If overrightarrow{p} = 4hat{i} - hat{j} + hat | vedclass

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(A) The equation of a plane passing through a point $\overrightarrow{p}$ and perpendicular to a normal vector $\overrightarrow{q}$ is given by $\overr

Vedclass · Accepted Answer

$4$

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(A) The equation of a plane passing through a point $\overrightarrow{p}$ and perpendicular to a normal vector $\overrightarrow{q}$ is given by $\overrightarrow{q} \cdot (\overrightarrow{r} - \overrightarrow{p}) = 0$.Given $\overrightarrow{p} = 4\hat{i} - \hat{j} + \hat{k}$ and $\overrightarrow{q} = 9\hat{i} - 2\hat{j} + 6\hat{k}$.Substituting these into the equation: $(9\hat{i} - 2\hat{j} + 6\hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k} - (4\hat{i} - \hat{j} + \hat{k})) = 0$.$(9\hat{i} - 2\hat{j} + 6\hat{k}) \cdot ((x-4)\hat{i} + (y+1)\hat{j} + (z-1)\hat{k}) = 0$.$9(x-4) - 2(y+1) + 6(z-1) = 0$.$9x - 36 - 2y - 2 + 6z - 6 = 0$.$9x - 2y + 6z - 44 = 0$.The perpendicular distance $d$ from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$.$d = \frac{|-44|}{\sqrt{9^2 + (-2)^2 + 6^2}} = \frac{44}{\sqrt{81 + 4 + 36}} = \frac{44}{\sqrt{121}} = \frac{44}{11} = 4$.

If $\overrightarrow{p} = 4\hat{i} - \hat{j} + \hat{k}$ is a point and $\overrightarrow{q} = 9\hat{i} - 2\hat{j} + 6\hat{k}$ is a normal vector,then the perpendicular distance of the origin from the plane passing through $\overrightarrow{p}$ and perpendicular to $\overrightarrow{q}$ is

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