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Question
Find the foot of the perpendicular drawn: from the point (5, 4, 2) to the line `(x + 1)/2 = (y - 3)/3 = (z - 1)/(-1)`. Also, find the equation of the perpendicular
Solution
`vec"r" = vec"a" + "t"vec"b"`
`vec"a" = -vec"i" + 3vec"j" + vec"k"`
`vec"b" = 2vec"i" + 3vec"j" - vec"k"`
Given points D(5, 4, 2) to the point A.
If P is the foot of the perpendicular from to the straight line.
F is of the form
(2t – 1, 3t + 3, -t + 1) and
`vec"DF" = vec"OF" - vec"OD" = (2"t" - 6)vec"i" + (3"t" - 1)vec"j" + (-"t" - l)vec"k"`
`vec"b"` is perpendicular to `vec"DF"`, we have
4t- 12 + 9t – 3 + t + 1 = 0
14t – 14 = 0
14t = 14
t = 1
∴ F(2 – 1, 3 + 3, -1 + 1) = F(1, 6, 0) is foot point.
Equation of the perpendicular.
(x1, y1, z1) = (5, 4, 2), (x2, y2, z2) = (1, 6, 0).
`(x - x_1)/(x_2 - x_1) = (y - y_1)/(y_2 - y_1) = (z - z_1)/(z_2 - z_1)`
`(x - 5)/(1 - 5) = (y - 4)/(6 - 4) = (z - 2)/(0 - 2)`
⇒ `(y - 4)/2 = (z - 2)/(-2)`
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