From a point `P(lambda, lambda, lambda)`, perpendicular PQ and PR are drawn respectively on the lines `y=x, z=1 and y=-x, z=-1`. If P is such tthat `angleQPR` is a right angle , then the possible value(s) of `lambda` is (are)

Options

A. `sqrt(2)`
B. 1
C. -1
D. `-sqrt(2)`

Answer

C

Explanation

The equation of the given lines are
`L_(1):y=x,z=1` or `x/1-y/1=(z-1)/0` …………..i
`L_(2):y=-x,z=-1` or `x/1=y/(-1)=(z+1)/0`……………ii
Let the coordinates `Q` and `R` be given by
`x/1=y/1=(z-1)/0=r_(1)` and `x/1=y/(-1)=(z+1)/0=r_(2)` respectively.
Thus, the coordinates of `Q` an `R` be `(r_(1),r_(1),1)` and `(r_(2),-r_(2),-1)` respectively.
`:.vec(PQ)=(r_(2)-lamda)hati+(r_(2)-lamda)hatj+(1-lamda)hatk`,
and `vec(PR)=(r_(2)-lamda)hati+(r_(2)-lamda)hatj+(-1-lamda)hatk`
It is given that `PQ` is perpendicular to `L_(1)` and `PR` is perpendicular to `L_(2)`.
`:.vec(PQ).(hati+hatj+0hatk)=0` and `vec(PR).(hati-hatj+0hatk)=0`
`impliesr_(1)-lamda+r_(1)-lamda+0(1-lamda)=0`
and `(r_(2)-lamda)-(r_(2)-lamda)+0(-1-lamda)=0`
`impliesr_(1)=lamda` and `r_(2)=0`
`:.vec(PQ)=(1-lamda)hatk` and `vec(PR)=-lamda(hati-lamdahatj+(-1-lamda)hatk`
Also `vec(PQ)_|_vec(PR)`
`:.vec(PQ).vec(PR)=0`
`implies(1-lamda)(-1-lamda)=lamda=+-1`
For `lamda=1,vec(PQ)=vec0`. So P and Q coincide.
`:.lamda=-1`
Hence, there is only one value of `lamda` equal to `-1`.