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Test: Skew Lines - JEE MCQ


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10 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Skew Lines

Test: Skew Lines for JEE 2025 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Skew Lines questions and answers have been prepared according to the JEE exam syllabus.The Test: Skew Lines MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Skew Lines below.
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Test: Skew Lines - Question 1

Two lines whose direction ratios are a1, b1, c1 and a2, b2, c2 are parallel, if

Test: Skew Lines - Question 2

For which value of a lines  and  are perpendicular?

Detailed Solution for Test: Skew Lines - Question 2

Test: Skew Lines - Question 3

The shortest distance between the lines whose equations are  and  is:

Test: Skew Lines - Question 4

Two lines whose direction ratios are a1,b1,c1 and a2,b2,c2 are perpendicular, if

Detailed Solution for Test: Skew Lines - Question 4

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 = (a1, 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

Test: Skew Lines - Question 5

The shortest distance between the parallel lines whose equations are and 

Test: Skew Lines - Question 6

The angle between the pair of lines given byand  is:

Test: Skew Lines - Question 7

The angle between the lines x = 2y = – 3z and – 4x = 6y = – z is:​

Detailed Solution for Test: Skew Lines - Question 7

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

Test: Skew Lines - Question 8

The angle between the lines whose direction cosines are given by the equations 3l + m + 5n = 0, 6nm - 2nl + 5lm = 0 is:

Detailed Solution for Test: Skew Lines - Question 8

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)

Test: Skew Lines - Question 9

The angle between the lines  and  is:

Test: Skew Lines - Question 10

The length of the shortest distance between the lines  and  is:

Detailed Solution for Test: Skew Lines - Question 10

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)

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