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.!