Question

find the shortest distance from the point (1 0 -3) to the plane x+2y+z=9

Answer 1

Formula :

Let us take any point as P (x₁, y₁, z₁) and any plane is

ax + by + cz + d = 0 ...(i)

Then, the shortest distance, i.e., the perpendicular distance of the plane (i) from the point P be

$= | \frac{ax_1 + by_1 + cz_1 + d}{ \sqrt{ {a}^{2} + {b}^{2} + {c}^{2} } } | \: \: units$

Answer :

Here, the given point is (1, 0, - 3) and the plane is

x + 2y + z - 9 = 0

Hence, the required shortest distance be

$= | \frac{1 + 2(0) + ( - 3) - 9}{ \sqrt{ {1}^{2} + {2}^{2} + {1}^{2} } } | \: \: units\\ \\ = | \frac{1 + 0 - 12}{ \sqrt{6} } | \: \: units \\ \\ = \frac{11}{ \sqrt{6} } \: \: units$

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