Explanation of an answer of a GRE review question Chapter 3, #1: Find the angle between the diagonals of the back and left faces of the cube

Question

The question is given below:

And its answer is given below:

My answer was:

Since the faces of the cube are squares and since The diagonals of a square are perpendicular bisectors, then they bisect the angle of the square which is $90^{\circ}$, hence the angle is $45^{\circ}$ + $45^{\circ}$ = $90 ^{\circ} $. But I realized my imagination mistake now, this is not the angle required, am I correct?

So, regarding to the given answer, I do not understand what does it mean "first octant of xyz-space" why octant?, Also "is it allowable to assume that each side of the cube has unit length?"

Also still I do not know exactly why the book used the given way in the above answer to answer the question, may be because the vectors are in 3 dimension, correct?

Answer 1

Question 1: What is an octant?

Answer: An octant is like a 2d cartesian quadrant, but in the 3d plane.

Question 2: Why can we assume the length of the diagonals?

Answer: The length of the two diagonals does not change the angle between them. Think about similar triangles -- they aren't congruent, but they have the same angles

Question 3: How did they get that answer?

Firstly, the book placed the cube's "left-hand bottom corner" at the origin because it makes it easier to think about vectors when their initial point is at the origin. Now, because we assumed that the cube had a length of 1, and we know that the vectors are diagonals, we can write out the coordinates of the terminal points of each vector: $$<1, 0, 1>$$ $$<0, 1, 1>$$

Next, you have to know the formula for the angle between two vectors:

$$\cos(\theta)=\frac{u*v}{||u||*||v||}$$

Now, we just have to plug values into the formula:

$$\frac{<1, 0, 1>*<0, 1, 1>}{\sqrt{1^2+0^2+1^2}*\sqrt{0^2+1^2+1^2}}=\frac{1}{2}$$

$$\cos^{-1}(\frac{1}{2})=\theta$$ $$\theta=60^\circ$$

Note: To multiply vectors, multiply the x, y, and z coordinates and add the products (how we got the numerator). To multiply lengths, use the Pythagorean theorem (how we solved the denominator).

Answer 2

Draw one more line, the diagonal of the top (horizontal) face of the cube, the diagonal that goes between the two far ends of the diagonals that are already drawn. Now you have a triangle in which all three sides have length $\sqrt2s$, if the side-length of the cube is $s$. Equilateral triangle, all angles are $60^\circ$, and one of these is the angle being asked about.

Answer 3

In three dimensional space, three mutually perpendicular planes through the origin divide the space into eight parts. Since there are eight of them, these parts are called "octants". The first octant is simply the area of space where $x,y,z$ are all greater than or equal to $0$. So are meant to place one corner of the cube at $(0,0,0)$ and the opposite corner at $(1,1,1)$.

It is fairly obvious that the angle between the diagonals does not depend on the size of the cube. But if you want to avoid the assumption that each side of the cube has unit length, then just assume that each side has length $a$. Then the vectors of the diagonals become $(a,0,a)$ and $(0,a,a)$ and there is an extra factor of $a^2$ in the numerator and the denominator of the fraction. These factors cancel each other and you have $\cos \theta = \frac {a^2}{2a^2}= \frac 1 2$.