What is the parametric equation of the line through (1,1,1) that is parallel to i+j+k?
2 Answers
Explanation:
Given
consider the line:
that is parallel to
Now consider the coordinates of the generic point:
For
# { (x=1+lamda), (y=1+lamda), (z=1+lamda) :} #
Explanation:
The vector equation of a line passing through a point
# bb(ul(r)) = bb(ul(a)) + lamda bb(ul(d)) #
So the vector equation of the line through
# bb(ul(r)) = ( (1), (1), (1) ) + lamda ( (1), (1), (1) ) #
If we denote
# ( (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) :} #