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Q.
The equation of straight line passing through the point $(a, b, c)$ and parallel to $Z$-axis is
Three Dimensional Geometry
Solution:
The equation of line passing through $(a, b, c)$ is
$\frac{x-a}{1}=\frac{y-b}{m}=\frac{z-c}{n}=\lambda$
Since, the line is parallel to $Z$-axis, therefore any point on this line will be of the form $\left(a, b, z_1\right)$.
Now, from Eq. (i) any point on the line is $(\lambda+a, \lambda m+b, \lambda n+c)$
$ \therefore \lambda+a=a \rightarrow 1=0, \lambda=0 $
$ \rightarrow \lambda m+b=b \rightarrow m=0, \lambda=0$
$ \rightarrow \lambda n+c=z_1 \rightarrow z_1=c (\because \lambda=0) $
$\rightarrow 1^2+m^2+n^2=1 $
$\rightarrow 0+0+n^2=1$
$\rightarrow n=1$
Hence, the equation of line is
$\frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1}$