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Question

Mathematics Question on Three Dimensional Geometry

The distance of the point (1, 2, 1) from the line x12=y21=z32\frac{x-1}{2} = \frac{ y-2}{1}=\frac{ z-3}{2} is

A

53\frac{\sqrt{5}}{3}

B

235\frac{2 \sqrt{3}}{5}

C

203\frac{20}{3}

D

253\frac{2 \sqrt{5}}{3}

Answer

253\frac{2 \sqrt{5}}{3}

Explanation

Solution

x12=y21=z32=k\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}=k
any point on the line = (2k + 1, k + 2, 2k + 3)
drs of AB : 2k, k, 2k + 2
drs of line : 2, 1, 2
ABline4k+k+4k+4=0AB \bot line \Rightarrow4k+k+4k+4=0
9k=49k=-4
k=49k=\frac{-4}{9}
Distance=6481+1681+10081=1809=253Distance=\sqrt{\frac{64}{81}+\frac{16}{81}+\frac{100}{81}}=\frac{\sqrt{180}}{9}=\frac{2\sqrt{5}}{3}