Question 20 of 25advanced🔧 ApplyShort Answer4 marks

Find the shortest distance between the lines whose vector equations are r⃗=(1−t)i^+(t−2)j^+(3−2t)k^\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k} and r⃗=(s+1)i^+(2s−1)j^−(2s+1)k^\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} - (2s+1)\hat{k}

Correct Answer

The shortest distance between the given lines is d = 8/29\sqrt{29} units.

Exercise: EXERCISE 11.2 | Q: 15 | (Page Page 14)
For More Understanding

Explanation

This question asks for the shortest distance between two skew lines. The context provides Example 9 which demonstrates the formula d = |(b₁ × b₂) · (a₂ - a₁)| / |b₁ × b₂|. Students need to first rewrite both equations in standard form r = a + λb, then identify the position vectors and direction vectors, and finally apply the formula.

Solution Steps

  1. Step 1: Rewrite both equations in standard form r = a + λb. For first line: r = (î - 2ĵ + 3k̂) + t(-î + ĵ - 2k̂), so a₁ = î - 2ĵ + 3k̂ and b₁ = -î + ĵ - 2k̂. For second line: r = (î - ĵ - k̂) + s(î + 2ĵ - 2k̂), so a₂ = î - ĵ - k̂ and b₂ = î + 2ĵ - 2k̂.

  2. Step 2: Calculate a₂ - a₁ = (î - ĵ - k̂) - (î - 2ĵ + 3k̂) = ĵ - 4k̂.

  3. Step 3: Calculate b₁ × b₂ using determinant method: b₁ × b₂ = 2î - 4ĵ - 3k̂.

  4. Step 4: Find |b₁ × b₂| = 4+16+9\sqrt{4 + 16 + 9} = 29\sqrt{29}.

  5. Step 5: Calculate (b₁ × b₂) · (a₂ - a₁) = (2î - 4ĵ - 3k̂) · (ĵ - 4k̂) = -4 + 12 = 8.

  6. Step 6: Apply formula: d = |8|/29\sqrt{29} = 8/29\sqrt{29} units.

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