Answer :
Answer:
Step-by-step explanation:
Suppose u = i + j = [tex]\Big \langle 1,1,0 \Big\rangle[/tex] & v = j + k = [tex]\Big \langle 0,1,1 \Big\rangle[/tex]
The direction vector for the line u*v is:
[tex]u\times v = \left |\begin{array}{ccc}i&j&k\\1&1&0\\0&1&1\end {array} \right|[/tex]
= (1-0) i - (1 - 0)j + ( 1- 0) k
= i - j - k
= [tex]\Big \langle 1,-1,1 \Big\rangle[/tex]
Hence, the equation of the line via the point (5,4,0 ) and the direction vector [tex]\Big \langle 1,-1,1 \Big\rangle[/tex] is as follows:
r(t) = (5,4,0) + t[tex]\Big \langle 1,-1,1 \Big\rangle[/tex]
r(t) = (5+t, 4-t, t)
The symmetric equations of the line are:
[tex]\dfrac{x-5}{1}= \dfrac{y-4}{-1} = \dfrac{z}{1}[/tex]
x - 5 = -(y-4) = z
The parametric equation of the line is:
[tex]\dfrac{x-5}{1}= \dfrac{y-4}{-1} = \dfrac{z}{1}= t[/tex]
x = 5 + t , y = 4 - t , z = t