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Mathematics Question on Three Dimensional Geometry

If the points (1,2,3)(1, 2, 3) and (2,−1,0)(2, -1, 0) lie on the opposite sides of the plane 2x+3y−2z=k,2x + 3y - 2z = k, then

A

k<1k < 1

B

k>2 k > 2

C

k<1k < 1 or k>k >

D

1<k<21 < k < 2

Answer

1<k<21 < k < 2

Explanation

Solution

Since, The points (1,2,3)(1, 2, 3) and (2,−1,0)(2, -1, 0) lie on the opposite sides of the plane 2x+3y−2z−k=02x + 3y - 2z - k = 0
So, (2+6−6−k)(4−3−k)<0(2 + 6 - 6 - k) (4 - 3 - k) < 0
⇒(k−1)(k−2)<0\Rightarrow (k - 1) (k - 2) < 0 ... (i)
∴1<k<2\therefore 1 < k < 2