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Three-Dimensional Geometry: Distance and Coordinate Problems
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The distance of origin from the image of (1, 2, 3) in plane x - y + z = 5 is
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A
$\sqrt{34}$
💡 Explanation:
$P(1,2,3),$ Plane :$x-y+z=5$
$F$ is foot of perpendicular form $P$ to plane and $I$ is image,then $PF=FI$
$\therefore$ If $(x,y,z)=(r+1,-r+2,r+3)$ are foot of perpendicular.
$ \Rightarrow (r+1)-(-r+2)+r+3=5\quad \quad \Rightarrow r=1\ \therefore F=(2,1,4)\ \therefore I=(3,0,5)$
$ \therefore$ distance of $I$ from origin $=\sqrt { { 3 }^{ 2 }+{ 0 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 34 } $