Question

If distance of $(-1,2,3)$ from x, y, z axis are $d_1,d_2,d_3$ respectively and the distance from xy, yz, zx, planes are $d_4,d_5,d_6$ then the value of${\sum }_{i=1}^6{d}_i$ is

• A. $\sqrt{5}+\sqrt{13}+\sqrt{10}+6$
• B. $\sqrt{2}+\sqrt{3}+6$
• C. $\sqrt{5}+\sqrt{13}+7$
• D. $\sqrt{4}+\sqrt{13}+7$

Answer 1

The correct option is A $\sqrt{5}+\sqrt{13}+\sqrt{10}+6$

Equation of,

x-axis$\Rightarrow \frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}$

y-axis$\Rightarrow \frac{x}{0}=\frac{y}{1}=\frac{z}{0}$

z-axis$\Rightarrow \frac{x}{0}=\frac{y}{0}=\frac{z}{1}$

$xy$ plane $\Rightarrow z=0$, $yz$ plane $\Rightarrow x=0$ ,$zx$ plane $\Rightarrow y=0$

if,$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$ is equation of line.

$(x_2,y_2,z_2)$ is a point through which $\perp$ is drawn on the line.

Let $F$ be foot of $\perp$ it will be $(lr+x,mr+y,nr+z)$

$(lr+x_1-x_2,mr+y_1-y_2+nr+z_1-z_2)$ is $\perp$ to $(l,m,n)$

$\therefore (lr+x_1-x_2)l+(mr+y_1-y_2)m+(nr+z_1-z_2)n=0\therefore r=\frac{(x_2-x_1)l+(y_2-y_1)m+(z_2-z_1)n}{l^2+m^2+n^2}$

$\therefore$ For x-axis,

$(x_1,y_1,z_1)=(0,0,0)(l,m,n)=(1,0,0)(x_2,y_2,z_2)=(-1,2,3)\therefore r=-1$

$\therefore$ foot of perpendicular $=(-1,0,0)$

Similarly,

for y-axis foot of $\perp$ $=(0,2,0)$

for x-axis foot of $\perp$ $=(0,0,3)$

$\therefore d_1=\sqrt{{(1+1)}^2+2^2+3^2}=\sqrt{13}\therefore d_2=\sqrt{10}{d}_3=\sqrt{5}{d}_4=3d_5=1d_6=2$

($d_1,d_2,d_3$ are distance between point pand feets of $\perp$).

$\therefore d_1+d_2+d_3+d_4+d_5+d_6=\sqrt{5}+\sqrt{13}+\sqrt{10}+6$