Q.Find the equations of the two lines through the origin which intersect the line $\dfrac{x-3}{2} = \dfrac{y-3}{1} = \dfrac{z}{1}$ at angles of $\dfrac{\pi}{3}$ each.

A line through the origin that intersects the given line meets it at a point $P$; the line is then $OP$, and the angle between $OP$ and the given line's direction must be $\dfrac{\pi}{3}$.

Point on the line. The given line $\dfrac{x-3}{2}=\dfrac{y-3}{1}=\dfrac{z}{1}=t$ gives $P=(3+2t,\ 3+t,\ t)$, with direction $\vec{d}=(2,1,1)$.

Angle condition. $\cos\dfrac{\pi}{3}=\dfrac{1}{2}=\dfrac{|\vec{OP}\cdot\vec{d}|}{|\vec{OP}|\,|\vec{d}|}$, where $\vec{OP}\cdot\vec{d}=9+6t$, $|\vec{d}|=\sqrt{6}$, $|\vec{OP}|^2=6t^2+18t+18$. …

The idea (do not forget the intersection condition)

The required line must (i) pass through the origin, (ii) actually intersect the given line, at (iii) an angle of $\tfrac{\pi}{3}$. The intersection condition is essential: without it the angle alone gives a whole cone of directions. The clean way to build in intersection is to let the line pass through a general point $P$ of the given line, so the required line is simply $OP$.

Set up

Write the given line in parameter form. With

a general point is

and the given line's direction is $\vec{d} = (2,1,1)$.

The line through the origin and $P$ has direction $\vec{OP} = (3+2t,\,3+t,\,t)$.

Apply the angle condition

We need the angle between $\vec{OP}$ and $\vec{d}$ to be $\tfrac{\pi}{3}$:

Compute each piece:

So

Solve for $t$

Square both sides:

$4(9+6t)^2 = 6\,(6t^2+18t+18).$ …

Method: A line through a point that meets a given line at a set angle

Use this whenever a required line must (a) pass through a fixed point, (b) actually intersect a given line, and (c) make a prescribed angle with it — the intersection condition is the part students skip.

Steps

Step 1: Force intersection by riding on the given line.

Write the given line in parameter form and take a general point $P(t)$ on it. Any line joining your fixed point to $P(t)$ is automatically guaranteed to intersect the given line — this single trick builds in the intersection condition that the angle alone cannot.

Step 2: Write the unknown direction.

The required line's direction is the join from the fixed point to $P(t)$; it carries the single unknown $t$.

Step 3: Impose the angle with the acute-angle formula. …

Common Mistakes

Mistake 1: Using only the angle condition and forgetting the line must intersect.

Why it's wrong: the angle alone is satisfied by a whole cone of directions through the origin, not two specific lines. Correct approach: force intersection by taking the required line through a general point $P(t)$ of the given line, so $OP$ automatically meets it.

Mistake 2: Stopping at one line.

Why it's wrong: squaring the angle equation gives a quadratic in $t$ with two roots — the problem literally asks for two lines. Correct approach: solve the quadratic fully ($t=-1,-2$ here) and report both directions. …