Answer:
the equation of the line L₃(t) is L₃(t)= (5,1,2)+(12,20,30)*t
Step-by-step explanation:
for the lines
L₁(t) = (5,4,9)+(-5,7,-5)*t
L₂(t) = (0,4,9)+(-5,1,-1)*t
then the vector V perpendicular to both L₁ and L₂ can be found through the vectorial product of its parallel vectors
[tex]V=\left[\begin{array}{ccc}i&j&k\\-5&7&-5\\-5&1&-1\end{array}\right] = \left[\begin{array}{ccc}7&-5\\1&-1\end{array}\right]*i + \left[\begin{array}{ccc}-5&-5\\-5&-1\end{array}\right]*j + \left[\begin{array}{ccc}-5&7\\-5&1\end{array}\right]*j = -2*i -20*j + 30*k= (-2,-20,30)\\[/tex]
then the equation of the line L₃(t) that goes through (5 ,1,2) in t=0 and has V as parallel vector is
L₃(t)= (5,1,2)+(-2,-20,30)*t
