If two vectors a → and b → are such that, a → = 2 , b → = 3 and a → . b → = 4 , then find a → - b → .

We have, a → = 2 , b → = 3 , a → . b → = 4 We know, a → - b → 2 = ( a → - b → ) . ( a → - b → ) = a → . a → - a → . b → - b → . a → + b → . b → = a → 2 - 2 a → . b → + b → 2 = 2 2 - 2 4 + 3 2 = 4 - 8 + 9 = 5 ⇒ a → - b → = 5

If a → = 2 i ^ + 3 j ^ + 6 k ^ and b → = 3 i ^ - 6 j ^ + 2 k ^ then evaluate a → . b → .

Given vectors a → = 2 i ^ + 3 j ^ + 6 k ^ and b → = 3 i ^ - 6 j ^ + 2 k ^ . The dot product of two vectors X → = p i ^ + q j ^ + r k ^ and Y → = c i ^ + d j ^ + e k ^ is X → · Y → = p c + q d + r e . The resultant of the dot product of vectors is a scalar quantity. Thus, the dot product is also known as a scalar product. The dot product of given vectors a → and b → is a → . b → = 2 i ^ + 3 j ^ + 6 k ^ . 3 i ^ - 6 j ^ + 2 k ^ ⇒ a → . b → = 2 × 3 + 3 × - 6 + 6 × 2 ⇒ a → . b → = 6 - 18 + 12 ⇒ a → . b → = 0

Find a → × b → if a → = 2 i ^ + j ^ + 3 k ^ and b → = 3 i ^ + 5 j ^ - 2 k ^ .

Given: a → = 2 i ^ + j ^ + 3 k ^ b → = 3 i ^ + 5 j ^ - 2 k ^ a → × b → = i ^ j ^ k ^ 2 1 3 3 5 - 2 = ( - 2 - 15 ) i ^ - ( - 4 - 9 ) j ^ + ( 10 - 3 ) k ^ = - 17 i ^ + 13 j ^ + 7 k ^ | a → × b → | = - 17 2 + 13 2 + ( 7 ) 2 = 289 + 169 + 49 = 507 square units

Show that the points P - 2 i ^ + 3 j ^ + 5 k ^ , Q i ^ + 2 j ^ + 3 k ^ and R 7 i ^ - k ^ are collinear.

To prove: the points P ( - 2 i ^ + 3 j ^ + 5 k ^ ) , Q ( i ^ + 2 j ^ + 3 k ^ ) and R ( 7 i ^ - k ^ ) are collinear. For collinear P Q → + Q R → = P R → P Q → = ( 1 - ( - 2 ) ) i ^ + ( 2 - 3 ) j ^ + ( 3 - 5 ) k ^ ⇒ P Q → = 3 i ^ - 1 j ^ - 2 k ^ Q R → = ( 7 - 1 ) i ^ + ( 0 - 2 ) j ^ + ( - 1 - 3 ) k ^ ⇒ Q R → = 6 i ^ - 2 j ^ - 4 k ^ P R → = ( 7 - ( - 2 ) ) i ^ + ( 0 - 3 ) j ^ + ( - 1 - 5 ) k ^ ⇒ P R → = 9 ı ^ - 3 ȷ ^ - 6 k ^ P Q → + Q R → = 9 i ^ - 3 j ^ - 6 k ^ = P R → Hence, points P , Q & R are collinear.

Find the angle between the vectors i ^ - 2 j ^ + 3 k ^ and 3 i ^ - 2 j ^ + k ^ .

Use the formula to find the angle between a → and b → , i.e., cos θ = a → · b → a → b → Let a → = i ^ - 2 j ^ + 3 k ^ and b → = 3 i ^ - 2 j ^ + k ^ Now, a → · b → = i ^ - 2 j ^ + 3 k ^ · 3 i ^ - 2 j ^ + k ^ = 1 × 3 + - 2 × - 2 + 3 × 1 = 3 + 4 + 3 = 10 Similarly, a → = 1 2 + ( - 2 ) 2 + 3 2 = 1 + 4 + 9 = 14 and b → = 3 2 + ( - 2 ) 2 + 1 2 = 9 + 4 + 1 = 14 Therefore, cos θ = 10 14 × 14 = 10 14 = 5 7 ⇒ θ = cos - 1 5 7 Hence, the angle between i ^ - 2 j ^ + 3 k ^ and 3 i ^ - 2 j ^ + k ^ is cos - 1 5 7 .

Find a unit vector perpendicular to the vectors 2 i ^ - j ^ + k ^ and 3 i ^ + 4 j ^ - k ^ .

Let the given vectors be a → = 2 i ^ - j ^ + k ^ and b → = 3 i ^ + 4 j ^ - k ^ ∴ a → × b → = i ^ j ^ k ^ 2 - 1 1 3 4 - 1 = ( 1 - 4 ) i ^ - ( - 2 - 3 ) j ^ + ( 8 + 3 ) k ^ = - 3 i ^ + 5 j ^ + 11 k ^ Therefore, | a → × b → | = - 3 i ^ + 5 j ^ + 11 k ^ = - 3 2 + 5 2 + 11 2 = 9 + 25 + 121 = 155 Hence, the unit vector perpendicular to a → and b → = ± a → × b → | a → × b → | = ± 1 155 - 3 i ^ + 5 j ^ + 11 k ^ .

E M B I B E

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

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

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

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

MEDIUM

12th Uttarakhand Board

IMPORTANT

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

If two vectors $\vec{a}$ and $\vec{b}$ are such that, $|\vec{a}| = 2, |\vec{b}| = 3$ and $\vec{a} . \vec{b} = 4$ , then find $|\vec{a} - \vec{b}|$ .

EASY

12th Uttarakhand Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Uttarakhand Board-2019>Q 5

If $\vec{a} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ and $\vec{b} = 3 \hat{i} - 6 \hat{j} + 2 \hat{k}$ then evaluate $\vec{a} . \vec{b}$ .

MEDIUM

12th Uttarakhand Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Uttarakhand Board-2019>Q 6

Find the area of a parallelogram whose adjacent sides are determined by the vectors $\vec{a} = \hat{i} - 2 \hat{j} + 2 \hat{k}$ and $\vec{b} = 2 \hat{i} - 6 \hat{j} + 3 \hat{k}$ .