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Question

Mathematics Question on Three Dimensional Geometry

What will be the distance of (1,0,2)(1, 0, 2) from the point of intersection of plane xy+z=16x - y + z = 16 and the line (x23)=(y+14)=(z212)\left(\frac{x-2}{3}\right) = \left(\frac{y+1}{4}\right) = \left(\frac{z-2}{12}\right) ?

A

1313 units

B

1717 units

C

2525 units

D

1919 units

Answer

1313 units

Explanation

Solution

Given line is x23\frac{x-2}{3}
=y+14=\frac{y+1}{4}
=z212=r=\frac{z-2}{12}=r(say) (i) \dots(i)
Any point on (i) is P(3r+2,4r1,12r+2)P(3r+2, 4r-1, 12r+2)
The point P lies on xy+z=16x - y + z = 16 (ii)\dots (ii)
3r+2(4r1)+12r+2=16\therefore 3r+2-(4r-1)+12r+2=16
11r=11\Rightarrow 11 r=11
r=1\Rightarrow r=1
Thus, the line cuts the plane at P(5,3,14)P(5, 3, 14).
The distance of Q(1,0,2)Q(1, 0,2) from P(5,3,14)P (5,3,14) is given by
PQ=(51)2+(30)2+(142)2PQ=\sqrt{\left(5-1\right)^{2}+\left(3-0\right)^{2}+\left(14-2\right)^{2}}
=16+9+144=\sqrt{16+9+144}
=13=13 units