Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

1) Write the vector equation of the line in $\mathbb R^3$ given by:

\begin{align} x+y+z&=6 \\ x+2y+z&=1 \end{align}

2) Write the vector equation of the plane in $\mathbb R^3$ that passing through (1,0,-1) and is perpendicular to the line of exercise 1)

share|cite|improve this question

Subtracting the first equation from the second gives $y=-5$. In this plane, you want all the points satisfying $x+z=11$, which is a line.

Parametrically, we can give it as $(t,-5,11-t)$, for all real $t$.

share|cite|improve this answer

For your second question, a plane is defined by a point and a normal direction (a direction that is perpendicular to every vector in your plane). In this exercise, this direction is the direction vector of your line which was given by vadim123: $\vec{d}=(1,0,-1)$. Using that and the given point, the equation of the plane is $$((x,y,z) - (1,0,-1))\bullet (1,0,-1)=0$$ Also, you can write your equation as $$x-z =2$$

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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