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Question Number 131733 by LYKA last updated on 07/Feb/21
find the minimum distance  between the point (1,1,1) and  the plane x+2y+3z=6
findtheminimumdistancebetweenthepoint(1,1,1)andtheplanex+2y+3z=6
Answered by physicstutes last updated on 07/Feb/21
 d = ∣((ax_1 +by_1  + cz_1 −D)/( (√(a^2 +b^2 +c^2 ))))∣  ⇒ d = ∣(((1)(1) +2(1) +3(1)−6)/( (√(1^2 +2^2 +3^2 ))))∣ = 0  since the point in question (1,1,1) lies on the plane  hence it has a minimum distance of 0 units from the   given plane.
d=ax1+by1+cz1Da2+b2+c2d=(1)(1)+2(1)+3(1)612+22+32=0sincethepointinquestion(1,1,1)liesontheplanehenceithasaminimumdistanceof0unitsfromthegivenplane.

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