Check if the line #x+1=1-y=-5z# lies on the plane #2x+3y-5z=1#?

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Answer 1 · 2017-06-21T17:53:37

See below.

Given the line with #p=(x,y,z)#

#L->p=p_0+lambda vec v equiv {x+1=1-y=-5z}#

with

#p_0 = (-1,1,0)#
#vec v = (1,-1,-1/5)#

and the plane

#Pi-> << vec n, p-p_1 >> = 2x+3y-5z-1=0#

with

#p_1 = (0,0,-1/5)#
#vec n = (2,3,-5)#

We say that #L in Pi hArr << vec n, p_0+lambda vec v - p_1 >> = 0#

or

#<< vec n, p_0-p_1 >> +lambda << vec n, vec v >> = 0#

Now, substituting values we can conclude.

As can be verified, #<< vec n, p_0-p_1 >> = 0# and #<< vec n, vec v >> = 0# so #L in Pi#