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  • The distance of the point (1, -2, 3) from the plane  measured parallel to the line 

The distance of the point (1, -2, 3) from the plane x-y+z=5 measured parallel to the line \frac{x}{2}=\frac{y}{3}=\frac{z}{-6} is:
Option: 1 \frac{7}{5}
Option: 2 1
Option: 3 \frac{1}{7}
Option: 4 7

Answers (1)

best_answer

\text { equation of line parallel to } \frac{x}{2}=\frac{y}{3}=\frac{z}{-6} \text { passes through }(1,-2,3) \text { is }

\\\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=r \\ x=2 r+1 \\ y=3 r-2 \\ z=-6 r+3 \\ \text { so } 2 r+1-3 r+2-6 r+3=5 \\ r=\frac{1}{7}

\\\mathrm{x}=\frac{9}{7}, \mathrm{y}=\frac{-11}{7}, \mathrm{z}=\frac{15}{7} \\ \text { Distance is }=\sqrt{\left(\frac{9}{7}-1\right)^{2}+\left(2-\frac{11}{7}\right)^{2}+\left(3-\frac{15}{7}\right)^{2}} \\ =\sqrt{\left(\frac{2}{7}\right)^{2}+\left(\frac{3}{7}\right)^{2}+\left(\frac{6}{7}\right)^{2}} \\ =\frac{1}{7} \sqrt{4+9+36}=1

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himanshu.meshram

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