Find a unit vector perpendicular to each of the vectors ( a → + b → ) and ( a → - b → ) , where a → = i ^ + j ^ + k ^ , b → = i ^ + 2 j ^ + 3 k ^ .

The given two vectors are a → = i ^ + j ^ + k ^ and b → = i ^ + 2 j ^ + 3 k ^ . So, a → + b → = i ^ + j ^ + k ^ + i ^ + 2 j ^ + 3 k ^ = 2 i ^ + 3 j ^ + 4 k ^ And a → - b → = i ^ + j ^ + k ^ - i ^ + 2 j ^ + 3 k ^ = - j ^ - 2 k ^ A vector which is perpendicular to both a → + b → and a → - b → is a → + b → × a → - b → So, a → + b → × a → - b → = i ^ j ^ k ^ 2 3 4 0 - 1 - 2 = - 2 i ^ + 4 j ^ - 2 k ^ And a → + b → × a → - b → = ( - 2 ) 2 + ( 4 ) 2 + ( - 2 ) 2 = 24 = 2 6 So, the unit vector perpendicular to both a → + b → and a → - b → is = ± a → + b → × a → - b → a → + b → × a → - b → = ± - 2 i ^ + 4 j ^ - 2 k ^ 2 6 = ± - i ^ + 2 j ^ - k ^ 6 Which is the required unit vector.

Show that the four points with position vectors 4 i ^ + 8 j ^ + 12 k ^ , 2 i ^ + 4 j ^ + 6 k ^ , 3 i ^ + 5 j ^ + 4 k ^ and 5 i ^ + 8 j ^ + 5 k ^ are coplanar.

Let A → = 4 i ^ + 8 j ^ + 12 k ^ , B → = 2 i ^ + 4 j ^ + 6 k ^ , C → = 3 i ^ + 5 j ^ + 4 k ^ & D → = 5 i ^ + 8 j ^ + 5 k ^ Now, A B → = Position vector of B - Position vector of A = 2 i ^ + 4 j ^ + 6 k ^ - 4 i ^ + 8 j ^ + 12 k ^ = - 2 i ^ - 4 j ^ - 6 k ^ Similarly, A C → = 3 i ^ + 5 j ^ + 4 k ^ - 4 i ^ + 8 j ^ + 12 k ^ = - i ^ - 3 j ^ - 8 k ^ and A D → = 5 i ^ + 8 j ^ + 5 k ^ - 4 i ^ + 8 j ^ + 12 k ^ = i ^ - 7 k ^ If A , B , C , D are coplanar, A B → , A C → , A D → lie in the same plane and are coplanar and A B → A C → A D → = 0 Now, A B → A C → A D → = - 2 - 4 - 6 - 1 - 3 - 8 1 0 - 7 = - 2 21 - 0 + 4 7 - - 8 - 6 0 - - 3 = - 42 + 60 - 18 = 0 Hence, A , B , C , D are coplanar.

Find the projection of the vector a → = i ^ + 3 j ^ + 7 k ^ on the vector b → = 7 i ^ - j ^ + 8 k ^ .

Given vectors are a → = i ^ + 3 j ^ + 7 k ^ and b → = 7 i ^ - j ^ + 8 k ^ To find: Projection of the vector a → on the vector b → We know that, the projection of vector a → on the vector b → is given by a → · b → b → Now, a → · b → = i ^ + 3 j ^ + 7 k ^ · 7 i ^ - j ^ + 8 k ^ = 1 7 + 3 - 1 + 7 8 = 7 - 3 + 56 = 60 And b → = 7 i ^ - j ^ + 8 k ^ = 7 2 + - 1 2 + 8 2 = 49 + 1 + 64 = 114 Therefore, the projection of vector a → on the vector b → is 60 114 .

Find a unit vector perpendicular to each of the vectors ( a → + b → ) and ( a → - b → ) , where a → = 3 i ^ + 2 j ^ + 2 k ^ , b → = i ^ + 2 j ^ - 2 k ^ .

Given, a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ Now, a → + b → = 4 i ^ + 4 j ^ , a → - b → = 2 i ^ + 4 k ^ And ( a → + b → ) × ( a → - b → ) = i ^ j ^ k ^ 4 4 0 2 0 4 = i ^ ( 16 - 0 ) - j ^ ( 16 - 0 ) + k ^ ( 0 - 8 ) = 16 i ^ - 16 j ^ - 8 k ^ | ( a → + b → ) × ( a → - b → ) | = 16 2 + ( - 16 ) 2 + ( - 8 ) 2 = 8 2 2 + 2 2 + 1 = 8 9 = 8 × 3 = 24 So, the unit vector in the direction of ( a → + b → ) × ( a → - b → ) = ± ( a → + b → ) × ( a → - b → ) | ( a → + b → ) × ( a → - b → ) | = ± 16 i ^ - 16 j ^ - 8 k ^ 24 = ± 2 i ^ - 2 j ^ - k ^ 3 = ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ Hence, the unit vector perpendicular to each of the vector a → + b → and a → - b → is given by ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ .

E M B I B E

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

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

Author:Embibe Experts

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

Attempt the free practice questions on Chapter 10: Vector Algebra, Exercise 1: Karnataka 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: Karnataka Board-2018 with Hints & Solutions

HARD

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Karnataka Board-2018>Q 4

Show that the position vector of the point $\Rightarrow n \vec{c} - n \vec{a} = m \vec{b} - m \vec{c}$ , which divides the line joining the points $A and B$ having position vectors $\vec{a} and \vec{b}$ internally in the ratio $m : n is \frac{m \vec{b} + n \vec{a}}{m + n}$ .

HARD

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Karnataka Board-2018>Q 5

Find $x$ such that the four points $A (3, 2, 1), B (4, x, 5), C (4, 2, - 2)$ and $D (6, 5, - 1)$ are coplanar.

MEDIUM

12th Karnataka Board

IMPORTANT

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

Find a unit vector perpendicular to each of the vectors $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ , where $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ .

MEDIUM

12th Karnataka Board

IMPORTANT

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

Show that the four points with position vectors $4 \hat{i} + 8 \hat{j} + 12 \hat{k}$ , $2 \hat{i} + 4 \hat{j} + 6 \hat{k}, 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$ and $5 \hat{i} + 8 \hat{j} + 5 \hat{k}$ are coplanar.

MEDIUM

12th Karnataka Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Karnataka Board-2020>Q 12

Find the projection of the vector $\vec{a} = \hat{i} + 3 \hat{j} + 7 \hat{k} on the vector \vec{b} = 7 \hat{i} - \hat{j} + 8 \hat{k}$ .

MEDIUM

12th Karnataka Board

IMPORTANT

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

Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a} = \hat{i} - \hat{j} + 3 \hat{k} and \vec{b} = 2 \hat{i} - 7 \hat{j} + \hat{k}$ .

MEDIUM

12th Karnataka Board

IMPORTANT

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

Find a unit vector perpendicular to each of the vectors $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ , where $\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}, \vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ .