Question

Find the distance of the point $(2,5,-3)$ from the plane \[ \vec{r}\cdot(6\hat{i}-3\hat{j}+2\hat{k})=4. \]

Hint:

Always reduce the vector plane equation $\vec{r}\cdot\vec{n}=d$ into Cartesian form to directly apply the distance formula.

Answer 1

Step 1: Write plane equation in Cartesian form.
\[ 6x - 3y + 2z = 4 \]

Step 2: Distance formula.
For a point $(x_1,y_1,z_1)$ and plane $ax+by+cz+d=0$, \[ D = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}} \] Here, $a=6$, $b=-3$, $c=2$, $d=-4$, and point $(2,5,-3)$.

Step 3: Substitute values.
\[ D = \frac{|6(2) - 3(5) + 2(-3) - 4|}{\sqrt{6^2+(-3)^2+2^2}} \] \[ = \frac{|12 - 15 - 6 - 4|}{\sqrt{36+9+4}} \] \[ = \frac{|-13|}{\sqrt{49}} = \frac{13}{7} \]

Final Answer: \[ \boxed{\dfrac{13}{7}} \]