Consider a point P with position vector `vec a` and a plane `π_1` whose equation is `vec r . hat n = d` Fig.
Consider a plane `π_2` through P parallel to the plane `π_1`. The unit vector normal to `π_2` is
Hence , its equation is `(vec r - vec a) . hat n = 0`
i.e., `(vec r . hat n = vec a . hat n)`
Thus, the distance ON′ of this plane from the origin is `|vec a . hat n|.` Therefore , the distance PQ from the plane `π_1` is above fig.
i.e. `ON - ON' = |d - vec a. hat n|`
which is the length of the perpendicular from a point to the given plane.
Let `P(x_1, y_1, z_1)` be the given point with position vector`vec a`
and Ax + By + Cz = D
be the Cartesian equation of the given plane. Then
`vec a = x_1 hat i + y_1 hat j + z_1 hat k`
`vec N = A hat i + B hat j + C hat k`
Hence, the perpendicular from P to the plane is
`|((x_1 hat i + y_1 hat j + z_1 hat k) . (A hat i + B hat j + C hat k) - D)/ sqrt (A^2 + B^2 +C^2)|`
`=|(Ax_1 + By_1 + Cz_1 - D)/sqrt(A^2 + B^2 +C^2)|`
Shaalaa.com | Three Dimensional Geometry Part 7 - The Plane
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Solve the following :
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