Q.161. Shortest distance, (minima)

Question 161.
Find the point closest to the origin and on the line of intersection of the planes
2x + 3y + z = 9 and x + 2y + 3z = 8.

Answer 161.
Normals of the planes 2x + 3y + z = 9 and x + 2y + 3z = 8 are
n1 = (2, 3, 1) and n2 = (1, 2, 3).

The direction of the line of intersection is
n1 x n2
= (2, 3, 1) x ( 1, 2, 3)
= (7, -5, 1)

If one of the point on this line has z-coordinate = 0, plugging z = 0 in the equations of the planes,
2x + 3y = 9 ... ( 1 ) and
x + 2y = 8 ... ( 2 )
Solving, x = - 6 and y = 7
=> (-6, 7, 0) is a point on the line of intersection of the given planes.

The parametric equation of the line of intersection is
x = - 6 + 7t, y = 7 - 5t and z = t, t belongs to R.

Distance from origin to the t-point on the line is given by
s^2 = (- 6 + 7t)^2 + (7 - 5t)^2 + t^2
=> s^2 = 75t^2 - 154t + 85 ... ( 3 )

For s to be minimum, s^2 has to be minimum,
=> d/dt(s^2) = 0 and d^2/dt^2 (s^2) > 0

d/dt(s^2) = 0 => 150t - 154 = 0 => t = 77/75
d^2(s^2)/dt^2 = 150 > 0

Plugging t = 77/75 in equation ( 3 ),
s^2 = 75 * (77/75)^2 - 154 * (77/75) + 85 ≈ 5.9467
=> closest distance, s ≈ 2.4386.

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