1 ч. - 1
* *
1 м = 0,
= 0, prove that
(1+1)* dy
+ 1 = 0
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Answer:
We know that the distance of the point (x
1
,y
1
,z
1
) from the plane ax+by+cz+d=0 is given by
a
2
+b
2
+c
2
∣ax
1
+by
1
+cz
1
+d∣
Distance of the point (1,1,1) from the plane 3x+4y−12z+13=0
The required distance =
(3)
2
+(4)
2
+(−12)
2
∣3(1)+4(1)−12(1)+13∣
=
9+16+144
∣3+4−12+13∣
=
13
8
units ---- ( 1 )
Distance of the point (−3,0,1) from the plane 3x+4y−12z+13=0
The required distance =
(3)
2
+(4)
2
+(−12)
2
∣3(−3)+4(0)−12(1)+13∣
=
9+16+144
∣−9+0−12+13∣
=
13
8
units ---- ( 2 )
So, from ( 1 ) and ( 2 ) we can say that the given points are equidistant from the given plane.
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