Answer with Step-by-step explanation:
We are given that a point (3,1,0)
Two vectors are
A=<1,1,0>
B=<0,1,1>
[tex]A\times B=\begin{vmatrix}i&j&k\\1&1&0\\0&1&1\end{vmatrix}[/tex]
[tex]A\times B=i-j+k[/tex]
Let v[tex]=A\times B=i-j+k[/tex]
[tex]v=<a,b,c>=<1,-1,1>[/tex]
[tex]r_0=<x_0,y_0,z_0>=<3,1,0>[/tex]
[tex]r=r_0+vt[/tex]
Substitute the values then we get
[tex]r=<3,1,0>+t<1,-1,1>[/tex]
[tex]r=<3+t,1-t,t>[/tex]
The parametric equation of the line
[tex]x=x_0+at,y=y_0+bt,z=z_0+ct[/tex]
Using the formula
The parametric equation of the line which is passing through the point (3,1,0) and perpendicular to both i+j and j+k is given by
[tex]x=3+t,y=1-t,z=t[/tex]
The symmetric equation of the line is given by
[tex]\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}[/tex]
Using the formula
The symmetric equation of the line which is passing through the point (3,1,0) and perpendicular to both i+j and j+k is given by
[tex]\frac{x-3}{1}=\frac{y-1}{-1}=z[/tex]