Question

The foot of the perpendicular from the origin to the join of $A(-9,4,5)$ and $(11,0,-1)$ is

• A. a point of trisection of AB
• B. A point dividing $\overline{AB}$ in the ratio $2=$
• C. The midpoint of AB
• D. $(6,1,1/2)$

Answer 1

The correct option is C The midpoint of AB

From fig, Let $A(x_1,y_1,z_1)$, $M(x,y,z)$ and $B(x_2,y_2,z_2)$

We have the equation for the line,

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

$⟹\frac{x+9}{20}=\frac{y-4}{-4}=\frac{z-5}{6}$

$⟹\frac{x+9}{-10}=\frac{y-4}{2}=\frac{z-5}{3}$

Say,

$⟹\frac{x+9}{-10}=\frac{y-4}{2}=\frac{z-5}{3}=r$

The direction ratio of line AB is, $(a_1,b_1,c_1)=(-10,2,3)$

$x=-10r-9$

$y=2r+4$

$z=3r+5$

The direction ratio of line OM is,

$(a_2,b_2,c_2)=(x-0,y-0,z-9)$

$(a_2,b_2,c_2)=(-10r-9,2r+4,3r+5)$

If $OM$ and $AB$ are perpendicular the,

$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$-10(-10r-9)+2(2r+4)+3(3r+5)=0$

$⟹r=-1$

$\therefore M(x,y,z)=(-10r-9,2r+4,3r+5)=\left(10\right(-1)-9,2(-1)+4,3(-1)+5)=(1,2,2)$

Now, from distance formula, to find the mid point of $AB$,

$(\frac{x_2+x_1}{2},\frac{y_2+y_1}{2},\frac{z_2-z_1}{2})=(\frac{11-9}{2},\frac{4+0}{2},\frac{-1+5}{2})=(1,2,2)$

This is equal to M. Hence the foot of the perpendicular from the origin to the join of $A,B$ is the midpoint of $AB$

Option C