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  3. Find the cartesian and vector equation of the plane passing through the  points A(2,5,-3),B(-2,-3.5) and C(5,3,-3).      
# School

Find the cartesian and vector equation of the plane passing through the 
points A(2,5,-3),B(-2,-3.5) and C(5,3,-3).

 

 

 

 
 
 
 
 

Answers (1)
R Ravindra Pindel

A:  we know that the general equation of the plane passing 
through three points\left ( x_{1} y_{1}z_{1}\right )\left ( x_{1} ,y_{1},z_{1}\right )\left ( x_{2}, y_{2},z_{2}\right )\left ( x_{3} ,y_{3},z_{3}\right )
\begin{vmatrix} x-x_{1}&y-y_{1} &z-z_{1} \\ x_{2}-x_{1}&y_{2}-y_{1} &z_{2}-z_{1} \\ x_{3}-x_{1}&y_{3}-y_{1} &z_{3}-z_{1} \end{vmatrix}= 0
Then the plane pasing through A\left ( 2,5,-3 \right )B\left ( -2,-3,5 \right ) and \: C\left ( 5,3,3 \right )
\begin{vmatrix} x-2 & y-5 & z+3\\ -2-2&-3-5 & 5+3\\ 5-2& 3-5 & -3+3 \end{vmatrix}= 0\Rightarrow \begin{vmatrix} x-2 & y-5 & z+3\\ -4&-8 & 8\\ 3& -2 & 0\end{vmatrix}= 0\Rightarrow \left ( x-2 \right )\left ( 0+16 \right )-\left ( y-5 \right )\left ( 0-24 \right )+\left ( z+3 \right )\left ( 8+24 \right )= 0\Rightarrow 16\left ( x-2 \right )+24\left ( y-5 \right )+32\left ( z+3 \right )= 0
\Rightarrow 8\left [ 2x-4+3y-15+4z+12 \right ]= 0
\Rightarrow 2x+3y+4z-7= 0
\Rightarrow 2x+3y+4z=7
This is the required cartesian equation of the plane.
Now,
The required plane passes through the point A\left ( 2,5,-3 \right )whose position vector is a= 2\hat{i} +5\hat{j}-3\hat{k} and is normal 
to the vector \vec{n} given by \vec{n}= \overrightarrow{AB}\times \overrightarrow{AC}
\therefore \overrightarrow{AB}= -2\hat{i}-3\hat{j}+5\hat{k}-\left ( 2\hat{i} +5\hat{j}-3\hat{k}\right )
\overrightarrow{AB}= -4\hat{i}-8\hat{j}+8\hat{k}+8\hat{k}
\overrightarrow{AC}= \left [ 5\hat{i}+3\hat{j}-3\hat{k} -\left ( 2\hat{i} +5\hat{j}-3\hat{k}\right )\right ]
        = 3\hat{i}-2\hat{j}
Now,
 \vec{n}= \overrightarrow{AB}\times \overrightarrow{AC}= \begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ -4 & \ -8 &\8 \\ 3 & \-2 &\0 \end{vmatrix}
= 16\hat{i}-\left ( -24 \right )\grave{j}+\left ( 8+24 \right )\hat{k}
= 16\hat{i}+24\hat{j}+32\hat{k}
The vector equation of the plane is
\vec{r}\cdot \vec{n}= \vec{a}\cdot \vec{n}
\Rightarrow \vec{r}\left ( 16\hat{i}+24\hat{j}+32\hat{k} \right )= 56
\Rightarrow \vec{r}\left ( 2\hat{i}+3\hat{j}+4\hat{k} \right )= 7
this is the required vector equation
        
 

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