Find the length and the foot of the perpendicular from the point `(7,\ 14 ,\ 5)` to the plane `2x+4y-z=2.` Also, the find image of the point `P` in the plane.

Explanation

The required length `=(2(7)+4(14)-(5)-2)/(sqrt(2^(2)+4^(2)+1^(2)))=(14+56-5-2)/(sqrt(4+16+1))=(63)/(sqrt(21))`
Let the coordinates of the foot of the perpendicular from the point `P(7, 14, 5)` be `M(alpha, beta, gamma)`.
Then the direction ratios of PM are `alpha-7, beta-14 and gamma-5`.
Therefore, the direction ratios of the normal to the plane are `alpha-7, beta-14, and gamma-5`.
But the direction ratios of normal to the given plane `2x+4y-z=2` are 2, 4 and -1. Hence,
`" "(alpha-7)/(2)=(beta-14)/(4)=(gamma-5)/(-1)=k`
`therefore" "alpha=2k+7, beta=4k+14 and gamma=-k+5" "`(i)
Since`alpha, beta and gamma` lie on the plane `2x+4y-z=2, 2alpha+4beta-gamma=2`
or `" "2(7+2k)+4(14+4k)-(5-k)=2`
or `" "14+4k+56+16k-5+k=2`
or `" "21k=-63`
or `" " k=-3`
Now, putting `k=-3` in (i), we get
`" "alpha=1, beta=2, gamma=8`
Hence, the foot of the perpendicular is (1, 2, 8).