Posted by sophie on Saturday, May 21, 2011 at 4:31pm.
Pick any point on the first plane, say,
(0,13,0)
distance from that point to
2x + y - 3z - 35 = 0
= |2(0) + 13 - 3(0) - 35|/√(2^2 + 1^2 + (-3)^2)
= 22/√14 or after rationalizing , 11√14/7
simple. Find a point on the first plane. When x and z are zero, y=13.
Now we find the distance between the point (0,13,0) and the second plane.
d= (2*0+1*13-3*0 -35)/sqrt(4+1+9)
d= 13/sqrt(14)
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