3danalytic geometry
Find the equations of the two lines through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at angles of $\frac{\pi}{3}$
Now our required line should be $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$. But I don't see how we are going to get three equations to obtain values of a,b,c. Any hint?
Here is a small diagram representing question:
You have to find equations of circles.
There centres are $(1,\alpha) \ and (1,-\alpha)$ by symmetry .
And the radius is $2$ units.
let equation of circles are : $x^2+y^2-2x\pm2\alpha y=0$ [c=0; as circles pass through (0,0)] ; then $R=\sqrt{\alpha^2+1}=2$
Hint:
Write the equation of the line in parametric form, so that an arbitrary point is expressed in terms of just one letter. Then you can use the dot product to form a quadratic equation in the parameter giving you two solutions.
So we have$$\frac{x-3}{2}=\frac{y-3}{1}=\frac z1=\lambda$$ $$\Rightarrow \underline{r}=\left(\begin{matrix}x\\y\\z\end{matrix}\right)=\left(\begin{matrix}2\lambda+3\\ \lambda+3\\ \lambda\end{matrix}\right)$$
Best Answer
It's easier to do this with vectors.
A general point on the line has position vector $\underline r=(3+2 \lambda)\underline i + (3+\lambda)\underline j + (0+\lambda)\underline k$, so the line passing through the originand that point has direction vector $(3+2 \lambda)\underline i + (3+\lambda)\underline j + (0+\lambda)\underline k$
The direction vector of the line is $2\underline i + \underline j + \underline k$
To find the angle they make, use the scalar product and the $\cos$ relationship.
$\cos \theta = \frac {\left((3+2 \lambda)\underline i + (3+\lambda)\underline j + (0+\lambda)\underline k \right). \left (2\underline i + \underline j + \underline k \right)}{\left|(3+2 \lambda)\underline i + (3+\lambda)\underline j + (0+\lambda)\underline k \right|. \left |2\underline i + \underline j + \underline k \right|}$
$\frac 12 = \frac {6+4 \lambda+3 + \lambda+\lambda}{\sqrt{\left((3+2 \lambda)^2+ (3+\lambda)^2+\lambda^2 \right)\left (4+1+1 \right)}}$