0

Find the equation of the plane passing points $A = (3,5,1)$ and $B = (0,3,−1)$ and parallel with the line $x=1-2t, y=2+t ,z=3t+2$

I don't really know what the line look like? what is the variable t here? THank you

Mr.Gandalf Sauron
  • 13,906
Alex
  • 113
  • 5
  • 2
    For every value of t you will get points (x,y,z) on the line – Tojra Feb 03 '22 at 07:24
  • 2
    The line $$\begin{bmatrix}x\y\z\end{bmatrix}=\begin{bmatrix}1\2\2\end{bmatrix}+t\begin{bmatrix}-2\1\3\end{bmatrix}.$$ This line passes through the point $(1,2,2)$ and has direction vector $(-2,1,3)$. So now you have a vector that is parallel to the plane. From the information given, can you find another vector in the plane? – Anurag A Feb 03 '22 at 07:25
  • can you give me more hints? sorry,my mathematics is bad. – Alex Feb 03 '22 at 07:31
  • 1
    Hint: Your math will get better if you do your own work. – John Douma Feb 03 '22 at 07:40
  • What definition of a line in 3D coordinate geometry are you familiar with? Does this help? $$\frac{x-1}{-2}=\frac{y-2}{1}=\frac{z-2}{3}$$ this is the cartesian equation for line in 3D. Are you familiar with direction ratios and concepts like that? – Mr.Gandalf Sauron Feb 03 '22 at 12:56
  • You wrote "I don't really know what the line look like?" But this isn't a question, it's a disclaimer. You ask what the variable $t$ is, but you can't expect Readers to tell you what something you said means. We need you to be the expert about what you are asking. Then we can help with an Answer. – hardmath Feb 03 '22 at 19:34

1 Answers1

1

$t$ is just a parameter. as $t$ varies over all real numbers , you will get the coordinates of the points of the line. Try and relate with what happens in 2D.

Technically a line in 2d say $y=mx+c$ is can be written in a set builder form as :-

$$\{(t,at+c)\,,-\infty<t<\infty\}$$.

The coefficients of the parameter $t$ in each of the expressions for $x,y$ and $z$ are called the direction ratios and when viewed as a tripple , they give a vector which is parallel to the line.

Compare this with the technique you actually use to draw a line with a ruler and pencil. Once you are given a point and a direction , or two points, you join them using the ruler and extend it on both sides. Isn't it analogous to considering a point of starting point $O$ with it's position vector given by $\vec{p}$ and a direction given by $\vec{v}$ and just saying that all the points in your line are the same as those with coordinates $\vec{p}+t\vec{v}\,\,$ for all $-\infty<t<\infty$?

For a hint on trying to solve a problem:-

Are you aware that for two vectors lying in a plane, their cross product is perpendicular to it. If yes then what can you say about the cross product of the pair of vectors $\vec{BA}=(3,2,2)=3\hat{i}+5\hat{j}+2\hat{k}$ and the vector parallel to the line given by it's direction ratios $-2\hat{i}+\hat{j}+3\hat{k}$? Will it be perpendicular to the plane?. If yes then don't you already have a point on the line and a vector perpendicular to it ?.

Another hint:- A tripple of vectors in $\mathbb{R}^{3}$ are coplanar iff their scalar tripple product is $0$. Then for an arbitrary point $(x,y,z)$ in the plane , what can you say about the tripple ?

1.$(x,y,z)-(3,5,1)$

2.$(-2,1,3)$

3.$\vec{BA}=(3,2,2)$

If $(x,y,z)$ lies on the plane then are they not coplanar?. If yes then using the fact that the scalar tripple product is $0$? can you come up with an equation of the plane?

Mr.Gandalf Sauron
  • 13,906