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  • Need explanation for: Equation of plane passing through points (0,0,0), (1,-1,1) and (2,2,-2)

Equation of plane passing through points (0,0,0), (1,-1,1) and (2,2,-2)

  • Option 1)

    x+y+z=0

  • Option 2)

    y-z=0

  • Option 3)

    y+z=0

  • Option 4)

    x+y=0

 

Answers (1)

best_answer

As we have learned

Plane passing through three points (cartesian form) -

Let the plane passes through

A(x_{1},y_{1},z_{1}),B(x_{2},y_{2},z_{2})\: and \: C(x_{3},y_{3},z_{3})

then the plane is given by

\begin{vmatrix} x-x_{1} &y-y_{1} &z-z_{1} \\ x-x_{2} & y-y_{2} & z-z_{2}\\ x-x_{3}&y-y_{3} & z-z_{3} \end{vmatrix}=0
 

 

- wherein

\underset{AB}{\rightarrow} = \left ( x_{2}-x_{1} \right )\hat{i}+ \left ( y_{2}-y_{1} \right )\hat{j}+ \left ( z_{2}-z_{1} \right )\hat{k}

\underset{AC}{\rightarrow} = \left ( x_{3}-x_{1} \right )\hat{i}+ \left ( y_{3}-y_{1} \right )\hat{j}+ \left ( z_{3}-z_{1} \right )\hat{k}

\vec{n}=\underset{AB}{\rightarrow} \times\underset{AC}{\rightarrow}

\left ( \vec{r} -\vec{a}\right )\cdot\underset{AB}{\rightarrow} \times\underset{AC}{\rightarrow}= 0

 

 Equation of plane will be \begin{vmatrix} x-0 & y-0 &z-0 \\ x-1& y+1 & z-1\\ x-2 &y-2 & z+2 \end{vmatrix}=0

on opening the determinant we get  y+z=0


Option 1)

x+y+z=0

Option 2)

y-z=0

Option 3)

y+z=0

Option 4)

x+y=0

Posted by

Himanshu

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