Consider the line which passes through the point P(-2, 2, -1), and which is parallel to the line x=1+7t, y=2+5t, z=3+6t. How do you find the point of intersection of this new line with each of the coordinate planes?

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Answer 1 · 2017-04-26T17:21:56

$L nn XY"-plane"={(-5/6,17/6,0)}.$

$L nn YZ"-plane"={(0,24/7,5/7)}, and,$

$L nn ZX"-plane"={(-24/5,0,-17/5)}.$

Observe that the direction vector $vecl$ of the given line is $(7,5,6).$

The reqd. line, say $L,$ is parallel to the given line, so, the

direction vector of $L$ can be taken as $vecl$ .

Further, $P(-2,2,-1) in L.$ Hence, its cartesian eqn. is,

$:. L : (x-(-2))/7=(y-2)/5=(z-(-1))/6=k, k in RR, or,$

$L : x=7k-2, y=5k+2, z=6k-1, k in RR.......(ast).$

Recall that the eqns. of the co-ordinate planes, namely, the $XY,$

$YZ and ZX$ planes are, resp.,

$z=0, x=0, and, y=0,.....(star)$

To find the points of intersection of $L$ with these planes, we need

solve the eqns. $(ast) and (star).$

$z=6k-1=0 rArr k=1/6 rArr x=7k-2=-5/6, y=17/6.$

$rArr L nn XY"-plane"={(-5/6,17/6,0)}.$

Similarly, $L nn YZ"-plane"={(0,24/7,5/7)}, and,$

$L nn ZX"-plane"={(-24/5,0,-17/5)}.$

Enjoy Maths.!