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Find the distance of the point $P(-1,-5,-10)$ from the point of intersection of the line joining the points $A(2,-1,2)$ and $B(5,3,4)$with the plane $x-y+z=5$


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Given: $A(2,-1,2)$, $B(5,3,4)$
Direction ratio of the line $AB=(5-2,3+1,4-2)$=(3,4,2)$
Equation of the line $AB$ is given by
Any point on this line is given in terms of $\lambda$ by
Let the line $AB$ intersect the plane $x-y+z=5$ at this point $C$
$\Rightarrow\:$ $C$ lies on the plane too.
$\therefore$ $C$ satisfies the equation of the plane.$
Step 2
We have to find the distance of the point $P(-1,-5,-10)$ from $C(2,-1,2)$
which is given by $\sqrt {2+1)^2+(-1+5)^2+(2+10)^2}=\sqrt {9+16+144}=13$
answered Mar 21, 2014 by rvidyagovindarajan_1

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