Find the vector equation of the plane passing | vedclass

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(A) Let the points be $A(\hat{i} + \hat{j} - 2\hat{k})$,$B(2\hat{i} - \hat{j} + \hat{k})$,and $C(\hat{i} + 2\hat{j} + \hat{k})$.The vectors lying in t

Vedclass · Accepted Answer

$\vec{r} \cdot (9\hat{i} + 3\hat{j} - \hat{k}) = 14$

Vedclass · Answer

(A) Let the points be $A(\hat{i} + \hat{j} - 2\hat{k})$,$B(2\hat{i} - \hat{j} + \hat{k})$,and $C(\hat{i} + 2\hat{j} + \hat{k})$.The vectors lying in the plane are:$\vec{AB} = (2\hat{i} - \hat{j} + \hat{k}) - (\hat{i} + \hat{j} - 2\hat{k}) = \hat{i} - 2\hat{j} + 3\hat{k}$$\vec{AC} = (\hat{i} + 2\hat{j} + \hat{k}) - (\hat{i} + \hat{j} - 2\hat{k}) = 0\hat{i} + \hat{j} + 3\hat{k}$The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{AB} 	imes \vec{AC}$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 0 & 1 & 3 \end{vmatrix} = \hat{i}(-6 - 3) - \hat{j}(3 - 0) + \hat{k}(1 - 0) = -9\hat{i} - 3\hat{j} + \hat{k}$.The vector equation of the plane passing through point $\vec{a} = \hat{i} + \hat{j} - 2\hat{k}$ with normal $\vec{n} = -9\hat{i} - 3\hat{j} + \hat{k}$ is $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$,which implies $\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.Calculating $\vec{a} \cdot \vec{n} = (\hat{i} + \hat{j} - 2\hat{k}) \cdot (-9\hat{i} - 3\hat{j} + \hat{k}) = -9 - 3 - 2 = -14$.Thus,$\vec{r} \cdot (-9\hat{i} - 3\hat{j} + \hat{k}) = -14$,or $\vec{r} \cdot (9\hat{i} + 3\hat{j} - \hat{k}) = 14$.

Find the vector equation of the plane passing through the points $\hat{i} + \hat{j} - 2\hat{k}$,$2\hat{i} - \hat{j} + \hat{k}$,and $\hat{i} + 2\hat{j} + \hat{k}$.

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