Q.Find the co-ordinates of the point where the line $\vec r = (\vec i+2\vec j-5\vec k) + t(2\vec i-3\vec j+4\vec k)$ meets the plane $\vec r\cdot(2\vec i+4\vec j-\vec k)=3$.

A line in $\mathbb R^3$ is uniquely fixed by (i) one point plus a direction, or (ii) two points (whose difference gives the direction). Every form below is built from these two ingredients.

Point + direction, through $A(\vec a)$ parallel to $\vec b$:

• Parametric vector: $\vec r=\vec a+t\vec b,\ t\in\mathbb R$.
• Non-parametric vector: $(\vec r-\vec a)\times\vec b=\vec 0$.
• Cartesian (symmetric): $\dfrac{x-x_1}{b_1}=\dfrac{y-y_1}{b_2}=\dfrac{z-z_1}{b_3}$, where $b_1,b_2,b_3$ are the direction ratios of $\vec b$ (or, scaled to unit length, the line's direction cosines).

Through two points $A(\vec a),B(\vec b)$: identical forms with $\vec b-\vec a$ in place of $\vec b$ — direction ratios $x_2-x_1,\,y_2-y_1,\,z_2-z_1$.

Reading off direction ratios/cosines is the master skill: whatever form a line is given in, the numbers under $x,y,z$ in the symmetric form (or the coefficients of $t$ in the parametric form) are its direction ratios; dividing by their magnitude $\sqrt{b_1^2+b_2^2+b_3^2}$ gives the direction cosines $l,m,n$ (with $l^2+m^2+n^2=1$). …