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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?

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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.

$\cos \theta = \frac {\left((3+2 \lambda)\underline i + (3+\lambda)\underline j + (0+\lambda)\underline k \right). \left (2\underline i + \underline j + \underline k \right)}{\left|(3+2 \lambda)\underline i + (3+\lambda)\underline j + (0+\lambda)\underline k \right|. \left |2\underline i + \underline j + \underline k \right|}$

$\frac 12 = \frac {6+4 \lambda+3 + \lambda+\lambda}{\sqrt{\left((3+2 \lambda)^2+ (3+\lambda)^2+\lambda^2 \right)\left (4+1+1 \right)}}$

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