# “Show” that the direction cosines of a vector satisfies…

"Show" that the direction cosines of a vector satisfies

$$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$$

I am stumped on these things:

1. "SHOW" that the direction cosines corresponds to a given vector to satisfy the relation above. ----> How do you "show" this? What does this mean? Does this mean to use the direction cosines of a vector?
1. I'm sure this is a proof but I don't know what the end result would look like or better, what I am expected to learn from this proof.

I am not looking for a mere answer but really an in-depth explanation of the problem.

Greatly appreciated. :)

• "Show" means exactly the same as "prove". Some authors use it in exercises in an attempt to sound less scary; others alternate between "show" and "prove" simply for linguistic variation. – Henning Makholm Sep 2 '12 at 23:51

## 3 Answers

If $$x^2+y^2+z^2=r^2,$$ then $$\frac{x^2}{r^2}+\frac{y^2}{r^2}+\frac{z^2}{r^2}=1,$$ or $$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma=1.$$

• +1 for its simplicity! I was searching for proof like this – Fawad Oct 9 '16 at 13:38

Given two vectors $A$, $B$, I'll use $\angle AB$ for the angle between them. So, the direction cosines of a vector $V=(x,y,z)$ are

$$\cos\alpha=\cos\angle (x,y,z)(1,0,0)=\frac{(x,y,z)\cdot(1,0,0)}{\|V\|\cdot\|(1,0,0)\|}$$

$$\cos\beta=\cos\angle (x,y,z)(0,1,0)=\frac{(x,y,z)\cdot(0,1,0)}{\|V\|\cdot\|(0,1,0)\|}$$

$$\cos\gamma=\cos\angle (x,y,z)(0,0,1)=\frac{(x,y,z)\cdot(0,0,1)}{\|V\|\cdot\|(0,0,1)\|}$$

but this is

$$\cos\alpha=\frac{x}{\sqrt{x^2+y^2+z^2}}$$

$$\cos\beta=\frac{y}{\sqrt{x^2+y^2+z^2}}$$

$$\cos\gamma=\frac{z}{\sqrt{x^2+y^2+z^2}}$$

so that

$$\cos^2\alpha+\cos^2\beta+\cos^2\gamma=\frac{x^2+y^2+z^2}{x^2+y^2+z^2}=1$$

Note it is easier to work with the normalized vectors since they are "the same" in terms of angles (you could have used $(x,0,0)$ or $(10,0,0)$, or any parallel vector), but these make the calculations much easier.

• If the typographical difference between $\|V\|$ and $||V||$ is not conspicuous enough to you, look at $\|V\|\|U\|$ versus $||V|| ||U||$. The latter is coded as ||V|| ||U|| and the former as \|V\|\|U\|. (I edited accordingly.) – Michael Hardy Oct 9 '16 at 20:02
• @MichaelHardy I posted this about 4 years ago, so by now I know the typographical difference between those. =) – Pedro Tamaroff Oct 9 '16 at 20:12
• ok, I'm convinced. $\qquad$ – Michael Hardy Oct 9 '16 at 22:55

Suppose the vector is $(x,y,z)$. The direction cosine $\alpha$ is the cosine of the angle between this vector and the $x$-axis. The vector $(x,0,0)$ points in the direction of the $x$-axis. So you want the cosine of the angle between the two vectors $(x,y,z)$ and $(x,0,0)$. The law of cosines tells us that $$\| (x,y,z) \|^2 + \|(x,0,0)\|^2 - 2\|(x,y,z)\|\|(x,0,0)\|\cos\alpha = \|(x,y,z)-(x,0,0)\|^2$$ (since $(x,y,z)-(x,0,0)$ corresponds to the third side of the triangle). Hence $$(x^2+y^2+z^2) + x^2 - 2\sqrt{x^2+y^2+z^2}\,\cdot |x|\cos\alpha = y^2+z^2.$$ And then $$x^2 - \sqrt{x^2+y^2+z^2}\,\cdot |x|\cos\alpha = 0.$$ Solve this for $\cos\alpha$. Similarly find $\cos\beta$ and $\cos\gamma$. Then add their squares.

(And remember that $x^2/|x| = |x|$.)

Later edit: The comment by "Blue" below gives a simplification: The points $(0,0,0)$, $(x,0,0)$, and $(x,y,z)$ are the vertices of a right triangle. From $(0,0,0)$ to $(x,y,z)$ is the hypotenuse; from $(0,0,0)$ to $(x,0,0)$ is the side that is "adjacent" to the angle $\alpha$. So $$\cos\alpha = \frac{\text{adjacent}}{\text{hypontenuse}} = \frac{|x|}{\sqrt{x^2+y^2+z^2}}.$$ The other two cosines are found in the same way, with $|y|$ and $|z|$ in their numerators instead of $|x|.$

• Dropping a perpendicular from point $P(x,y,z)$ to the $x$-axis gives point $P_x(x,0,0)$. (This is effectively part of the definition of the Cartesian coordinate system.) Angle $\alpha$ (or its supplement, if $x<0$) is between leg $OP_x$ and hypotenuse $OP$ of right triangle $OPP_x$, so $\cos\alpha = x/\sqrt{x^2+y^2+z^2}$. – Blue Sep 3 '12 at 0:15
• @Blue : That is simpler than the law of cosines. – Michael Hardy Sep 3 '12 at 0:25