Mathematics

Mathematics

Problem

A line ll passing through the origin is perpendicular to the lines l1:(3+t)i^+(−1+2t)j^+(4+2t)k^,−∞<t<∞l_1: (3+t)\hat{i} + (-1+2t)\hat{j} + (4+2t)\hat{k}, -\infty < t < \infty l2:(3+2s)i^+(3+2s)j^+(2+s)k^,−∞<s<∞l_2: (3+2s)\hat{i} + (3+2s)\hat{j} + (2+s)\hat{k}, -\infty < s < \infty Find the coordinate(s) of the point(s) on l2l_2 at a distance of 17\sqrt{17} from the point of intersection of ll and l1l_1. a. (73,73,53)\left(\frac{7}{3}, \frac{7}{3}, \frac{5}{3}\right) b. (−1,−1,0)(-1, -1, 0) c. (1,1,1)(1, 1, 1) d. (79,79,89)\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)
Intuitive explanation

The key is to recognize that a line perpendicular to both l1l_1 and l2l_2 must be parallel to their common perpendicular direction, found via the cross product of their direction vectors. Once you have line ll, find where it intersects l1l_1, then use the distance condition to locate point(s) on l2l_2 that are exactly 17\sqrt{17} units away from that intersection point. This combines vector geometry (perpendicularity via dot product) with distance calculations in 3D space.

Intuitive explanation

The key is to recognize that a line perpendicular to both l1l_1 and l2l_2 must be parallel to their common perpendicular direction, found via the cross product of their direction vectors. Once you have line ll, find where it intersects l1l_1, then use the distance condition to locate point(s) on l2l_2 that are exactly 17\sqrt{17} units away from that intersection point. This combines vector geometry (perpendicularity via dot product) with distance calculations in 3D space.

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