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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 $ .

Answer:

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?

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

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  • $\begingroup$ can you explain again the condition that the line is parallel to the xy-plane $\endgroup$ – Mabud Ali Sarkar May 10 '18 at 8:13
  • $\begingroup$ 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. $\endgroup$ – Bernard May 10 '18 at 8:37
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HINT: A vector normal to $yz$ plane is, e.g., $\vec{n}=(1,0,0)=\vec{k}$.

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

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