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Question Number 126609 by abdullahquwatan last updated on 22/Dec/20

cube ABCD.EFGH side 4 cm point P is center BF and Q center AD. distance E to PQG

$$\mathrm{cube}\:\mathrm{ABCD}.\mathrm{EFGH}\:{side}\:\mathrm{4}\:{cm}\:{point}\:{P}\:{is}\:{center}\:{BF}\:{and}\:{Q}\:{center}\:{AD}.\:{distance}\:{E}\:{to}\:{PQG} \\ $$

Answered by liberty last updated on 22/Dec/20

the equation of plane passes trought  P(4,4,2),Q(2,0,0),G(0,4,4)  ⇒  determinant (((x−2       y          z)),((−2          4         4)),((    2          4          2)))= 0  ⇒(x−2)(8−16)−y(−4−8)+z(−8−8)=0  ⇒−8(x−2)+12y−16z=0  ⇒−2x+3y−4z+4=0  ⇒2x−3y+4z−4=0  the distance point E(4,0,4) to PQG is   ((∣8−0+16−4∣)/( (√(4+9+16)))) = ((20)/( (√(29)))) cm

$${the}\:{equation}\:{of}\:{plane}\:{passes}\:{trought} \\ $$$${P}\left(\mathrm{4},\mathrm{4},\mathrm{2}\right),{Q}\left(\mathrm{2},\mathrm{0},\mathrm{0}\right),{G}\left(\mathrm{0},\mathrm{4},\mathrm{4}\right) \\ $$$$\Rightarrow\:\begin{vmatrix}{{x}−\mathrm{2}\:\:\:\:\:\:\:{y}\:\:\:\:\:\:\:\:\:\:{z}}\\{−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\mathrm{4}}\\{\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:\mathrm{4}\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}=\:\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{2}\right)\left(\mathrm{8}−\mathrm{16}\right)−{y}\left(−\mathrm{4}−\mathrm{8}\right)+{z}\left(−\mathrm{8}−\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow−\mathrm{8}\left({x}−\mathrm{2}\right)+\mathrm{12}{y}−\mathrm{16}{z}=\mathrm{0} \\ $$$$\Rightarrow−\mathrm{2}{x}+\mathrm{3}{y}−\mathrm{4}{z}+\mathrm{4}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{x}−\mathrm{3}{y}+\mathrm{4}{z}−\mathrm{4}=\mathrm{0} \\ $$$${the}\:{distance}\:{point}\:{E}\left(\mathrm{4},\mathrm{0},\mathrm{4}\right)\:{to}\:{PQG}\:{is} \\ $$$$\:\frac{\mid\mathrm{8}−\mathrm{0}+\mathrm{16}−\mathrm{4}\mid}{\:\sqrt{\mathrm{4}+\mathrm{9}+\mathrm{16}}}\:=\:\frac{\mathrm{20}}{\:\sqrt{\mathrm{29}}}\:{cm} \\ $$

Commented by abdullahquwatan last updated on 22/Dec/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

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