Question

Distance of the point $(x_1,y_1,z_1)$ from the line $\frac{x-x_2}{l}=\frac{y-y_2}{m}=\frac{z-z_2}{n}$, where $l,m$ and $n$ are the direction cosines of line is

• A. $\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2+(z_1-z_2{)}^2-[l(x_1-x_2)+m(y_1-y_2)+n(z_1-z_2){]}^2}$
• B. $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$
• C. $\sqrt{(x_2-x_1)l+(y_2-y_1)m+(z_2-z_1)n}$
• D. None of the above

Answer 1

The correct option is D None of the above

$P(x_1,y_1,z_1)$

$l=\frac{x-x_2}{l}=\frac{y-y_2}{m}=\frac{z-z_2}{n}=t$

$l,m,n\to$ Direction cosines

$n=\left(\begin{array}{c}l\\ m\\ n\end{array}\right)$

$\pi :n=\left(\begin{array}{c}l\\ m\\ n\end{array}\right)$

$n.r=n.p$

$\left(\begin{array}{c}l\\ m\\ n\end{array}\right)\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}l\\ m\\ n\end{array}\right)\left(\begin{array}{c}{x}_1\\ y_1\\ z_1\end{array}\right)$

$lx+my+nz=\left(d\right)\to l{x}_1+m{y}_1+n{z}_1$

$\hookrightarrow$ equation of $\pi$

$L\cap \pi$

$x=lt+x_2$

$y=mt+y_2$

$z=nt+z_2$

$l(lt+x_2)+m(mt+y_2)+n(nt+z_2)=0$

On solving we get some value for $t$

Which will give us a point $P.$

Distance can be calculated.

$d=\sqrt{{(lt+x_2-x_1)}^2+{(nt+z_2-z_1)}^2+{(mt+y_2-y_1)}^2}$

Hence, the answer is none of the above.