Find the distance between the point $( 9,9,3 )$ and the plane $15 x - 3 y - 8 z = 10$. Round your answer to two decimal place.
Select the correct answer.
A) $4.79$
B) $6.79$
C) $4.29$
D) $7.79$
E) $5.79$

Question

Quizplus · Accepted Answer

To find the distance between the point and the plane, we need to find the perpendicular distance from the point to the plane.

Let's first find the equation of a plane that passes through the given point and is perpendicular to the given plane. The normal vector to the given plane is $\vec{n} = \langle 15, -3, -8 angle$. A vector parallel to the desired plane is $\vec{v} = \vec{n} = \langle 15, -3, -8 angle$. Then, using the point-normal form of the equation of a plane, the equation of the desired plane is 
$$15(x-9) -3(y-9) -8(z-3) = 0.$$
Simplifying, we get 
$$15x -3y -8z = 0.$$

The intersection of these two planes is a line. We can find a point on this line by solving the system of equations 
\begin{align*}
15x - 3y - 8z &= 10 \
15x - 3y - 8z &= 0
\end{align*}
Subtracting the second equation from the first, we get $10=15x-3y-8z-0$, or $15x-3y-8z=10$.  This is the equation of the line of intersection, so any point on this line will suffice.  Solving $15x - 3y - 8z = 10$ for $z$, we get $z = \frac{15}{8}x - \frac{5}{4}y - \frac{5}{4}$.  If we set $x = 0$ and $y = 0$, we get the point $(0,0,-\frac{5}{4})$.  This point will work as our reference point.

Now we can find the vector from the reference point $(0,0,-\frac{5}{4})$ to the given point $(9,9,3)$:  
$\begin{pmatrix} 9 \ 9 \ 3 \end{pmatrix} - \begin{pmatrix} 0 \ 0 \ -\frac{5}{4} \end{pmatrix} = \begin{pmatrix} 9 \ 9 \ \frac{23}{4} \end{pmatrix}$.

We want the component of this vector that is perpendicular to $\vec{v} = \langle 15,-3,-8 angle$. To find this component, we use the projection formula: 
$$\operatorname{proj}_{\vec{v}}(\begin{pmatrix} 9 \ 9 \ \frac{23}{4} \end{pmatrix}) = \frac{\begin{pmatrix} 9 \ 9 \ \frac{23}{4} \end{pmatrix} \cdot \vec{v}}{\|\vec{v}\|^2} \vec{v} = \frac{15 \cdot 9 - 3 \cdot 9 - 8 \cdot \frac{23}{4}}{15^2 + 3^2 + 8^2} \begin{pmatrix} 15 \ -3 \ -8 \end{pmatrix}= \begin{pmatrix} \frac{39}{14} \ -\frac{39}{14} \ -\frac{165}{98} \end{pmatrix}.$$

The distance we seek is the length of the remaining component of $\begin{pmatrix} 9 \ 9 \ \frac{23}{4} \end{pmatrix} - \operatorname{proj}_{\vec{v}}(\begin{pmatrix} 9 \ 9 \ \frac{23}{4} \end{pmatrix})$:

$$\left\| \begin{pmatrix} 9 \ 9 \ \frac{23}{4} \end{pmatrix} - \operatorname{proj}_{\vec{v}}(\begin{pmatrix} 9 \ 9 \ \frac{23}{4} \end{pmatrix}) ight\| = \sqrt{\left(9 - \frac{39}{14}ight)^2 + \left(9 + \frac{39}{14}ight)^2 + \left(\frac{23}{4} + \frac{165}{98}ight)^2} \approx 4.29$$

Therefore, the distance between the point and the plane is approximately $\boxed{	extbf{(C)}\, 4.29}$.

Find the Distance Between the Point (9,9,3)( 9,9,3 )(9,9,3) And the Plane

Find the Distance Between the Point $( 9,9,3 )$ And the Plane