Skew Lines - Free MCQ Practice Test with solutions, JEE Maths

Condition for Perpendicularity

Let the direction ratios of the first line be a1, b1, c1, and the direction ratios of the second line be a2, b2, c2. The lines are perpendicular if the dot product of their direction vectors is zero.

Mathematical Expression:
a₁ · a₂ + b₁ · b₂ + c₁ · c₂ = 0

Explanation:

1. Direction Vectors:

   • First line: v₁ = (a_1, b₁, c₁)
   • Second line: v₂ = (a₂, b₂, c₂)

2. Dot Product:

   • The dot product of v1 and v2 is calculated as:
     v₁ · v₂ = a₁ a₂ + b₁ b₂ + c₁ c₂

3. Perpendicularity Condition:

   • For two vectors to be perpendicular, their dot product must be zero:
     v₁ · v₂ = 0 → ₂ = a₁a₂ + b₁b₂ + c₁c₂ = 0

Conclusion:
Two lines whose direction ratios are a1, b1, c1 and a2, b2, c2 are perpendicular if and only if:a₁a₂ + b₁b₂ + c₁c₂ = 0

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)