Does the line passing through $(3,4,-1)$ which is normal to $x+4y-z = -2$, intersect any of the coordinate axes?

I'm not sure how to go about this question. Any help would be greatly appreciated. Thank you in advance!


closed as off-topic by Saad, Vinyl_cape_jawa, Lee David Chung Lin, Parcly Taxel, Shailesh Mar 16 at 4:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Vinyl_cape_jawa, Lee David Chung Lin, Parcly Taxel, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.


The line passing through $(3,4,-1)$ and normal to the plane $x+4y-z = -2$ has the following parametric equation: $$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}3\\\color{blue}{4}\\\color{blue}{-1}\end{pmatrix}+\lambda\begin{pmatrix}1\\\color{blue}{4}\\\color{blue}{-1}\end{pmatrix}$$ It intersects a coordinate axis if two coordinates are (simultaneously) zero. Can you find a value of $\lambda$ for which this happens? Hint: look at the blue coordinates.

  • $\begingroup$ Thanks, I get that, but is there a specific method to deduce this instead of simply using logic and observation? $\endgroup$ – Mayuri Gupta Mar 15 at 13:43
  • $\begingroup$ If it's not that obivous, you can always look for points of intersection with the axes by solving the corresponding systems of equations - and see if they have a solution. In this case, setting $y=0$ and $z=0$ would yield the solution $\lambda=-1$. $\endgroup$ – StackTD Mar 15 at 13:47

Not the answer you're looking for? Browse other questions tagged or ask your own question.