(i) Let 1 & 2 be two skew lines. If P, Q are two distinct points on 1 and R, S are two distinct points on 2 ,then prove that PR can not be parallel to QS.
(ii) A line with direction cosines proportional to (2, 7 – 5) is drawn to intersect the lines
and 
. Find the coordinate of the points of intersection
and thelength intercepted on it. Also find the equation of intersecting straight line.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(2, 8, – 3) & (0, 1, 2) ; 
; 
Sol. (i) Let equation of the line 1 be 
= 
and equation of the line 2 be 
= 
,
where 
, 
and 
are non-coplanar.
Let the position vectors of points P and Q be 
and 
respectively.
Let the position vectors of points R and S be 
and 
respectively.
Then the lines PR and QS are parallel if and only if 
i.e. (1 – k) 
+ ( μ 1 – k μ 2 ) 
– ( λ 1 – k λ 2 ) 
= 
∴ 1 – k = 0, μ 1 – k μ 2 = 0, λ 1 – k λ 2 = 0
i.e. μ 1 = μ 2 and λ 1 = λ 2 which is not possible ∴ PR can not be parallel to QS.
(ii) The given equation
... (i) and 
... (ii)
any point P on (i) is (3r 1 + 5, – r 1 + 7, r 1 – 2) and any point Q on (ii) is
(– 3r 2 – 3, 2r 2 + 3, 4r 2 + 6)
the direction ratios of PQ are
(3r 1 + 3r 2 + 8, – r 1 – 2r 2 + 4, r 1 – 4r 2 – 8) ... (iii)
suppose the line with d.r’s 2, 7, – 5 will be proportional to the d.r.’s given by (iii)
∴ 
... (iv)
Solving (iv), we get r 1 = r 2 = – 1
So point of intersection are P(2, 8, – 3) and Q(0, 1, 2)
and intercepted length = PQ = 
and equation of PQ is 
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