Find the direction cosines of the unit vector | vedclass

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(B) The given equation of the plane is $\vec{r} \cdot (6 \hat{i} - 3 \hat{j} - 2 \hat{k}) = -1$. To find the unit normal vector,we rewrite the equatio

Vedclass · Accepted Answer

$\frac{-6}{7}, \frac{3}{7}, \frac{2}{7}$

Vedclass · Answer

(B) The given equation of the plane is $\vec{r} \cdot (6 \hat{i} - 3 \hat{j} - 2 \hat{k}) = -1$. To find the unit normal vector,we rewrite the equation as $\vec{r} \cdot (-6 \hat{i} + 3 \hat{j} + 2 \hat{k}) = 1$. The normal vector is $\vec{n} = -6 \hat{i} + 3 \hat{j} + 2 \hat{k}$. The magnitude of the normal vector is $|\vec{n}| = \sqrt{(-6)^2 + 3^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$. The unit normal vector $\hat{n}$ is given by $\frac{\vec{n}}{|\vec{n}|} = \frac{-6}{7} \hat{i} + \frac{3}{7} \hat{j} + \frac{2}{7} \hat{k}$. The direction cosines are the components of the unit normal vector,which are $\frac{-6}{7}, \frac{3}{7}, \frac{2}{7}$.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

Find the direction cosines of the unit vector perpendicular to the plane $\vec{r} \cdot (6 \hat{i} - 3 \hat{j} - 2 \hat{k}) + 1 = 0$.

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