JEE Main & Advanced Mathematics Three Dimensional Geometry Intersection of Straight Line and a Sphere

Let the equations of the sphere and the straight line be   

\[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2vy+2wz+d=0\]               ……(i)

and         \[\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}=r\],  (say)                    …..(ii)

Any point on the line (ii) is \[(\alpha +lr,\beta +mr,\gamma +nr)\].

If this point lies on the sphere (i) then we have,

\[{{(\alpha +lr)}^{2}}+{{(\beta +mr)}^{2}}+{{(\gamma +nr)}^{2}}+2u(\alpha +lr)+2v(\beta +mr)\]\[+2w(\gamma +nr)+d=0\]

or, \[{{r}^{2}}[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}]+2r[l(u+\alpha )+m(v+\beta )]+n(w+\gamma )]\] \[+\,({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+2u\alpha +2v\beta +2w\gamma +d)=0\] ……(iii)

This is a quadratic equation in r and so gives two values of r and therefore the line (ii) meets the sphere (i) in two points which may be real, coincident and imaginary, according as root of (iii) are so.

If l, m, n are the actual d.c.’s of the line, then \[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\] and then the equation (iii) can be simplified.