Question

Ratio in which the xy-plane divides the join of (1, 2, 3) and (4, 2, 1) is
(a) 3 : 1 internally
(b) 3 : 1 externally
(c) 1 : 2 internally
(d) 2 : 1 externally

Answer 1

$$\begin{gathered} \left(\mathrm{b}\right)3:1\mathrm{externally} \\ \mathrm{Suppose}\mathrm{the}XY-\mathrm{plane}\mathrm{divides}\mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{joining}\mathrm{the}\mathrm{points}P\mathit{}\left(1,2,3\right)\mathrm{and}Q\mathit{}\left(4,2,1\right)\mathit{}\mathrm{in}\mathrm{the}\mathrm{ratio}\mathit{}k:1. \\ \mathrm{Using}\mathrm{the}\mathrm{section}\mathrm{formula},\mathrm{the}\mathrm{coordinates}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{intersection}\mathrm{are}\mathrm{given}\mathrm{by} \\ \left(\frac{k\left(4\right)+1}{k+1},\frac{k\left(2\right)+2}{k+1},\frac{k\left(1\right)+3}{k+1}\right) \\ \mathrm{The}Z-\mathrm{coordinate}\mathrm{of}\mathrm{any}\mathrm{point}\mathrm{on}\mathrm{the}XY-\mathrm{plane}\mathrm{is}\mathrm{zero}. \\ \Rightarrow \frac{k\left(1\right)+3}{k+1}=0 \\ \Rightarrow k+3=0 \\ \Rightarrow k=-3=-\frac{3}{1} \\ \mathrm{Thus},\mathrm{the}XY-\mathrm{plane}\mathrm{divides}\mathrm{the}\mathrm{line}\mathrm{segment}\mathrm{joining}\mathrm{the}\mathrm{given}\mathrm{points}\mathrm{in}\mathrm{the}\mathrm{ratio}3:1\mathrm{externally}. \end{gathered}$$