# Find the parametric equation of the line passing through the point $\ (-3,4,1) \$

Find the parametric equation of the line passing through the point $\ (-3,4,1) \$ parallel to the $\ xy-plane \$ and perpendicular to the $\ yz-plane$ .

Let the equation of the line through $\ (-3,4,1) \$ is

$\frac{x+3}{l}=\frac{y-4}{m}=\frac{z-1}{n} \$

But how to use the given conditions?

These equations suppose you have a directing vector of the line, and if some coordinates of the directing vector are $0$, the corresponding numerator is $0$.

• Perpendicular to the $yz$-plane: a directing vector is $\vec u=(1,0,0)$.
• Parallel to the $xy$-plane: that is implied by what the directing vector is.

So the parametric equations are $\;\begin{cases} x=-3+t,\\y=4,\\z=1. \end{cases}$

• can you explain again the condition that the line is parallel to the xy-plane – Mabud Ali Sarkar May 10 '18 at 8:13
• Well, a line is parallel to a plane if it is parallel to a line in the plane, or a directing vector of the line is collinear to a vector of the directing plane. It happens that here, the directing vector $\vec u$ is a vector of the $xy$-plane. – Bernard May 10 '18 at 8:37

HINT: A vector normal to $yz$ plane is, e.g., $\vec{n}=(1,0,0)=\vec{k}$.

• so $\ l \cdot 1+m \cdot 0 +n \cdot 0=0 \ \Rightarrow l=0 \$ . But how to use the parallel condition – Mabud Ali Sarkar May 10 '18 at 8:04
• @MabudAliSarkar $l=0$ is not going to be helpful in $\frac{x+3}{l}$ – Henry May 10 '18 at 8:05
• @MabudAliSarkar In this case it is only to check. You know, that the line is described by $P_0+\vec{k}t$. – Przemysław Scherwentke May 10 '18 at 8:05
• we have to use both the conditions – Mabud Ali Sarkar May 10 '18 at 8:09
• @MabudAliSarkar But a vector normal to $yz$ plane is parallel to $x$ ahis, so to $hy$ plane. – Przemysław Scherwentke May 10 '18 at 8:21