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