What is the parametric equation of the line through (1,1,1) that is parallel to i+j+k?

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Answer 1 · 2017-08-08T22:39:31

#{(x = t), (y= t),(z= t):}#

Given #vecu = veci+vecj+veck#

consider the line:

#P = tvecu#

that is parallel to #vecu#.

Now consider the coordinates of the generic point:

#{(x_p= tvecu * veci = t), (y_p= tvecu * vecj = t),(z_p= tvecu * veck = t):}#

For #t=1# thus we have #P = (1,1,1)# as required.

Answer 2 · 2017-08-08T22:45:24

# { (x=1+lamda), (y=1+lamda), (z=1+lamda) :} #

The vector equation of a line passing through a point #bb(ul(a))# in the direction of a vector #bb(ul(d))# is given by:

# bb(ul(r)) = bb(ul(a)) + lamda bb(ul(d)) #

So the vector equation of the line through #(1,1,1)# that is parallel to # bb(ul(hat(i))) + bb(ul(hat(j))) + bb(ul(hat(k))) # is:

# bb(ul(r)) = ( (1), (1), (1) ) + lamda ( (1), (1), (1) ) #

If we denote #bb(ul(r))# by #(x,y,z)# tghen we can write this as:

# ( (x), (y), (z) ) = ( (1), (1), (1) ) + ( (lamda), (lamda), (lamda) ) #
# " " = ( (1+lamda), (1+lamda), (1+lamda) ) #

And so by equating terms, the parametric equations would be:

# { (x=1+lamda), (y=1+lamda), (z=1+lamda) :} #