Vector Algebra — Part 2: Scalar and vector products: visual lesson, ISC Class 12 Mathematics

Lesson

Vector Algebra, Part 2

Two ways to multiply vectors.

In this lesson you will

Find the scalar product in components and use it for angles
Test vectors for perpendicularity
Find scalar and vector projections
Find the vector product and unit vectors perpendicular to two vectors
Find areas of triangles and parallelograms and test collinearity with the vector product

Part 2 at a glance.

Concept

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = a_1b_1 + a_2b_2 + a_3b_3$ .

The answer is a number (a scalar).
Property	Statement
Definition	$\vec{a} \cdot \vec{b} = \|\vec{a}\|\|\vec{b}\|\cos\theta$
In components	$a_1b_1 + a_2b_2 + a_3b_3$
Commutative	$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$
Self product	$\vec{a} \cdot \vec{a} = \|\vec{a}\|^2$
Unit vectors	$\hat{i} \cdot \hat{i} = 1$ , $\hat{i} \cdot \hat{j} = 0$
Perpendicular	$\vec{a} \cdot \vec{b} = 0$ for non-zero $\vec{a}, \vec{b}$

Worked example

Find the angle between $\hat{i} + 2\hat{j} + 2\hat{k}$ and $2\hat{i} - \hat{j} + 2\hat{k}$ .

Divide the dot product by the product of the magnitudes.
Step	Working
Dot product	$1(2) + 2(-1) + 2(2) = 4$
Magnitudes	$\|\vec{a}\| = 3$ , $\|\vec{b}\| = 3$
Angle	$\cos\theta = \dfrac{4}{9}$ , $\theta = \cos^{-1}\dfrac49$