If a straight line makes an angle of $60^{\circ}$ with each of the $X$ and $Y$ axes, what is the angle of the line with the $Z$ axis?
MY WORK:
I assumed a vector $a\hat{i}+b\hat{j}+c\hat{k}$ which meets the above criteria . Now, from property of dot product, I get:
$$\frac{a}{ \sqrt{a^2+b^2+c^2}}=\frac{1}{2}$$
and $$\frac{b}{\sqrt{a^2+b^2+c^2}}=\frac{1}{2}$$
Again , $$a^2+b^2=c^2$$
So, from dot product of the Z component :
$$\cos^2\theta_z=\frac{a^2+b^2}{a^2+b^2+c^2}$$
$$=2\times \frac{1}{4}$$
$$=\frac{1}{\sqrt{2}}$$
So , $$\theta_z=45^{\circ}$$
And the answer is correct. Is it valid to use a vector? But I cannot understand it geometrically. I mean, it is from a chapter of straight lines. But I did not use any straight line properties. I think that'll be lengthy. Any suggestions?
Best Answer
Your argument is indeed good: in space you can write the equation of a straight line with only a point and a vector called the direction vector. Your problem doesn't ask for a specific line so, with only the direction vector, you've generated a set of parallel lines which all satisfy your requirement on the angles.
For completness, the parametric equation of a line crossing a point $P(x_p,y_p,z_p)$ with direction vector $\vec{v}=(a,b,c)$ is: $$\left\{\begin{matrix}x=x_p+at\\y=y_p+bt\\z=z_p+ct\end{matrix}\right.$$