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Equation of plane passing through points A(1,1,1), B(-1,0,1) and C(1,2,3) is

Option 1)
$x+2y+z=0$

Option 2)
$x+2y-z=0$

Option 3)
$x-2y+z=0$

Option 4)
$x-2y-z=0$

Answers (1)

best_answer

As we have learned in

Plane passing through three points (vector form) -

Let the plane passes through $A(a),B(b)\: and\, C(c)$

then the plane is given by $\left [ r\, b\, c \right ]+\left [ r\, a\, b \right ]+\left [ r\, c\, a \right ]= \left [ a\, b\, c \right ]$

- wherein

$\vec{n}= \vec{AB}\times \vec{BC}$

$\vec{n}= \left ( \vec{b}- \vec{a} \right )\times \left ( \vec{c}- \vec{a} \right )$

$\left ( \vec{r}- \vec{a} \right )\cdot \vec{n}=0$

$\left ( \vec{r}- \vec{a} \right )\cdot \left ( \vec{b}- \vec{a} \right )\times \left ( \vec{c}- \vec{a} \right )= 0$

$\vec{AB}=-2\hat{i}-\hat{j}+0\hat{k}; \vec{AC}=0\hat{i}+\hat{j}+2\hat{k}$

Vector perpendicular to plane will be along $\vec{AB}\times\vec{AC}$

$\therefore\vec{AB}\times\vec{AC}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ -2 &-1 &0 \\ 0& 1 &2 \end{vmatrix}= -2\hat{i}+4\hat{j}-2\hat{k}$

$\vec{n}=-2\hat{i}+4\hat{j}-2\hat{k}$

$\therefore$ equation of plane will be

$(\vec{r}-(\hat{i}+\hat{j}+\hat{k}))\cdot-(2\hat{i}+4\hat{j}-2\hat{k})=0$

$\Rightarrow\vec{r}\cdot(-2\hat{i}+4\hat{j}-2\hat{k})=0$

2x-4y+2z=0

$\Rightarrow$ x-2y+z=0

Option 1)

$x+2y+z=0$

Option 2)

$x+2y-z=0$

Option 3)

$x-2y+z=0$

Option 4)

$x-2y-z=0$

Posted by

Himanshu

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