The direction cosines of a normal to the plan | vedclass

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(C) Let the points be $A(4,2,3)$,$B(-1,4,2)$,and $C(3,2,1)$.Two vectors lying on the plane are $\vec{AB} = (-1-4)\hat{i} + (4-2)\hat{j} + (2-3)\hat{k}

Vedclass · Accepted Answer

$\frac{-4}{\sqrt{101}}, \frac{-9}{\sqrt{101}}, \frac{2}{\sqrt{101}}$

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(C) Let the points be $A(4,2,3)$,$B(-1,4,2)$,and $C(3,2,1)$.Two vectors lying on the plane are $\vec{AB} = (-1-4)\hat{i} + (4-2)\hat{j} + (2-3)\hat{k} = -5\hat{i} + 2\hat{j} - 1\hat{k}$ and $\vec{AC} = (3-4)\hat{i} + (2-2)\hat{j} + (1-3)\hat{k} = -1\hat{i} + 0\hat{j} - 2\hat{k}$.The normal vector $\vec{n}$ is given by $\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -5 & 2 & -1 \ -1 & 0 & -2 \end{vmatrix} = \hat{i}(-4-0) - \hat{j}(10-1) + \hat{k}(0 - (-2)) = -4\hat{i} - 9\hat{j} + 2\hat{k}$.The magnitude of $\vec{n}$ is $|\vec{n}| = \sqrt{(-4)^2 + (-9)^2 + 2^2} = \sqrt{16 + 81 + 4} = \sqrt{101}$.The direction cosines are $\frac{-4}{\sqrt{101}}, \frac{-9}{\sqrt{101}}, \frac{2}{\sqrt{101}}$.

The direction cosines of a normal to the plane passing through $(4,2,3)$,$(-1,4,2)$ and $(3,2,1)$ are .....

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