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Find-the-perpendicular-distance-from-1-7-1-to-3x-2y-2z-6-




Question Number 56696 by Tawa1 last updated on 21/Mar/19
Find the perpendicular distance from  (1, 7, 1)  to  3x − 2y + 2z  =  6
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{distance}\:\mathrm{from}\:\:\left(\mathrm{1},\:\mathrm{7},\:\mathrm{1}\right)\:\:\mathrm{to}\:\:\mathrm{3x}\:−\:\mathrm{2y}\:+\:\mathrm{2z}\:\:=\:\:\mathrm{6} \\ $$
Commented by mr W last updated on 21/Mar/19
d=((∣3×1−2×7+2×1−6∣)/( (√(3^2 +(−2)^2 +2^2 ))))=((15)/( (√(17))))
$${d}=\frac{\mid\mathrm{3}×\mathrm{1}−\mathrm{2}×\mathrm{7}+\mathrm{2}×\mathrm{1}−\mathrm{6}\mid}{\:\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }}=\frac{\mathrm{15}}{\:\sqrt{\mathrm{17}}} \\ $$
Commented by Tawa1 last updated on 22/Mar/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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