The position vector of three points A , B , C are given to be i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ respectively. Find A B → and A C → .

Given the position vectors of three points A , B , C as i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ . Now, A B → = Position vector of B - Position vector of A . ⇒ A B → = 4 i ^ + 4 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A B → = 3 i ^ - 3 j ^ + k ^ and A C → = Position vector of C - Position vector of A ⇒ A C → = - 2 i ^ + 4 j ^ + 2 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A C → = - 3 i ^ + j ^ - k ^

The position vector of three points A , B , C are given to be i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ respectively. Find the angle between A B → and A C → .

Given the position vectors of three points A , B , C as i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ . Now, A B → = Position vector of B - Position vector of A . ⇒ A B → = 4 i ^ + 4 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A B → = 3 i ^ - 3 j ^ + k ^ and A C → = Position vector of C - Position vector of A ⇒ A C → = - 2 i ^ + 4 j ^ + 2 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A C → = - 3 i ^ + j ^ - k ^ Now, A B → = 3 2 + - 3 2 + 1 2 = 19 and A C → = - 3 2 + 1 2 + - 1 2 = 11 and A B → · A C → = 3 i ^ - 3 j ^ + k ^ · - 3 i ^ + j ^ - k ^ ⇒ A B → · A C → = - 9 - 3 - 1 = - 13 The angle between A B → and A C → is given by: cos ⁡ θ = A B → ⋅ A C → | A B → | ⋅ | A C → | ⇒ cos ⁡ θ = - 13 19 · 11 ⇒ θ = cos - 1 - 13 19 · 11

The position vector of three points A , B , C are given to be i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ respectively. Find a vector which is perpendicular to both A B → and A C → having magnitude 9 units.

Given the position vectors of three points A , B , C as i ^ + 3 j ^ + 3 k ^ , 4 i ^ + 4 k ^ and - 2 i ^ + 4 j ^ + 2 k ^ . Now, A B → = Position vector of B - Position vector of A . ⇒ A B → = 4 i ^ + 4 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A B → = 4 i ^ + 4 k ^ - i ^ - 3 j ^ - 3 k ^ ⇒ A B → = 3 i ^ - 3 j ^ + k ^ and A C → = Position vector of C - Position vector of A ⇒ A C → = - 2 i ^ + 4 j ^ + 2 k ^ - i ^ + 3 j ^ + 3 k ^ ⇒ A C → = - 2 i ^ + 4 j ^ + 2 k ^ - i ^ - 3 j ^ - 3 k ^ ⇒ A C → = - 3 i ^ + j ^ - k ^ Now, A B → × A C → = i ^ j ^ k ^ 3 − 3 1 − 3 1 − 1 = 2 i ^ - 6 k ^ Also, | A B → × A C → | = 2 2 + - 6 2 = 2 10 Let r → be the vector which is perpendicular to A B → × A C → then we have, r → = ± k ⋅ A B → × A C → | A B → × A C → | , where k is a scalar. Now, A B → × A C → | A B → × A C → | = 2 i ^ − 6 k ^ 2 10 = i ^ − 3 k ^ 10 Thus, r → is given by = 9 i ^ − 3 k ^ 10

If a → , b → , c → are coplanar, prove that a → + b → , b → + c → , c → + a → are coplanar.

If a → , b → , c → are coplanar, then a → b → c → = 0 To prove: a → + b → , b → + c → , c → + a → are coplanar, which is [ a → + b → b → + c → c → + a → ] = 0 Now, [ a → + b → b → + c → c → + a → ] = ( a → + b → ) · ( b → + c → ) × ( c → + a → ) = ( b → + c → ) × ( c → + a → ) · ( a → + b → ) = b → × c → + b → × a → + c → × c → + c → × a → · ( a → + b → ) = b → × c → + b → × a → + c → × a → · ( a → + b → ) ∵ c → × c → = 0 = ( b → × c → ) · a → + ( b → × a → ) · a → + ( c → × a → ) · a → + ( b → × c → ) · b → + ( b → × a → ) · b → + ( c → × a → ) · b → = [ b → c → a → ] + 0 + 0 + 0 + 0 + [ c → a → b → ] = 2 [ a → b → c → ] {Since, [ a → b → c → ] = [ b → c → a → ] = [ c → a → b → ] } = 2 × 0 = 0 Hence, proved that a → + b → , b → + c → , c → + a → are coplanar.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Kerala Board-2018

Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 1: Kerala Board-2018

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 1: Kerala Board-2018

Attempt the free practice questions on Chapter 10: Vector Algebra, Exercise 1: Kerala Board-2018 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 1: Kerala Board-2018 with Hints & Solutions

EASY

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Kerala Board-2019>Q 7

The position vector of three points $A, B, C$ are given to be $\hat{i} + 3 \hat{j} + 3 \hat{k}, 4 \hat{i} + 4 \hat{k}$ and $- 2 \hat{i} + 4 \hat{j} + 2 \hat{k}$ respectively. Find $\vec{A B}$ and $\vec{A C}$ .

MEDIUM

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Kerala Board-2019>Q 8

The position vector of three points $A, B, C$ are given to be $\hat{i} + 3 \hat{j} + 3 \hat{k}, 4 \hat{i} + 4 \hat{k}$ and $- 2 \hat{i} + 4 \hat{j} + 2 \hat{k}$ respectively. Find the angle between $\vec{A B}$ and $\vec{A C}$ .

HARD

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Kerala Board-2019>Q 9

The position vector of three points $A, B, C$ are given to be $\hat{i} + 3 \hat{j} + 3 \hat{k}, 4 \hat{i} + 4 \hat{k}$ and $- 2 \hat{i} + 4 \hat{j} + 2 \hat{k}$ respectively. Find a vector which is perpendicular to both $\vec{A B}$ and $\vec{A C}$ having magnitude $9$ units.

EASY

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Kerala Board-2019>Q 10

If $\vec{a}, \vec{b}, \vec{c}$ are coplanar vectors, write the vector perpendicular to $\vec{a}$ .

MEDIUM

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Kerala Board-2019>Q 11

If $\vec{a}, \vec{b}, \vec{c}$ are coplanar, prove that $\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}$ are coplanar.

EASY

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Kerala Board-2020>Q 13

If $\vec{a}, \vec{b}, \vec{c}$ are three coplanar vectors, then $[\vec{a} \vec{b} \vec{c}]$ is

EASY

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Kerala Board-2020>Q 14

If $|\vec{a}| = 2, |\vec{b}| = 3$ and $θ$ is the angle between $\vec{a}$ and $\vec{b}$ . Then maximum value of $\vec{a} \cdot \vec{b}$ occurs when $θ = . . . . . . . . . . . . . . . .$ .

MEDIUM

12th Kerala Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Kerala Board-2020>Q 15

If $\vec{b} = 2 i + j - k, \vec{c} = i + 3 k$ and $\vec{a}$ is a unit vector. Find the maximum value of Scalar triplet product $[\vec{a} \vec{b} \vec{c}]$ .

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