# Thornton & Marion (5th Edition), Chapter 01, Exercise Problem 07

## Chapter 1. Matrices, Vectors, and Vector Calculus

### Problem 07. Scalar Product of two (unit) vectors

##### The problem asks you to
1) Find the vectors describing the diagonals
2) The angle between diagonal vectors
##### This problem assumes
A unit cube in the Cartesian (Rectangular) coordinate system
1) Position vectors
2) Scalar Product of vectors
##### Solution
1) Find diagonal vectors

First, we can define the vector $\vec{A}$$\vec{A}$ from the origin to (1,1,1).

And, we can also define the vector $\vec{B}$$\vec{B}$ from (0,0,1) to (1,1,0).

Thus, a pair of diagonal vectors can be expressed as

$\vec{A} = \hat{i} + \hat{j} + \hat{k}$$\vec{A} = \hat{i} + \hat{j} + \hat{k}$
$\vec{B} = \hat{i} + \hat{j} - \hat{k}$$\vec{B} = \hat{i} + \hat{j} - \hat{k}$

2) The angle between diagonal vectors

From the scalar product,

$\vec{A} \cdot \vec{B} = AB \cos\theta$$\vec{A} \cdot \vec{B} = AB \cos\theta$

Since,

$\vec{A} \cdot \vec{B} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) = 1 + 1 - 1 = 1$$\vec{A} \cdot \vec{B} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) = 1 + 1 - 1 = 1$

$AB \cos\theta = \sqrt{3}\sqrt{3} \cos\theta$$AB \cos\theta = \sqrt{3}\sqrt{3} \cos\theta$

Therefore,

$\theta = \cos^{-1} (\frac{1}{3}) = 70.53^\circ$$\theta = \cos^{-1} (\frac{1}{3}) = 70.53^\circ$