Question
Download Solution PDFWhat is the distance of the point (1, 2, 3) form the plane x – 3y + 2z + 13 = 0?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Perpendicular Distance of a Point from a Plane
Let us consider a plane given by the Cartesian equation, Ax + By + Cz = d
And a point whose coordinate is, (x1, y1, z1)
Now, distance = \(\left| {\frac{{A{x_1}\; + \;B{y_1}\; + \;C{z_1} - \;d}}{{\sqrt {{A^2}\; + \;\;{B^2}\; + \;{C^2}} }}} \right|\)
Calculation:
We have to find the distance of the point (1, 2, 3) form the plane x – 3y + 2z + 13 = 0
Distance \(= \left| {\frac{{1\; \times \;1\; + \left( { - 3} \right)\; \times \;2\; + \;2\; \times \;3\; + \;13}}{{\sqrt {{1^2}\; + \;\;{{\left( { - 3} \right)}^2}\; + \;{{\left( 2 \right)}^2}} }}} \right|\)
\(= \left| {\frac{{1\; - 6\; + \;6\; + \;13}}{{\sqrt {1\; + \;\;9\; + \;4} }}} \right|\)
\(= \;\left| {\frac{14}{{\sqrt {14} }}} \right|\) = √14
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