# Find the shortest distance for Point P (2, 5, 2) Information gi

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Find the shortest distance for Point P (2,-5, -2)
Information given:
x = 4+2t
y = -4-3t
z = 5+4t
Solution
taking a general point on the vector in terms of \'t\'.
x = 4+2t, y = -4-3t, z = 5+4t
distance(D) of this general points from P (2,-5, -2) is given
by:
[ (4+2t-2)^2 + (-4-3t+5)^2 + (5+4t+2)^2 ]
= [ (2+2t)^2 + (3t-1)^2 + (4t+7)^2 ]
minimising D is the same as minimising D^2
hence, D^2 = (2+2t)^2 + (3t-1)^2 + (4t+7)^2
d(D^2)/dt = 2.(2+2t).2 + 2.(3t-1).3 + 2.(4t+7).4
= 8t+8 + 18t -6 + 32t+56
= 58t+58
= 58(t+1)

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Find the shortest distance for Point P (2, -5, -2) Information given: x = 4+2t y = -4-3t z = 5+4t Solution taking a general point on the vector in terms of \'t\'. x = 4+2t, y = -4-3t, z = 5+4t distance(D) of this general points from P (2, -5, -2) is given by: [ (4+2t-2)^2 + (-4-3t+5)^2 + (5+4t+2)^2 ] = [ (2+2t)^2 + (3t-1)^2 + (4t+7)^2 ] minimising D is the same as minimising D^2 hence, D^2 = (2+2t)^2 + (3t-1)^2 + (4t+7)^2 d(D^2)/dt = 2.(2+2t).2 + 2.(3t-1).3 + 2.(4t+7).4 = 8t+8 + 18t -6 + 32t+56 = 58t+58 = 58(t+1) it is minimum at t=-1 as d^2(D^2)/dt^2 > 0 and hence is minimum at t=-1 => D ...
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