Answer the following question:

Asked by nkrbasak | 15th Mar, 2013, 09:13: PM

Expert Answer:

By reflection of the point in a plane is meant a point which is at the same distance from the plane but on the other side of the plane and the line joining the point and its reflection is perpendicular to the plane.

Normal to the plane 2x + 4y - z = 2 is (2,4,-1)
So, the relection of the point (7,14,5) in the plane 2x + 4y - z = 2 lies on the line
(x - 7)/2 = (y - 14)/4 = (z -5)/-1 = k, k ? R.
So, x = 7 + 2k,
y = 14 + 4k and
z = 5 -k

Let for some k, it represent the foot of perpendicular from the point (7.14.5) to the plane
2x + 4y - z = 2
2(7+2k) + 4(14+4k) -(5-k) = 2
4k +14 +56 +16k -5+k = 2
20k = -63
k = -63/20
 
=> foot of perpendicular is (7/10, 7/5, 163/20)

If (a, b, c) is the reflection of the point (7,14,5), the above foot of perpendicular is its midpoint.
 
=> (a+7)/2 = 7/10, (b+14)/2 = 7/5 and (c+5)/2 = 163/20
=> a = - 28/5, b = -56/5 and c = 113/10
=> the image of the point is
(-28/5, -56/5, 113/20)
 
Since you are appearing for board exams, you should not leave any topic. Generally the ques which are not similar to those given in ncert are not asked. But you never know, the person who sets the paper might ask some diff ques too.

Answered by  | 17th Mar, 2013, 07:14: PM

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