Let point A divide the line segment joining P(−1,−1,2) and Q(5,5,10) internally in the ratio r:1 (r>0). If O is the origin and (OQ⋅OA)−51∣OP×OA∣2=10, then the value of r is:
Solutions · 1
Official solution
A=(r+15r−1,r+15r−1,r+110r+2).
Computing OQ⋅OA=r+110(15r+1) and ∣OP×OA∣2=(r+1)2800r2, substituting into the given equation:
\[\frac{10}{r+1}(15r+1) - \frac15\cdot\frac{800r^2}{(r+1)^2} = 10\]
This simplifies to 2r2−14r=0⇒r(r−7)=0. Since r>0, r=7.