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Prove-that-points-4-1-3-amp-5-1-4-lies-on-same-side-of-the-plane-x-y-z-7-




Question Number 45555 by rahul 19 last updated on 14/Oct/18
Prove that points (4,−1,3) & (5,−1,4)  lies on same side of the plane x+y+z=7.
$${Prove}\:{that}\:{points}\:\left(\mathrm{4},−\mathrm{1},\mathrm{3}\right)\:\&\:\left(\mathrm{5},−\mathrm{1},\mathrm{4}\right) \\ $$$${lies}\:{on}\:{same}\:{side}\:{of}\:{the}\:{plane}\:{x}+{y}+{z}=\mathrm{7}. \\ $$
Commented by rahul 19 last updated on 14/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18
4−1+3−7=−1  5−1+4−7=+1  opposite side  mid point of (4,−1,3) and(5,−1,4)  ((4+5)/2),((−1−1)/2),((3+7)/2)  ((9/2),−1,(7/2))  now put ((9/2),−1,(7/2)) in x+y+z−7=0  (9/2)−1+(7/2)−7  =8−8  =0   that confirm points are opposite to plsnes  and are mirrir image...
$$\mathrm{4}−\mathrm{1}+\mathrm{3}−\mathrm{7}=−\mathrm{1} \\ $$$$\mathrm{5}−\mathrm{1}+\mathrm{4}−\mathrm{7}=+\mathrm{1} \\ $$$${opposite}\:{side} \\ $$$${mid}\:{point}\:{of}\:\left(\mathrm{4},−\mathrm{1},\mathrm{3}\right)\:{and}\left(\mathrm{5},−\mathrm{1},\mathrm{4}\right) \\ $$$$\frac{\mathrm{4}+\mathrm{5}}{\mathrm{2}},\frac{−\mathrm{1}−\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}+\mathrm{7}}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{9}}{\mathrm{2}},−\mathrm{1},\frac{\mathrm{7}}{\mathrm{2}}\right) \\ $$$${now}\:{put}\:\left(\frac{\mathrm{9}}{\mathrm{2}},−\mathrm{1},\frac{\mathrm{7}}{\mathrm{2}}\right)\:{in}\:{x}+{y}+{z}−\mathrm{7}=\mathrm{0} \\ $$$$\frac{\mathrm{9}}{\mathrm{2}}−\mathrm{1}+\frac{\mathrm{7}}{\mathrm{2}}−\mathrm{7} \\ $$$$=\mathrm{8}−\mathrm{8} \\ $$$$=\mathrm{0}\:\:\:{that}\:{confirm}\:{points}\:{are}\:{opposite}\:{to}\:{plsnes} \\ $$$${and}\:{are}\:{mirrir}\:{image}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18
yes question is wrong...
$${yes}\:{question}\:{is}\:{wrong}… \\ $$

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