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Mathematics Question on introduction to three dimensional geometry

Perpendicular are drawn from points on the line x+22=y+1−1=z3 \frac{x+2}{2}=\frac{y+1}{-1}= \frac{z}{3} to the plane x+y+z=3x+ y + z = 3. The feet of perpendiculars lie on the line

A

x5=y−18=z−2−13 \frac{x}{5} = \frac{y-1}{8}=\frac{z-2}{-13}

B

x2=y−13=z−2−5 \frac{x}{2} = \frac{y-1}{3}=\frac{z-2}{-5}

C

x4=y−13=z−2−7 \frac{x}{4} = \frac{y-1}{3}=\frac{z-2}{-7}

D

x2=y−1−7=z−25 \frac{x}{2} = \frac{y-1}{-7}=\frac{z-2}{5}

Answer

x2=y−1−7=z−25 \frac{x}{2} = \frac{y-1}{-7}=\frac{z-2}{5}

Explanation

Solution

The correct answer is D:x2=y−1−7=z−25\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 (x1,y1,z1)(x_1,y_1,z_1) to
        \, \, \, \, \, \, \, \, ax+by+cz+d=0 be (x2,y2,z2)ax+by+cz+d = 0\, be \, (x_2,y_2,z_2) then
  25mmx2−x1a=y2−y1b=z2−z1c\space25mm \frac{x_2-x_1}{a} = \frac{y_2-y_1}{b} = \frac{z_2-z_1}{c}
  35mm=−(ax1+by1+cz1+d)a2+b2+c2\space35mm = \frac{-(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}
Any point on x+22=y+1−1=z3=λ\frac{x+2}{2} = \frac{y+1}{-1} =\frac{z}{3} = \lambda
⇒\Rightarrow   30mmx=2λ−2,y=−λ−1,z=3λ\space30mm x = 2 \lambda - 2, y = - \lambda - 1, z = 3 \lambda
Let foot of perpendicular from (2λ−2,−λ−1,3λ)(2 \lambda -2,-\lambda-1,3\lambda)
to x+ y + z = 3 be (x2,y2,z2)(x_2,y_2,z_2).
∴\therefore    x2−(2λ−2)1=y2−(−λ−1)1=z2−(3λ)1\, \, \, \frac{x_2-(2 \lambda-2)}{1} = \frac{y_2-(- \lambda-1)}{1} =\frac{z_2-(3 \lambda)}{1}
  30mm=(2λ−2−λ−1+3λ−3)1+1+1\space30mm =\frac{(2 \lambda-2 - \lambda - 1 + 3\lambda-3 )}{1+1+1}
⇒    x2−2λ+2=y2+λ+1=z2−3λ=2−4λ3\Rightarrow \, \, \, \, x_2 - 2 \lambda + 2= y_2 + \lambda + 1 = z_2 - 3\lambda = 2 -\frac{4\lambda}{3}
∴\therefore   20mmx2=2λ3,y2=1−7λ3,+2=z2=2+5λ3\space20mm x_2 =\frac{2 \lambda}{3}, y_2 = 1 - \frac{7 \lambda}{3}, + 2= z_2 = 2 + \frac{5 \lambda}{3}
⇒\Rightarrow   20mmλ=x2−02/3,=1−y2−1−7/3,=+z2−25/3\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
=x2/3=y−1−7/3=z−25/3   ⇒  x2=y−1−7=z−25=\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}