3D Geometry
Shortest Distance Between Lines
Grade 12
Question:
<p><strong>Paragraph for Question nos. 640 and 641</strong><br>Consider two lines \(L_1: 2\hat{i}+\hat{j}-\hat{k}+\lambda(\hat{i}+2\hat{k})\) and \(L_2: 3\hat{i}+\hat{j}-\hat{k}+\mu(\hat{i}+\hat{j}-\hat{k})\). Let \(\Pi\) be the plane which contains the line \(L_1\) and parallel to \(L_2\) and intersecting coordinate axes at \(A\), \(B\) and \(C\) respectively.</p><p>The shortest distance between \(L_1\) and \(L_2\) is:</p>
<p>(a) \(\dfrac{1}{\sqrt{5}}\)</p>
<p>(b) \(\dfrac{1}{\sqrt{6}}\)</p>
<p>(c) \(\dfrac{1}{\sqrt{8}}\)</p>
<p>(d) \(\dfrac{1}{\sqrt{14}}\)</p>
Step-by-Step Solution
Key Concept: The shortest distance between two skew lines is found using the formula: d = |((P₂-P₁)·(d₁×d₂))|/|d₁×d₂|, where P₁, P₂ are points on the lines and d₁, d₂ are direction vectors. First verify the lines are skew by checking they're non-parallel and non-intersecting.
Step 1: Identify position vectors and direction vectors. L_1: P_1 = 2î + ĵ - k̂, d_1 = î + 2k̂ L_2: P_2 = 3î + ĵ - k̂, d_2 = î + ĵ - k̂ Step 2: Compute P_2 - P_1 = (3î + ĵ - k̂) - (2î + ĵ - k̂) = î Step 3: Calculate d_1 × d_2: d_1 × d_2 = |î ĵ k̂| = î(−2−0) − ĵ(−1−0) + k̂(1−0) = −2î + ĵ + k̂ |0 1 −1| |d_1 × d_2| = √(4 + 1 + 1) = √6 Step 4: Calculate (P_2 - P_1)·(d_1 × d_2) = î·(−2î + ĵ + k̂) = −2 Step 5: Apply distance formula: d = |−2|/√6 = 2/√6 = 2√6/6 = √6/3 ∴ Answer: B
Correct Answer: B