Find the vector and cartesian equations of the plane passing through the points (2,5,-3),(-2,-3,5) and (5,3,-3). Also ,find the point of intersection of this plane with the line passing through points (3,1,5) and (-1,-3,-1)

 

 

 

 
 
 
 
 

Answers (1)

\vec{a}= 2\hat{i}+5\hat{j}-3\hat{k}
\vec{b}= -2\hat{i}-3\hat{j}+5\hat{k}
\vec{c}= 5\hat{i}+3\hat{j}-3\hat{k}
Equation of plane passing through three points
\left ( \vec{r} -\vec{a}\right )\cdot \left ( \vec{b}- \vec{a}\right )\times \left ( \vec{c}-\vec{a} \right )= 0
\vec{b}-\vec{a}= -4\hat{i}-8\hat{j}+8\hat{k}
\vec{c}-\vec{a}= 3\hat{i}-2\hat{j}
\left ( \vec{b}-\vec{a} \right)\times\left(\vec{c}-\vec{a} \right )= \begin{bmatrix} i & j&k \\ -4& -8 & 8\\ 3& -2 & 0 \end{bmatrix}
                                      = i\left ( 0+16 \right )-j\left ( 0-24 \right )+k\left ( 8+24 \right )
                                      \Rightarrow 16\hat{i}+24\hat{j}+32\hat{k}
Vector Equation of plane
\left ( \vec{r}-\left ( 2\hat{i}+5\hat{j}-3\hat{k} \right ) \cdot \left ( 16\hat{i}+24\hat{j}+32\hat{k} \right )\right )= 0
\Rightarrow \vec{r}.\left ( 16\hat{i}+24\hat{j} +32\hat{k}\right )-\left ( 32+120-36 \right )= 0
\Rightarrow \vec{r}.8\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right )= -56
\Rightarrow \vec{r}.\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right )= 7
cartesian equation
\vec{r}= x\hat{i}+y\hat{j}+z\hat{k}
2x+3y+4z= 7
hence going through 2 points (3,1,5) and (-1,-3,-1)
Difference between  2points (-4,-4,-6)
so  x= 3-4t
      y= 1-4t
      z= 5-6t
and plane 16x+24y+32z= 56
Intersection points
16\left ( 3-4t \right )+24\left ( 1+4t \right )+32\left ( 5+6t \right )= 56
48-64t+24-96t+160-192t= 56
\Rightarrow 232-56= 192t+64t+96t
\Rightarrow176= 352k
\Rightarrow t= \frac{1}{2}
x= 3-4\left ( \frac{1}{2} \right )= 3-2= 1\: \: \: \:\: x= 1,y= -1,z= 2
y= 1-4\left ( \frac{1}{2} \right )= 1-2= -1
z= 5-6\left ( \frac{1}{2} \right )= 5-3= 2
 

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