If A , B , C are three non-collinear points with position vectors a → , b → , c → , respectively, show that the length of the perpendicular from C on A B is a → × b → + b → × c → + c → × a → b → - a →

Consider △ A B C with side A B So, length of the perpendicular from C on A B is the height of the triangle, h We know that area of the triangle = 1 2 × base × height ⇒ h = Area of the triangle 1 2 A B ⇒ h = 1 2 A B → × A C → 1 2 A B → ⇒ h = b → - a → × c → - a → b → - a → ⇒ h = b → × c → - b → × a → - a → × c → + a → × a → b → - a → ⇒ h = a → × b → + b → × c → + c → × a → b → - a → Hence proved.

If a → = i ^ - 2 j ^ + 3 k ^ , b → = 2 i ^ + 3 j ^ - 5 k ^ , prove that a → and a → × b → are perpendicular.

Given, a → = i ^ - 2 j ^ + 3 k ^ , b → = 2 i ^ + 3 j ^ - 5 k ^ Now, let us find a → × b → = i ^ j ^ k ^ 1 - 2 3 2 3 - 5 = i ^ 10 - 9 - j ^ - 5 - 6 + k ^ 3 + 4 = i ^ + 11 j ^ + 7 k ^ . . . . i Now, let us find: a → · a → × b → a → · a → × b → = i ^ - 2 j ^ + 3 k ^ · i ^ + 11 j ^ + 7 k ^ = 1 - 22 + 21 = 0 Hence, a → , and a → × b → are perpendicular.

If a → and b → are non-collinear vectors, find the value of x such that the vectors α → = x - 2 a → + b → and β → = 3 + 2 x a → - 2 b → are collinear.

We know that if two vectors p → and q → are collinear vectors, p → = k q → Given that vectors α → = x - 2 a → + b → , β → = 3 + 2 x a → - 2 b → are collinear. a → and b → are non-collinear vectors. So, α → = k β → ⇒ x - 2 a → + b → = k 3 + 2 x a → - 2 b → ⇒ x - 2 a → + b → = k 3 + 2 x a → - 2 k b → Equating the coefficient of a → and b → , x - 2 = k 3 + 2 x And 1 = - 2 k ⇒ k = - 1 2 Now, x - 2 = k 3 + 2 x ⇒ x - 2 = - 1 2 3 + 2 x ⇒ 2 x - 4 = - 3 - 2 x ⇒ 4 x = 1 ⇒ x = 1 4

Write the angle between the two lines: x + 1 2 = y − 2 5 = z + 3 4 and x − 1 5 = y + 2 2 = z − 1 − 5 .

Given lines are: x + 1 2 = y − 2 5 = z + 3 4 and x − 1 5 = y + 2 2 = z − 1 − 5 . Direction ratios of two lines are: 2 , 5 , 4 and 5 , 2 , - 5 . Let θ be the angle between these two lines. So, cos θ = 2 i ^ + 5 j ^ + 4 k ^ · 5 i ^ + 2 j ^ - 5 k ^ 2 2 + 5 2 + 4 2 · 5 2 + 2 2 + - 5 2 ⇒ cos θ = 10 + 10 - 20 4 + 25 + 16 · 25 + 4 + 25 ⇒ cos θ = 0 45 · 54 ⇒ cos θ = 0 So, θ = cos - 1 0 = 90 ° Hence, the given two lines are perpendicular to each other.

Prove that a → ⋅ [ ( b → + c → ) × ( a → + 3 b → + 4 c → ) ] = [ a → b → c → ]

Consider, a → ⋅ [ ( b → + c → ) × ( a → + 3 b → + 4 c → ) ] = a → ⋅ [ b → × a → + 3 ( b → × b → ) + 4 ( b → × c → ) + c → × a → + 3 ( c → × b → ) + 4 ( c → × c → ) ] = a → ⋅ [ b → × a → + 4 ( b → × c → ) + c → × a → + 3 ( c → × b → ) ] ∵ b → × b → = 0 , c → × c → = 0 = a → ⋅ [ b → × a → + 4 ( b → × c → ) + c → × a → − 3 ( b → × c → ) ] ∵ b → × c → = - c → × b → = a → ⋅ [ b → × a → + b → × c → + c → × a → ] = a → ( b → × a → ) + a → ( b → × c → ) + a → ( c → × a → ) ] = 0 + a → b → c → + 0 = a → b → c → = RHS Hence proved.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Vectors>ICSE-2020

Embibe Experts Solutions for Chapter: Vectors, Exercise 1: ICSE-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Vectors, Exercise 1: ICSE-2018

Attempt the free practice questions on Chapter 1: Vectors, Exercise 1: ICSE-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: Vectors, Exercise 1: ICSE-2018 with Hints & Solutions

MEDIUM

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Vectors>ICSE-2018>Q 2

If $A, B, C$ are three non-collinear points with position vectors $\vec{a}, \vec{b}, \vec{c}$ , respectively, show that the length of the perpendicular from $C$ on $A B$ is $\frac{|(\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{a})|}{|\vec{b} - \vec{a}|}$

EASY

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Vectors>ICSE-2019>Q 4

If $\vec{a}$ and $\vec{b}$ are perpendicular vectors, $| \vec{a} + \vec{b} | = 13$ and $| \vec{a} | = 5,$ find the value of $|\vec{b}|$ .

MEDIUM

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Vectors>ICSE-2019>Q 5

If $\vec{a} = \hat{i} - 2 \hat{j} + 3 \hat{k}, \vec{b} = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ , prove that $\vec{a}$ and $\vec{a} \times \vec{b}$ are perpendicular.

MEDIUM

12th ICSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 1 - Vectors>ICSE-2019>Q 6

If $\vec{a}$ and $\vec{b}$ are non-collinear vectors, find the value of $x$ such that the vectors $\vec{α} = (x - 2) \vec{a} + \vec{b}$ and $\vec{β} = (3 + 2 x) \vec{a} - 2 \vec{b}$ are collinear.