Question

Perpendicular are drawn from points on the line $ \frac{x+2}{2}=\frac{y+1}{-1}= \frac{z}{3}$ to the plane $x+ y + z = 3$. The feet of perpendiculars lie on the line

• A. $ \frac{x}{5} = \frac{y-1}{8}=\frac{z-2}{-13}$
• B. $ \frac{x}{2} = \frac{y-1}{3}=\frac{z-2}{-5}$
• C. $ \frac{x}{4} = \frac{y-1}{3}=\frac{z-2}{-7}$
• D. $ \frac{x}{2} = \frac{y-1}{-7}=\frac{z-2}{5}$

Answer 1

The Correct Option is D

The correct answer is D:\(\frac{x}{2} = \frac{y-1}{-7}=\frac{z-2}{5}\)
PLAN To find the foot of perpendiculars and find its locus. 
 Foot of perpendicular from \((x_1,y_1,z_1)\) to 
\(\, \, \, \, \, \, \, \,\) \(ax+by+cz+d = 0\, be \, (x_2,y_2,z_2)\) then 
\(\space25mm \frac{x_2-x_1}{a} = \frac{y_2-y_1}{b} = \frac{z_2-z_1}{c}\)
\(\space35mm = \frac{-(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}\) 
Any point on \(\frac{x+2}{2} = \frac{y+1}{-1} =\frac{z}{3} = \lambda\)
\(\Rightarrow\) \(\space30mm x = 2 \lambda - 2, y = - \lambda - 1, z = 3 \lambda\) 
Let foot of perpendicular from \((2 \lambda -2,-\lambda-1,3\lambda)\) 
to x+ y + z = 3 be \((x_2,y_2,z_2)\). 
\(\therefore\) \(\, \, \, \frac{x_2-(2 \lambda-2)}{1} = \frac{y_2-(- \lambda-1)}{1} =\frac{z_2-(3 \lambda)}{1}\)
\(\space30mm =\frac{(2 \lambda-2 - \lambda - 1 + 3\lambda-3 )}{1+1+1}\)
\(\Rightarrow \, \, \, \, x_2 - 2 \lambda + 2= y_2 + \lambda + 1 = z_2 - 3\lambda = 2 -\frac{4\lambda}{3}\)
\(\therefore\) \(\space20mm x_2 =\frac{2 \lambda}{3}, y_2 = 1 - \frac{7 \lambda}{3}, + 2= z_2 = 2 + \frac{5 \lambda}{3}\)
\(\Rightarrow\) \(\space20mm \lambda =\frac{x_2 - 0 }{2/3}, = 1 - \frac{y_2-1}{-7/3}, = + \frac{z_2 - 2}{5/3}\) 
Hence, foot of perpendicular lie on 
\(=\frac{x }{2/3} = \frac{y-1}{-7/3} = \frac{z - 2}{5/3} \, \, \, \Rightarrow \, \, \frac{x }{2}= \frac{y-1}{-7} = \frac{z - 2}{5}\)