s + 8 t ⇒ 8 t + 10 s = 5 . . . . . ( 2 ) - 3 = - 1 - 9 s - 6 t ⇒ 6 t + 9 s = 2 . . . . . ( 3 ) Equation (1) × 3 ⇒ 6 t + 3 s = 9 . . . . . . . 4 Solving 3 and 4 ⇒ s = - 7 6 From (1) 2 t = 3 + 7 6 = 25 6 ⇒ t = 25 12 Substitute s , t in ( 2 ) , we have 8 25 12 + 10 - 7 6 = 5 ⇒ 50 3 - 35 3 = 15 3 ⇒ 5 = 5 Given points are collinear. ∴ Given point i ^ + 2 j ^ - 3 k ^ lines in the same plane

E M B I B E

Intermediate First Year Mathematics paper - 1A>Algebra, Vector Algebra, Trigonometry>Chapter 4 - Addition of Vectors>Exercise 4 (b)

Telangana Board Solutions for Exercise 2: Exercise 4 (b)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Exercise 2: Exercise 4 (b)

Attempt the practice questions from Exercise 2: Exercise 4 (b) with hints and solutions to strengthen your understanding. Intermediate First Year Mathematics paper - 1A solutions are prepared by Experienced Embibe Experts.

Questions from Telangana Board Solutions for Exercise 2: Exercise 4 (b) with Hints & Solutions

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 4 - Addition of Vectors>Exercise 4 (b)>Q I

Find the vector equation of the line joining the points $\bar{r} = 2 \hat{i} + \hat{j} + 3 \hat{k} + t [- 6 \hat{i} + 2 \hat{j} - 4 \hat{k}]$ and $- 4 \hat{i} + 3 \hat{j} - \hat{k}$ .

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 4 - Addition of Vectors>Exercise 4 (b)>Q I

Find the vector equation of the plane passing through the points $\hat{i} - 2 \hat{j} + 3 \hat{k}, - 5 \hat{j} - \hat{k}$ and $- 3 \hat{i} + 5 \hat{j}$ .

EASY

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 4 - Addition of Vectors>Exercise 4 (b)>Q I

Find the vector equation of the plane passing through the points $(0, 0, 0), (0, 5, 0)$ and $(2, 0, 1)$ .

MEDIUM

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 4 - Addition of Vectors>Exercise 4 (b)>Q II

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar find the point of intersection of the line passing through the points $2 \vec{a} + 3 \vec{b} - \vec{c}, 3 \vec{a} + 4 \vec{b} - 2 \vec{c}$ with the line joining the points $\vec{a} - 2 \vec{b} + 3 \vec{c}, \vec{a} - 6 \vec{b} + 6 \vec{c}$ .

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 4 - Addition of Vectors>Exercise 4 (b)>Q II

$A B C D$ is a trapezium in which $A B$ and $C D$ are parallel. Prove by vector methods that the midpoints of the sides $A B, C D$ and intersection of the diagonals are collinear.

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 4 - Addition of Vectors>Exercise 4 (b)>Q II

In a quadrilateral $A B C D$ , if the midpoints of one pair of opposite sides and points of intersection of diagonals are collinear, using vector methods prove that the quadrilateral $A B C D$ is a trapezium.

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 4 - Addition of Vectors>Exercise 4 (b)>Q III

Find vector equation of the plane which passes through the points $2 \hat{i} + 4 \hat{j} + 2 \hat{k}, 2 \hat{i} + 3 \hat{j} + 5 \hat{k}$ and parallel to the vector $3 \hat{i} - 2 \hat{j} + \hat{k}$ . Also find point where this plane meets line joining points $2 \hat{i} + \hat{j} + 3 \hat{k} and 4 \hat{i} - 2 \hat{j} + 3 \hat{k}$ .

HARD

11th Telangana Board

IMPORTANT

Intermediate First Year Mathematics paper - 1A>Chapter 4 - Addition of Vectors>Exercise 4 (b)>Q III

Find the vector equation of the plane passing through the points $4 \hat{i} - 3 \hat{j} - \hat{k}, 3 \hat{i} + 7 \hat{j} - 10 \hat{k} and 2 \hat{i} + 5 \hat{j} - 7 \hat{k}$ and show that the point $\hat{i} + 2 \hat{j} - 3 \hat{k}$ lies in the plane.

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