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Question

Mathematics Question on Three Dimensional Geometry

Direction cosines of the line x+22=2y−53,z=−1\frac{x+2}{2} = \frac{2y-5}{3}, z = -1 is

A

45,35,0\frac{4}{5}, \frac{3}{5}, 0

B

35,45,15\frac{3}{5}, \frac{4}{5}, \frac{1}{5}

C

−35,45,0-\frac{3}{5}, \frac{4}{5}, 0

D

45,−25,15\frac{4}{5}, -\frac{2}{5}, \frac{1}{5}

Answer

45,35,0\frac{4}{5}, \frac{3}{5}, 0

Explanation

Solution

Given equation of line is x+22=2y−53,z+1=0\frac{x+2}{2}=\frac{2y-5}{3}, z+1=0 or x+22=y−5232,z+1=0\frac{x+2}{2}=\frac{y-\frac{5}{2}}{\frac{3}{2}}, z+1=0 Hence, DRDR of a line are Now , $\sqrt{2^{2}+\left(\frac{3}{2}\right)^{2}+0}$ $=\sqrt{4+\frac{9}{4}}=\sqrt{\frac{25}{4}}=\frac{5}{2}$ $\therefore DC$ of a line are or $$