Find the length of the perpendicular drawn from the origin to the plane 2x-3y+6z+21=0
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ans for this question is 3 units
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Given a place 2X+3Y+6Z+21=02X+3Y+6Z+21=0.
The distance of a point P(x1,y1,z1)P(x1,y1,z1) from a plane Ax+BY+CZ+D=0Ax+BY+CZ+D=0 is given by Ax1+By1+Cz1+DA2+B2+C2−−−−−−−−−−√Ax1+By1+Cz1+DA2+B2+C2
We need to calculate the distance from the origin P(0,0,0)P(0,0,0) to the place.
⇒⇒ Distance =2(0)−3(0)+6(0)+2122+32+62−−−−−−−−√=2(0)−3(0)+6(0)+2122+32+62
⇒⇒ Distance =2149−−√=217=3=2149=217=3 units.
Answer = 3 unit..
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The distance of a point P(x1,y1,z1)P(x1,y1,z1) from a plane Ax+BY+CZ+D=0Ax+BY+CZ+D=0 is given by Ax1+By1+Cz1+DA2+B2+C2−−−−−−−−−−√Ax1+By1+Cz1+DA2+B2+C2
We need to calculate the distance from the origin P(0,0,0)P(0,0,0) to the place.
⇒⇒ Distance =2(0)−3(0)+6(0)+2122+32+62−−−−−−−−√=2(0)−3(0)+6(0)+2122+32+62
⇒⇒ Distance =2149−−√=217=3=2149=217=3 units.
Answer = 3 unit..
Hope this question's answer are helpful
And press the point of thanks
And please bainalest
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