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Question: From a point \( P(\lambda , \lambda , \lambda ) \) , perpendicular PQ and PR are drawn respectively ...

From a point P(λ,λ,λ)P(\lambda , \lambda , \lambda ) , perpendicular PQ and PR are drawn respectively on the lines y=x,z=1y = x, z = 1 and y=−x,z=−1y = - x, z = - 1 . If P is such that ∠QPR\angle QPR is a right angle, then the possible value(s) of λ\lambda is (are)
A. 2\sqrt 2
B. 1
C. -1
D. −2- \sqrt 2

Explanation

Solution

This is a three dimensional coordinate geometry of line question. To solve this question first we will derive the equation of the two lines from given data and their respective direction ratios. Then we will derive the direction ratios of line PQ and PR. Then as per given data and using perpendicular formula, we will get the value of λ\lambda

Complete step-by-step answer:
Line L1{L_1} is given by, y=x,z=1y = x,z = 1
Hence the equation of line L1{L_1} is x1=y1=z−10=q\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{{z - 1}}{0} = q (let)
(1,1,0)(1,1,0) is its direction ratios
⇒x=q,y=q,z=1\Rightarrow x = q,y = q,z = 1
Let the coordinates of point Q be Q(q,q,1)Q(q,q,1)
Line L2{L_2} is given by y=−x,z=−1y = - x,z = - 1
Hence the equation of line L2{L_2} is x1=y−1=z+10=r\dfrac{x}{1} = \dfrac{y}{{ - 1}} = \dfrac{{z + 1}}{0} = r (let)
(1,−1,0)(1, - 1,0) is its direction ratios
⇒x=r,y=−r,z=−1\Rightarrow x = r,y = - r,z = - 1
Let the coordinates of point R be R(r,−r,−1)R(r, - r, - 1)
We know that the direction ratios of a line joining two points (x1,y1,z1)({x_{1,}}{y_{1,}}{z_1}) and (x2,y2,z2)({x_{2,}}{y_{2,}}{z_2}) are x2−x1,y2−y1,z2−z1{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1} ………………… Formula 1
Hence, the direction ratios of line joining PQ are, λ−q,λ−q,λ−1\lambda - q,\lambda - q,\lambda - 1
Again we know that two lines having direction ratios (a1,b1,c1)({a_{1,}}{b_{1,}}{c_1}) and (a2,b2,c2)({a_{2,}}{b_{2,}}{c_2}) are perpendicular if a1a2+b1b2+c1c2=0{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0 …………………… Formula 2
Now according to the question, PQ⊥L1PQ \bot {L_1}
∴\therefore 1(λ−q)+1(λ−q)+0(λ−1)=01(\lambda - q) + 1(\lambda - q) + 0(\lambda - 1) = 0
Simplifying the above equation we get,
λ−q+λ−q=0\lambda - q + \lambda - q = 0
⇒2λ=2q\Rightarrow 2\lambda = 2q
⇒λ=q\Rightarrow \lambda = q
The direction ratios of line joining PQ are, λ−λ,λ−λ,λ−1\lambda - \lambda ,\lambda - \lambda ,\lambda - 1
Similarly, the direction ratios of line joining PR are, λ−r,λ+r,λ+1\lambda - r,\lambda + r,\lambda + 1
Now according to the question, PR⊥L2PR \bot {L_2}
∴\therefore 1(λ−r)+(−1)(λ+r)+0(λ+1)=01(\lambda - r) + ( - 1)(\lambda + r) + 0(\lambda + 1) = 0
Simplifying the above equation we get,
λ−r−λ−r=0\lambda - r - \lambda - r = 0
⇒r=o\Rightarrow r = o
Hence, coordinates of R is (0,0,−1)(0,0, - 1)
The direction ratios of line joining PR are, λ,λ,λ+1\lambda ,\lambda ,\lambda + 1
As ∠QPR=90∘\angle QPR = 90^\circ i.e. PQ⊥PRPQ \bot PR
Then putting their direction ratios in formula 2 we get,
(λ−λ)(λ−0)+(λ−λ)(λ−0)+(λ−1)(λ+1)=0(\lambda - \lambda )(\lambda - 0) + (\lambda - \lambda )(\lambda - 0) + (\lambda - 1)(\lambda + 1) = 0
Simplifying the above equation we get,
(λ−1)(λ+1)=0(\lambda - 1)(\lambda + 1) = 0
λ=1\lambda = 1 is rejected as P and Q are different points.
∴\therefore λ=−1\lambda = - 1

So, the correct answer is “Option C”.

Note: In 3-D coordinate geometry, the equation of line with direction vector l, m, n that passes through the point x’, y’, z’ is given by the formula, x−x′l=y−y′m=z−z′n\dfrac{{x - x'}}{l} = \dfrac{{y - y'}}{m} = \dfrac{{z - z'}}{n}
The direction ratios of a line joining two points (x1,y1,z1)({x_{1,}}{y_{1,}}{z_1}) and (x2,y2,z2)({x_{2,}}{y_{2,}}{z_2}) are x2−x1,y2−y1,z2−z1{x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1} .
Two lines having direction ratios (a1,b1,c1)({a_{1,}}{b_{1,}}{c_1}) and (a2,b2,c2)({a_{2,}}{b_{2,}}{c_2}) are perpendicular if a1a2+b1b2+c1c2=0{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0 .
You should remember all the rules, properties and formulas coordinate geometry.
You might mistakenly take both the values, but according to the question, P and Q are two different points.