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Question Number 9453 by Raja Naik last updated on 09/Dec/16

find dc′s dr′s of a mormal to the plane 2x+(5/2)y+(7/8)z=23

$$\mathrm{find}\:\mathrm{dc}'\mathrm{s}\:\mathrm{dr}'\mathrm{s}\:\mathrm{of}\:\mathrm{a}\:\mathrm{mormal}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{plane}\:\mathrm{2x}+\frac{\mathrm{5}}{\mathrm{2}}\mathrm{y}+\frac{\mathrm{7}}{\mathrm{8}}\mathrm{z}=\mathrm{23} \\ $$

Answered by mrW last updated on 10/Dec/16

d.r.′s of the normal are (2,(5/2),(7/8)) (√(2^2 +((5/2))^2 +((7/8))^2 ))=((√(705))/8) 2×(8/(√(705)))=((16)/(√(705))) (5/2)×(8/(√(705)))=((20)/(√(705))) (7/8)×(8/(√(705)))=(7/(√(705))) d.c.′s of the normal are (((16)/(√(705))),((20)/(√(705))),(7/(√(705))))

$$\mathrm{d}.\mathrm{r}.'\mathrm{s}\:\mathrm{of}\:\mathrm{the}\:\mathrm{normal}\:\mathrm{are}\:\left(\mathrm{2},\frac{\mathrm{5}}{\mathrm{2}},\frac{\mathrm{7}}{\mathrm{8}}\right) \\ $$$$\sqrt{\mathrm{2}^{\mathrm{2}} +\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{7}}{\mathrm{8}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{705}}}{\mathrm{8}} \\ $$$$\mathrm{2}×\frac{\mathrm{8}}{\sqrt{\mathrm{705}}}=\frac{\mathrm{16}}{\sqrt{\mathrm{705}}} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}×\frac{\mathrm{8}}{\sqrt{\mathrm{705}}}=\frac{\mathrm{20}}{\sqrt{\mathrm{705}}} \\ $$$$\frac{\mathrm{7}}{\mathrm{8}}×\frac{\mathrm{8}}{\sqrt{\mathrm{705}}}=\frac{\mathrm{7}}{\sqrt{\mathrm{705}}} \\ $$$$\mathrm{d}.\mathrm{c}.'\mathrm{s}\:\mathrm{of}\:\mathrm{the}\:\mathrm{normal}\:\mathrm{are}\:\left(\frac{\mathrm{16}}{\sqrt{\mathrm{705}}},\frac{\mathrm{20}}{\sqrt{\mathrm{705}}},\frac{\mathrm{7}}{\sqrt{\mathrm{705}}}\right) \\ $$

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