question_answer

The length and foot of the perpendicular from the point       (7, 14, 5) to the plane \[2x+4y-z=2,\]are [AISSE 1987]

A)            \[\sqrt{21},(1,\,2,\,8)\]

B)            \[3\sqrt{21},(3,\,2,\,8)\]

C)            \[21\sqrt{3},(1,\,2,\,8)\]

D)            \[3\sqrt{21},(1,\,2,\,8)\]

Correct Answer: D

Solution :

            Let M be the foot of perpendicular from (7, 14, 5) to the given plane, then PM is normal to the plane. So, its d.r.'s are 2, 4, ?1. Since PM passes through \[P(7,14,5)\] and has d.r.'s 2, 4, ?1.                    Therefore, its equation is          \[\frac{x-7}{2}=\frac{y-14}{4}=\frac{z-5}{-1}=r\], (Say)                    Þ \[x=2r+7\], \[y=4r+14\], \[z=-r+5\]                    Let co-ordinates of M be \[(2r+7,\,4r+14,\,-r+5)\]                    Since M lies on the plane \[2x+4y-z=2\], therefore \[2(2r+7)+4(4r+14)-(-r+5)=2\] Þ \[r=-3\]                    So, co-ordinates of foot of perpependicular are \[M(1,\,2,8)\]                    Now, PM = Length of perpendicular from P                                                                  = \[\sqrt{{{(1-7)}^{2}}+{{(2-14)}^{2}}+{{(8-5)}^{2}}}=3\sqrt{21}.\]