Find parametric equations and symmetric equations for the line. (Use the parameter t.) The line through (3, 5, 0) and perpendicular to both i + j and j + k x(t), y(t), z(t) = The symmetric equations are given by −(x − 3) = y − 5 = z. x + 3 = −(y + 5), z = 0. x − 3 = y − 5 = −z. x + 3 = −(y + 5) = z. x − 3 = −(y − 5) = z.

Answer:(x-3)=-(y-5)=zStep-by-step explanation:Giventhe point through which line passes is (3,5,0)so we need a vector along the line to get the equation of lineIt is given that line is perpendicular to both i+j & j+ktherefore their cross product will give us the vector perpendicular to bothv=(i+j)\times (j+k)=i-j+ktherefore we get direction vector of line so we can write [tex]\frac{x-3}{1}=\frac{y-5}{-1} =\frac{z-0}{1}[/tex]=ti.e.x=t+3,y=-t+5,z=t