Find the vector equation of a plane which is at a distance of 7 units

Question:

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3 \hat{i}+5 \hat{j}-6 \hat{k}$.

Solution:

The normal vector is, $\vec{n}=3 \hat{i}+5 \hat{j}-6 \hat{k}$

$\therefore \hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{3 \hat{i}+5 \hat{j}-6 \hat{k}}{\sqrt{(3)^{2}+(5)^{2}+(6)^{2}}}=\frac{3 \hat{i}+5 \hat{j}-6 \hat{k}}{\sqrt{70}}$

It is known that the equation of the plane with position vector $\vec{r}$ is given by, $\vec{r} \cdot \hat{n}=d$

$\Rightarrow \hat{r} \cdot\left(\frac{3 \hat{i}+5 \hat{j}-6 \hat{k}}{\sqrt{70}}\right)=7$

This is the vector equation of the required plane.

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