Question 17 of 25intermediate🔧 ApplyShort Answer4 marks

Find the shortest distance between the lines r⃗=(i^+2j^+k^)+λ(i^−j^+k^)\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) and r⃗=2i^−j^−k^+μ(2i^+j^+2k^)\vec{r} = 2\hat{i} - \hat{j} - \hat{k} + \mu(2\hat{i} + \hat{j} + 2\hat{k})

Correct Answer

The shortest distance between the given lines is 322\frac{3\sqrt{2}}{2} units.

Exercise: EXERCISE 11.2 | Q: 12 | (Page Page 14)
For More Understanding

Explanation

This question tests the application of the shortest distance formula between two skew lines. The student must correctly identify the position vectors and direction vectors from the vector equations, then apply the formula d = |(b₁ × b₂) · (a₂ - a₁)| / |b₁ × b₂|. The cross product calculation is crucial, and students should verify their arithmetic carefully.

Solution Steps

  1. Step 1: Identify the position vectors and direction vectors from the given equations.

  2. a₁ = î + 2ĵ + k̂, b₁ = î - ĵ + k̂

  3. a₂ = 2î - ĵ - k̂, b₂ = 2î + ĵ + 2k̂

  5. Step 2: Calculate a₂ - a₁

  6. a₂ - a₁ = (2î - ĵ - k̂) - (î + 2ĵ + k̂) = î - 3ĵ - 2k̂

  8. Step 3: Calculate b₁ × b₂ using determinant method

  9. b₁ × b₂ = |î ĵ k̂; 1 -1 1; 2 1 2| = î(-2-1) - ĵ(2-2) + k̂(1+2) = -3î + 3k̂

  11. Step 4: Calculate |b₁ × b₂|

  12. |b₁ × b₂| = 9+0+9\sqrt{9 + 0 + 9} = 18\sqrt{18} = 32\sqrt{2}

  14. Step 5: Calculate (b₁ × b₂) · (a₂ - a₁)

  15. (b₁ × b₂) · (a₂ - a₁) = (-3î + 3k̂) · (î - 3ĵ - 2k̂) = -3(1) + 0(-3) + 3(-2) = -3 - 6 = -9

  17. Step 6: Apply the shortest distance formula

  18. d = |(b₁ × b₂) · (a₂ - a₁)| / |b₁ × b₂| = |-9| / 32\sqrt{2} = 9/32\sqrt{2} = 3/2\sqrt{2} = 32\sqrt{2}/2

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