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October 20, 2014

October 20, 2014

Posted by **sophie** on Saturday, May 21, 2011 at 4:31pm.

- Math -
**Reiny**, Saturday, May 21, 2011 at 5:52pmPick 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

- Math -
**bobpursley**, Saturday, May 21, 2011 at 6:11pmsimple. 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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