Find the projection of a → = 2 i ^ + 3 j ^ + 2 k ^ on the vector b → = i ^ + 2 j ^ + k ^ .

The projection of a vector a → on b → is given by a → · b → b → . Therefore, a → · b → b → = 2 i ^ + 3 j ^ + 2 k ^ · i ^ + 2 j ^ + k ^ 1 2 + 2 2 + 1 2 = 2 × 1 + 3 × 2 + 2 × 1 1 + 4 + 1 = 2 + 6 + 2 6 = 10 6

If a → + b → + c → = 0 → , then prove that a → × b → = b → × c → = c → × a → .

Given that, a → + b → + c → = 0 → and To prove: a → × b → = b → × c → = c → × a → Here, a → = - ( b → + c → ) ⇒ a → × b → = - ( b → + c → ) × b → = - b → × b → - c → × b → = 0 → + b → × c → = b → × c → [Using a → × a → = b → × b → = c → × c → = 0 ] Again, b → = - ( a → + c → ) ⇒ a → × b → = - a → × ( a → + c → ) = - a → × a → - a → × c → = 0 - a → × c → = c → × a → Thus, a → × b → = b → × c → = c → × a → Hence, proved.

Find the angle between two vectors a → and b → with magnitudes 1 and 2 , respectively, and satisfying a → · b → = 1

The angle between two vectors a → and b → can be find out using the formula, cos θ = a → · b → a → b → In the question, a → · b → = 1 , a → = 1 and b → = 2 is given, so substituting these values, cos θ = 1 1 × 2 cos θ = 1 2 θ = cos - 1 1 2 θ = 60 ° or π 3 .

Find a vector in the direction of the vector a → = i ^ - 2 j ^ that has magnitude of 7 units.

To find a vector in the direction of vector a → with a magnitude of 7 units, we first need to find the unit vector in the direction of vector a → . The unit vector is given as, u a → = a → a → ⇒ u a → = i ^ - 2 j ^ 1 2 + - 2 2 ⇒ u a → = i ^ - 2 j ^ 1 + 4 ⇒ u a → = i ^ - 2 j ^ 5 ⇒ u a → = 1 5 i ^ - 2 5 j ^ Now, we have got a unit vector in the direction of a → . The vector of a magnitude of 7 units in the direction of a → can be obtained as, 7 u a → = 7 1 5 i ^ - 2 5 j ^ ⇒ 7 u a → = 7 5 i ^ - 14 5 j ^

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

The given two vectors are a → = i ^ + j ^ + k ^ and b → = i ^ + 2 j ^ + 3 k ^ . Now we need to find a → + b → and a → - b → . So, a → + b → = i ^ + j ^ + k ^ + i ^ + 2 j ^ + 3 k ^ a → + b → = 2 i ^ + 3 j ^ + 4 k ^ And, a → - b → = i ^ + j ^ + k ^ - i ^ + 2 j ^ + 3 k ^ a → - b → = - j ^ - 2 k ^ Now the cross product of these vectors will be a → + b → × a → - b → = i ^ j ^ k ^ 2 3 4 0 - 1 - 2 ⇒ a → + b → × a → - b → = - 2 i ^ + 4 j ^ - 2 k ^ And the corresponding magnitude will be a → + b → × a → - b → = ( - 2 ) 2 + ( 4 ) 2 + ( - 2 ) 2 ⇒ a → + b → × a → - b → = 4 + 16 + 4 ⇒ a → + b → × a → - b → = 24 ⇒ a → + b → × a → - b → = 6 × 4 ⇒ a → + b → × a → - b → = 2 6 Now we know vector the product a → + b → × a → - b → will be perpendicular to the plane of a → + b → and a → - b → . So the unit vector perpendicular to the plane of a → + b → and a → - b → is given as = a → + b → × a → - b → a → + b → × a → - b → Now substituting the corresponding value, we get = - 2 i ^ + 4 j ^ - 2 k ^ 2 6 = - i ^ + 2 j ^ - k ^ 6 Or, the unit vector can be written as = - 1 6 i ^ + 2 6 j ^ - 1 6 k ^ .

Show that three points A ( - 2 i ^ + 3 j ^ + 5 k ^ ) , B ( i ^ + 2 j ^ + 3 k ^ ) and C ( 7 i ^ - k ^ ) are collinear.

A B → = ( i ^ + 2 j ^ + 3 k ^ ) - ( - 2 i ^ + 3 j ^ + 5 k ^ ) = 3 i ^ - j ^ - 2 k ^ B C → = ( 7 i ^ - k ^ ) - ( i ^ + 2 j ^ + 3 k ^ ) = 6 i ^ - 2 j ^ - 4 k ^ Now , B C → = 6 i ^ - 2 j ^ - 4 k ^ = 2 ( 3 i ^ - j ^ - 2 k ^ ) B C → = λ A B → , where λ = 2 ∴ B C → and A B → are collinear Since , point B is common to both , A , B and C are collinear .

E M B I B E

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

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

Author:Embibe Experts

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

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

MEDIUM

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2019>Q 14

Find the projection of $\vec{a} = 2 \hat{i} + 3 \hat{j} + 2 \hat{k}$ on the vector $\vec{b} = \hat{i} + 2 \hat{j} + \hat{k}$ .

MEDIUM

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2019>Q 15

If $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , then prove that $\vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a}$ .

MEDIUM

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2019>Q 16

Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec{a} = 3 \hat{i} + \hat{j} + 4 \hat{k} and \vec{b} = \hat{i} - \hat{j} + \hat{k}$ .

HARD

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2019>Q 17

Show that the angle between any two diagonals of a cube is $\cos^{- 1} (\frac{1}{3}) .$

EASY

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2020>Q 19

Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $1$ and $2$ , respectively, and satisfying $\vec{a} \cdot \vec{b} = 1$

EASY

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2020>Q 20

Find a vector in the direction of the vector $\vec{a} = \hat{i} - 2 \hat{j}$ that has magnitude of $7$ units.

MEDIUM

12th Madhya Pradesh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Madhya Pradesh Board-2020>Q 21

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