question_answer

The distance of the point (3,8,2) from the line  \[\frac{\pi }{6}\]measured parallel to the plane \[\frac{\pi }{2}\] is

A)  2                            

B) 3

C)  6                            

D) 7

Correct Answer: D

Solution :

Let P (3, 8, 2) be the given point. Let the equation of a line through P (3, 8, 2) and parallel to the plane \[3x+2y-2z+15=0\] be \[\frac{x-3}{a}=\frac{y-8}{b}=\frac{z-2}{c}\]                                            ?(i) Then, normal to the plane is perpendicular to the parallel line. Then,       \[3a+2b-2c=0\]                               ?(ii) Line (i) intersects the line \[\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}\]at  Q. \[\therefore \]\[\left| \begin{matrix}    3-1 & 8-3 & 2-2  \\    a & b & c  \\    2 & 4 & 3  \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[\left| \begin{matrix}    2 & 5 & 0  \\    a & b & c  \\    2 & 4 & 3  \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[2(3b-4c)-5(3a-2c)=0\] \[\Rightarrow \]\[15a-6b-2c=0\]                                               ?(iii) On solving Eqs. (ii) and (iii) by cross-multiplication, we get \[\frac{a}{-4-12}=\frac{b}{-30+6}=\frac{c}{-18-30}\] \[\Rightarrow \]\[\frac{a}{-16}=\frac{b}{-24}=\frac{c}{-48}\Rightarrow \frac{a}{2}=\frac{b}{3}=\frac{c}{6}\] Substituting a, b and c in Eq. (i), we get \[\frac{x-3}{2}=\frac{y-8}{3}=\frac{z-2}{6}=\lambda \][say] Let    the    coordinates    of    Q    be \[(2\lambda +3,3\lambda +8,6\lambda +2).\]It lies on the line \[\frac{x-1}{2}=\frac{y-3}{4}=\frac{z-2}{3}\] \[\therefore \]\[\lambda +1=\frac{3\lambda +5}{4}=2\lambda \]\[\Rightarrow \]\[\lambda =1\] \[\therefore \]Coordinate of Q =(2 + 3, 3 + 8, 6 + 2) i.e. Q = (5,11,8). Hence, \[PQ=\sqrt{{{(5-3)}^{2}}+{{(11-8)}^{2}}+{{(8-2)}^{2}}}\] \[=\sqrt{4+9+36}=7\]