<p>The distance of the point having position vector \(-\hat{i} + 2\hat{j} + 6\hat{k}\) from the straight line passing through the point \((2, 3, -4)\) and parallel to the vector, \(6\hat{i} + 3\hat{j} - 4\hat{k}\) is ______.</p>
Step-by-Step Solution
Key Concept: The distance from a point to a line in 3D is found using the formula: d = ||(P₀P) × d||/||d||, where P₀ is a point on the line, P is the external point, and d is the direction vector of the line.
Step 1: Identify given information. Point P = (-1, 2, 6), Point on line P_0 = (2, 3, -4), Direction vector d = 6 î + 3 ĵ - 4 k̂ Step 2: Find vector P_0P. P_0P = P - P_0 = (-1-2) î + (2-3) ĵ + (6-(-4)) k̂ = -3 î - ĵ + 10 k̂ Step 3: Calculate cross product (P_0P) × d . (P_0P) × d = | î** **ĵ** **k̂ | |-3 -1 10| |6 3 -4| = î [(-1)(-4) - (10)(3)] - ĵ [(-3)(-4) - (10)(6)] + k̂ [(-3)(3) - (-1)(6)] = î [4 - 30] - ĵ [12 - 60] + k̂ [-9 + 6] = -26 î + 48 ĵ - 3 k̂ Step 4: Calculate magnitude of cross product. ||(P_0P) × d || = √(676 + 2304 + 9) = √2989 Step 5: Calculate magnitude of direction vector. || d || = √(36 + 9 + 16) = √61 Step 6: Apply distance formula. Distance = √2989/√61 = √(2989/61) = √49 = 7 ∴ Answer: 7
Correct Answer: 7