<p>The position vectors of the points <em>P</em> and <em>Q</em> are \(5\hat{i} + 7\hat{j} - 2\hat{k}\) and \(-3\hat{i} + 3\hat{j} + 6\hat{k}\), respectively. The vector \(\vec{A} = 3\hat{i} - \hat{j} + \hat{k}\) passes through the point <em>P</em> and the vector \(\vec{B} = -3\hat{i} + 2\hat{j} + 4\hat{k}\) passes through the point <em>Q</em>. A third vector \(2\hat{i} + 7\hat{j} - 5\hat{k}\) intersects vectors \(\vec{A}\) and \(\vec{B}\). Then the distance between the two position vectors of the points of intersection is __________ (up to three decimal places).</p>
Step-by-Step Solution
Key Concept: Find the intersection points of the third line with lines A and B by setting up parametric equations and solving for parameters where the third vector's point lies on each line. The distance between these two intersection points is the answer.
Step 1: Write parametric equations for the three lines. Line A through P(5,7,-2) with direction 3î - ĵ + k̂: r_1(s) = (5+3s, 7-s, -2+s) Line B through Q(-3,3,6) with direction -3î + 2ĵ + 4k̂: r_2(t) = (-3-3t, 3+2t, 6+4t) Line C through origin with direction 2î + 7ĵ - 5k̂: r_3(u) = (2u, 7u, -5u) Step 2: Find intersection of Line C with Line A by equating r_1(s) = r_3(u): 5 + 3s = 2u ... (i) 7 - s = 7u ... (ii) -2 + s = -5u ... (iii) From (ii): s = 7 - 7u. Substitute in (iii): -2 + 7 - 7u = -5u → 5 = 2u → u = 2.5, s = -10.5 Intersection point R_1 = (2(2.5), 7(2.5), -5(2.5)) = (5, 17.5, -12.5) Step 3: Find intersection of Line C with Line B by equating r_2(t) = r_3(u): -3 - 3t = 2u ... (iv) 3 + 2t = 7u ... (v) 6 + 4t = -5u ... (vi) From (v): t = (7u - 3)/2. Substitute in (iv): -3 - 3(7u-3)/2 = 2u → -6 - 21u + 9 = 4u → 3 = 25u → u = 0.12, t = -1.92 Intersection point R_2 = (2(0.12), 7(0.12), -5(0.12)) = (0.24, 0.84, -0.6) Step 4: Calculate distance between R_1 and R_2: d = √[(5-0.24)^2 + (17.5-0.84)^2 + (-12.5+0.6)^2] d = √[22.7176 + 275.9056 + 141.61] = √440.233 = 20.981 ∴ Answer: 20.981
Correct Answer: 20