Consider triangle `A O B` in the `x-y` plane, where `A-=(1,0,0),B-=(0,2,0)a n d O-=(0,0,0)dot` The new position of `O ,` when triangle is rotated about side `A B` by `90^0` can be a. `(4/5,3/5,2/(sqrt(5)))` b. `((-3)/5,(sqrt(2))/5,2/(sqrt(5)))` c. `(4/5,2/5,2/(sqrt(5)))` d. `(4/5,2/5,1/(sqrt(5)))`

Options

A. `((4)/(5),(3)/(5),(2)/(sqrt5))`
B. `((-3)/(5),(sqrt2)/(5),(2)/(sqrt5))`
C. `((4)/(5),(2)/(5),(2)/(sqrt5))`
D. `((4)/(5),(2)/(5),(1)/(sqrt5))`

Answer

c

Explanation

Equation of line `AB` is `(x-1)/(1)= (y)/(-2) = (z)/(0) = lamda`
Now `AB bot OC rArr 1 (lamda+1)+ (-2lamda)(-2)=0`
or `" "5lamda=-1 or lamda=-(1)/(5)`
`rArr" "C` is `((4)/(5), (2)/(5), 0)`. Now
`" "x_(1)^(2)+ (y_(1)-2)^(2)+ z_(1)^(2)= 4" "`(i)
and `" "(x_(1)-1)^(2)+ y_(1)^(2)+ z_(1)^(2)= 1" "`(ii)
Now `OC bot CD`
`rArr" "(x_(1)- (4)/(5))(4)/(5) + (y_(1)- (2)/(5)) (2)/(5) + (z_(1)-0 ) 0=0" "` (iii)
From (i) and (ii), we get
`" "-4y_(1)+ 2x_(1)=0 or x_(1) = 2y_(1)`
From (iii), putting `x_(1)= 2y_(1) rArr 2y_(1) =(4)/(5) or y_(1) = (2)/(5) rArr x_1= (4)(5)`. Putting this value of `x_1 and y_1` in (i), we get
`" "z_(1)= pm (2)/(sqrt5)`