Question

find parametric equations for the line. (use the parameter t.) the line through (3, 2, 0) and perpendicular to both i + j and j + k (x(t), y(t), z(t)) = ( ) find the symmetric equations. o x - 3 = -(y - 2) = z o x + 3 = -(y + 2), z = 0 o x + 3 = -(y + 2) = z o x - 3 = y - 2 = -z o -(x - 3) = y - 2 = z resources read it

Explanation:

Step1: Find the direction vector

The cross - product of $\vec{a}=\mathbf{i}+\mathbf{j}=(1,1,0)$ and $\vec{b}=\mathbf{j}+\mathbf{k}=(0,1,1)$ gives the direction vector $\vec{v}$ of the line. $\vec{v}=\vec{a}\times\vec{b}=

$$\begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&1&0\\0&1&1\end{vmatrix}$$

=\mathbf{i}(1 - 0)-\mathbf{j}(1 - 0)+\mathbf{k}(1 - 0)=(1,-1,1)$.

Step2: Write the parametric equations

The line passes through the point $(x_0,y_0,z_0)=(3,2,0)$ and has direction vector $\vec{v}=(1,-1,1)$. The parametric equations of the line are $x(t)=x_0 + vt_x=3 + t$, $y(t)=y_0+vt_y=2 - t$, $z(t)=z_0+vt_z=t$. So $(x(t),y(t),z(t))=(3 + t,2 - t,t)$.

Step3: Find the symmetric equations

From the parametric equations $t=x - 3$, $t=2 - y$, $t = z$. Then the symmetric equations are $x - 3=-(y - 2)=z$.

Answer:

$(x(t),y(t),z(t))=(3 + t,2 - t,t)$; A. $x - 3=-(y - 2)=z$