Test: Skew Lines - JEE MCQ

x = 2y = -3z     -4x = 6y = -z
x/1 = y/(½) = z(-⅓)                   x/(-¼) = y/(⅙) = z/(-1)
Cosθ = [(a1a2 + b1b2 + c1c2)/(a1 + b1 + c1)^½ * (a2 + b2 + c2)^½]
Cosθ ={[(1*(-¼)) + (½)(⅙) + (-⅓)(-1)]/[(1)² + (½)² + (-⅓)²]1/2 * [(-¼)² + (⅙)² + (-1)²]1/2}
= {[(-¼ + 1/12  - ⅓)]/[2 + 1 - ⅔]^1/2 * [ -½ + ⅓ -½]^½}
Cosθ = 0
θ = 90deg

3l + m + 5n = 0
m = - (3l + 5n) -----------(1)
6mn - 2nl + 5lm = 0 ----------(2)
Substitute m=-(3l+5n) in eq(2)
⇒ 6[- (3l + 5n)]n - 2nl + 5l[- (3l + 5n)] = 0
⇒( -18ln - 30n)n-2nl-15l^2+25ln=0
⇒ l(l + 2n) + n(l + 2n) = 0
⇒ (l + n) (l + 2n) = 0
∴ l = - n and l = -2n
( l / -1 ) = ( n / 1) and ( l / -2) = ( n / 1) -------(3)
Substitute l in equation 1, we get
m = - (3l + 5n)
m = -2n and m = n
( m / -2) = ( n / 1) and ( m / 1) = ( n / 1 ) --------(4)
From ( 3) and (4) we get
( l / -1 ) = ( m / -2) = ( n / 1),
( l / -2) = ( m / 1) = ( n / 1 )
l : m : n = -1 : -2 : 1
l : m : n = -2 : 1 : 1
i.e D.r's ( -1, -2, 1) and ( -2 , 1 , 1)
Angle between the lines whose direction cosines are
Cos θ = ( -1 × -2 + -2×1 + 1×1) / √ ((-1)^2+(-2)²+1²))*√((-2)²+1²+1²))
Cos θ = 1 / √6 √6
Cos θ = 1 / 6
∴ θ = cos inverse of (1/6)
∴Angle between the lines whose direction cosines is cos^-1(1/6)

let P and Q be the points on the given lines, respectively. then the general coordinates of P and Q are: 
P(k+3, -2k+5, k+7) and Q (7m-1, -6m-1, m-1)
therefore the direction ratios of PQ are (7m-k-4,-6m+2k-6, m-k-8)
now PQ will be the shortest distance if it is perpendicular to both the given lines, therefore by the condition of perpendicularity,
1(7m-k-4) -2(-6m+2k-6) + 1(m-k-8) = 0  (1)
7(7m-k-4) -6(-6m+2k-6) + 1(m-k-8) = 0  (2)
now solving (1) and (2),
m=0 and k = 0
hence the points are P(3,5,7) and Q (-1,-1,-1), therefore the shortest distance between the lines
PQ = sqrt((3+1)²+(5+1)² +(7+1)²) 
= sqrt(16+36+64) = sqrt(116) 
= 2sqrt(29)