11.  Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

Answers (1)
D Divya Prakash Singh

We know that the equation of the line that passes through the points (x_{1},y_{1},z_{1})and (x_{2},y_{2},z_{2}) is given by the relation;

\frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}

and the line passing through the points, \frac{x-5}{3-5} =\frac{y-1}{4-1} = \frac{z-6}{1-6}

\implies \frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} = k\ (say)

\implies x=5-2k,\ y=3k+1,\ z=6-5k  

And any point on the line is of the form (5-2k,3k+1,6-5k).

So, the equation of ZX plane is y=0

Since the line passes through YZ- plane,

we have then,

3k+1 = 0

\Rightarrow k =-\frac{1}{3}

or  5-2k = 5-2\left ( -\frac{1}{3} \right ) = \frac{17}{3}   and  6-5k= 6-5(\frac{-1}{3}) = \frac{23}{3}

So, therefore the required point is \left ( \frac{17}{3},0,\frac{23}{3} \right ).