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?

Question

7445 views around the world

You can reuse this answer
Creative Commons License

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.!