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Previous Year Question (PYQs)



The distance between the point $(3,4,5)$ and the point where the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ meets the plane $x+y+z=17$ is:





Solution

Line in parametric form Let parameter $t$ $x=3+t$ $y=4+2t$ $z=5+2t$ Substitute in plane $x+y+z=17$ $(3+t)+(4+2t)+(5+2t)=17$ $12+5t=17$ $t=1$ Point of intersection $x=4$ $y=6$ $z=7$ Distance from $(3,4,5)$ $d=\sqrt{(4-3)^2+(6-4)^2+(7-5)^2}$ $d=\sqrt{1+4+4}$ $d=3$


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