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

Answers (1)

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 the YZ plane is x=0

Since the line passes through YZ- plane,

we have then,

5-2k = 0

\Rightarrow k =\frac{5}{2}

or  3k+1 = 3(\frac{5}{2})+1 = \frac{17}{2}   and  6-5k= 6-5(\frac{5}{2}) = \frac{-13}{2}

So, therefore the required point is \left ( 0,\frac{17}{2},\frac{-13}{2} \right )

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