JEE Main & Advanced
JEE Main Paper (Held On 25 April 2013)
question_answer
Let ABC be a triangle with vertices at points A (2,3,5), B (-1,3,2) and C (\[\lambda ,5,\mu \]) in three dimensional space. If the median through A is equally inclined with the axes, the is\[\left( \lambda ,\mu \right)\]equal to:
JEE Main Online Paper ( Held On 25 April 2013 )
A) \[\left( 10,7 \right)\]
B) \[\left( 7,5 \right)\]
C) \[\left( 7,10 \right)\]
D) \[\left( 5,7 \right)\]
Correct Answer:
C
Solution :
Since AD is the median Now, dR's of AD is \[a=\left( \frac{\lambda -1}{2}-2 \right)=\frac{\lambda -5}{2}\] \[b=4-3=1,c=\frac{\mu +2}{2}-5=\frac{\mu -8}{2}\] Also, a,b, c are dR's \[\therefore \]\[a=kl,b=km,c=kn\] where \[l=m=n\]and\[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\]\[\Rightarrow \]\[l=m=n=\frac{1}{\sqrt{3}}\] Now,\[a=1,b=1\]and\[c=1\] \[\Rightarrow \]\[\lambda =7\]and\[\mu =10\]