<p><strong>182.</strong> In a tetrahedron \(OABC\), if \(\vec{OA} = \vec{i}\), \(\vec{OB} = \vec{i} + \vec{j}\) and \(\vec{OC} = \vec{i} + 2\vec{j} + \vec{k}\), if shortest distance between edges \(OA\) and \(BC\) is \(m\), then \(2m\) is equal to \(\ldots\) (Where \(O\) is the origin)</p>
Step-by-Step Solution
Key Concept: The shortest distance between two skew lines in 3D space 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.
Step 1: Identify the two lines and their direction vectors. Line OA passes through O(0,0,0) with direction vector d_1 = OA = i = (1, 0, 0) Line BC passes through B with direction vector d_2 = BC = OC - OB = ( i + 2 j + k ) - ( i + j ) = j + k = (0, 1, 1) Step 2: Find a vector connecting a point on OA to a point on BC. Point on OA: O = (0, 0, 0) Point on BC: B = (1, 1, 0) Vector from O to B: OB = (1, 1, 0) Step 3: Calculate the cross product d_1 × d_2 . d_1 × d_2 = | i j k | = i (0·1 - 0·1) - j (1·1 - 0·0) + k (1·1 - 0·0) |1 0 0| |0 1 1| d_1 × d_2 = 0 i - 1 j + 1 k = (0, -1, 1) Step 4: Calculate the magnitude of the cross product. | d_1 × d_2 | = √(0^2 + (-1)^2 + 1^2) = √2 Step 5: Calculate the scalar triple product. ( OB ) · ( d_1 × d_2 ) = (1, 1, 0) · (0, -1, 1) = 1(0) + 1(-1) + 0(1) = -1 Step 6: Apply the distance formula. m = |( OB ) · ( d_1 × d_2 )| / | d_1 × d_2 | = |-1| / √2 = 1/√2 Step 7: Calculate 2m. 2m = 2 · (1/√2) = 2/√2 = 2√2/2 = √2 ∴ Answer: 2m = √2
Correct Answer: 2