Three Dimensional Geometry Question 93
Question: If $ P\equiv (0, 1, 0),Q\equiv (0, 0, 1) $ , then projection of $ PQ $ on the plane $ x+y+z=3 $ is
[EAMCET 2002]
Options:
A) $ \sqrt{3} $
3
C) $ \sqrt{2} $
2
Show Answer
Answer:
Correct Answer: C
Solution:
Given plane is $ x+y+z-3=0 $ . From point P and Q draw PM and QN perpendicular on the given plane and QR ⊥ MP. $ |MP| =\frac{|0+1+0-3|}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{2}{\sqrt{3}} $ , $ |NQ| =\frac{2}{\sqrt{3}} $
$ |PQ| =\sqrt{{{(0-0)}^{2}}+{{(0-1)}^{2}}+{{(1-0)}^{2}}}=\sqrt{2} $
$ |RP|=|MP|-|MR| = |MP|-|NQ|=0 $
\ $ |NM|,=,|QR|,=\sqrt{PQ^{2}-RP^{2}}=\sqrt{{{(\sqrt{2})}^{2}}-0} $ $ =,\sqrt{2}. $
BETA
