if the length of the perpendicular from the point
`(beta,0,beta )(betane 0) " to the line " , (x)/1 =(y-1)/(0) =(z+1)/(-1) " is " sqrt((3)/(2)),` then `beta " is equal to "`

Options

A. 2
B. -2
C. -1
D. 1

Answer

C

Explanation

Equations of given line is
`(x)/(1)= (y-1)/(0) =(z+1)/(-1)`
Now one of the point of line is `P (0,1-1 ) ` and the given point is`Q(beta , 0 , beta)`

From the figure the length of the perpendicular
`QM = l = sqrt((3)/(2)) " ""(given)"`
`rArr (|PQ xx PM|)/(|PM|) = sqrt((3)/(2)) " " .....(ii)`
and PM=a vector along given line (1) `=hat(i) - hat(k)` ltbr. so `PQxx PM = |{:(hat(i) ,,hat(j) ,,hat(k)),( beta ,,-1,,beta+1),(1,,0,,-1):}|`
`= hat(i) - hat(j) (-beta- beta-1) + hat(k) = hat(i) + (2beta + 1)^(2) hat(k)`
` " Now " (|PQ xx PM|)/(|PM|) = (sqrt(1+(2beta+1)^(2)+1)/(sqrt(2))) " "......(iii)`
From Eqs . (ii) and (iii) we get
`sqrt((1+(2beta +1)^(2)+1)/(2)) = sqrt((3)/(2)) rArr .(1+(2beta +1)^(2) +1)/(2) =(3)/(2)`
`rArr (2beta +1)^(2) =1 rArr 2beta +1 = +-1`
`rArr 2beta + 1= 1 " or " 2beta +1= -1 rArr beta = 0 " or " beta =-1`