Q.
The foot of the perpendicular from (0,2,3) to the line 5x+3=2y−1=3z+4 is
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Introduction to Three Dimensional Geometry
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Solution:
Let N be the foot of the perpendicular from the point (0, 2, 3) on the given line 5x+3=2y−1=3z+4 ...(1)
Any pt. on (1) is (5r−3,2r+1,3r−4). If this point is N, then direction numbers of NP are <5r−3−0,2r+1−2,3r−4−3> i.e.,<5r−3,2r−1,3r−7>. Since PN is ⊥ to the given line. ∴5(5r−3)+2(2r−1)+3(3r−7) = 0 ⇒38r−38=0⇒r=1 ∴ N is (5 - 3, 2 + 1, 3 - 4) i.e.. (2, 3, -1),