The equation of the plane in normal form whic | vedclass

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(C) Given points are $A(-2,1,3), B(1,1,1), C(2,3,4)$.The vectors lying on the plane are:$\overrightarrow{AB} = (1 - (-2))\hat{i} + (1 - 1)\hat{j} + (1

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$\left(\frac{-4}{\sqrt{173}}\right) x+\left(\frac{11}{\sqrt{173}}\right) y+\left(\frac{-6}{\sqrt{173}}\right) z=\frac{1}{\sqrt{173}}$

Vedclass · Answer

(C) Given points are $A(-2,1,3), B(1,1,1), C(2,3,4)$.The vectors lying on the plane are:$\overrightarrow{AB} = (1 - (-2))\hat{i} + (1 - 1)\hat{j} + (1 - 3)\hat{k} = 3\hat{i} - 2\hat{k}$$\overrightarrow{BC} = (2 - 1)\hat{i} + (3 - 1)\hat{j} + (4 - 1)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$The normal vector $\vec{n}$ to the plane is given by the cross product $\overrightarrow{AB} 	imes \overrightarrow{BC}$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 0 & -2 \ 1 & 2 & 3 \end{vmatrix} = \hat{i}(0 - (-4)) - \hat{j}(9 - (-2)) + \hat{k}(6 - 0) = 4\hat{i} - 11\hat{j} + 6\hat{k}$The equation of the plane passing through $A(-2,1,3)$ with normal $\vec{n} = 4\hat{i} - 11\hat{j} + 6\hat{k}$ is:$4(x + 2) - 11(y - 1) + 6(z - 3) = 0$$4x + 8 - 11y + 11 + 6z - 18 = 0$$4x - 11y + 6z + 1 = 0$To convert to normal form $lx + my + nz = p$,we divide by $\sqrt{a^2 + b^2 + c^2} = \sqrt{4^2 + (-11)^2 + 6^2} = \sqrt{16 + 121 + 36} = \sqrt{173}$.Since the constant term is $1$,we rewrite $4x - 11y + 6z = -1$ as $-4x + 11y - 6z = 1$.Dividing by $\sqrt{173}$:$\left(\frac{-4}{\sqrt{173}}ight)x + \left(\frac{11}{\sqrt{173}}ight)y + \left(\frac{-6}{\sqrt{173}}ight)z = \frac{1}{\sqrt{173}}$.

The equation of the plane in normal form which passes through the points $(-2,1,3), (1,1,1)$ and $(2,3,4)$ is

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