logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  TN XII Math  >>  Vector Algebra
0 votes

Find the vector cartesian equation of the plane containing the line $\large\frac{x-2}{2}=\frac{y-2}{3}=\frac{z-1}{-2}$ and passing through the point $(-1 , 1 , -1 )$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Equation of a plane through points $A(x_1, y_1, z_1)(pv \overrightarrow a)$ and $B(x_2, y_2, z_2)(pv\: \overrightarrow b)$ and parallel to a vector $\overrightarrow u$ (with D.r.s $l, m, n)$ $ \overrightarrow r = (1-s) \overrightarrow a+s\overrightarrow b+t\overrightarrow v$ (vector form) $ \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l & m & n \end{vmatrix}=0 $ ( cartesian form) $ [ \overrightarrow r-\overrightarrow a\: \: \overrightarrow b-\overrightarrow a\: \: \overrightarrow u]=0$ ( Non parametre vector equation)
  • The line $ \large\frac{x-x_1}{l}=\large\frac{y-y_1}{m}=\large\frac{z-z_1}{n}$ contains the point $(x_1, y_1, z_1) $ and is parallel to the vector $ l\overrightarrow i+m\overrightarrow j + n\overrightarrow k$
Step 1
The required plane contains the line $ \large\frac{x-2}{2}=\large\frac{y-2}{3}=\large\frac{z-1}{-2}$
$ \therefore $ It contains the point $A(2, 2, 1)(\overrightarrow a=2\overrightarrow i+2\overrightarrow j+\overrightarrow k)$ which lies on the line and it is parallel to the vector $\overrightarrow v=2\overrightarrow i+3\overrightarrow j-2\overrightarrow k$ which is parallel to the line. The plane also contains the point $B(-1, 1, -1) (\overrightarrow b=-i+\overrightarrow j-\overrightarrow k)$
Step 2
The vector equation of the plane is
$ \overrightarrow r = (1-s)\overrightarrow a+s\overrightarrow b+t\overrightarrow v$
$ \overrightarrow r = (1-s)(2\overrightarrow i+2\overrightarrow j+\overrightarrow k)+s(-\overrightarrow i+\overrightarrow j-\overrightarrow k)+t(2\overrightarrow i+3 \overrightarrow j-2\overrightarrow k)$
Step 3
The cartesian equation is
$ \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l & m & n \end{vmatrix}=0$
where $ (x_1, y_1, z_1 )=(2,2,1), (x_2, y_2,z_2)=(-1, 1, -1), (l, m, n)=(2, 3, -2)$
$ \therefore \begin{vmatrix} x-2 & y-2 & z-1 \\ -1-2 & 1-2 & -1-1 \\ 2 & 3 & -2 \end{vmatrix}=0 \Rightarrow \begin{vmatrix} x-2 & y-2 & z-1 \\ -3 & -1 & -2 \\ 2 & 3 & -2 \end{vmatrix}=0$
$(x-2)(2+6)-(y-2)(6+4)+(z-1)(-9+2)=0$
$8x-16-10y+20-7z+7=0$
$ 8x-10y-7z=-11$
answered Jun 16, 2013 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...