Vector Algebra Question 81
Question: A vector $ \overset{\to }{\mathop{a}},=(x,y,z) $ of length $ 2\sqrt{3} $ which makes equal angles with the vectors $ \overset{\to }{\mathop{b}},=(y,-2z,3x) $ and $ \overset{\to }{\mathop{c}},=(2z,3x,-y) $ is perpendicular to $ \overset{\to }{\mathop{d}},=(1,-1,2) $ and makes an obtuse angle with y-axis is
Options:
A) (- 2, 2, 2)
B) $ (1,1,\sqrt{10}) $
C) (2, - 2, - 2)
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Since, $ \overset{\to }{\mathop{a}}, $ is $ \bot $ to $ \overset{\to }{\mathop{d}}, $ , so $ x-y+2z=0 $ ?(1)
Moreover, $ |\overset{\to }{\mathop{b}},|=|\overset{\to }{\mathop{c}},|, $ so $ \overset{\to }{\mathop{a}},.\overset{\to }{\mathop{b}},=\overset{\to }{\mathop{a}},.\overset{\to }{\mathop{c}}, $
as $ \overset{\to }{\mathop{a}}, $ makes equal angles with $ \overset{\to }{\mathop{b}}, $ and $ \overset{\to }{\mathop{c}}, $ . Thus $ xy-2yz+3xz=2xz+3xy-yz $
$ \Rightarrow xz-2xy-yz=0 $ ?(2)
Also $ x^{2}+y^{2}+z^{2}=12 $ ?(3)
and $ y<0 $
Put the value of y from eq. (1) in eq. (2),
We get, $ x^{2}+2xz+z^{2}=0; $ so, $ x=-z $ and $ y=z $
Again put these values in eq. (3), we get
$ z^{2}=4\Rightarrow z=\pm 2 $
But $ y<0 $ and $ y=z $ . Hence, $ z=-2=y $ and $ x=2 $
BETA
