Find the distance of the point (3, 3, 3) from the plane `vecr.(5hati+2hatj-7hatk)+9=0`.

Options

A. `(9)/(sqrt78)` units
B. `(5)/(sqrt78)` units
C. `(3)/(sqrt78)` units
D. `(4)/(sqrt78)` units

Answer

To find the distance of the point (3, 3, 3) from the plane given by the equation \(\vec{r} \cdot (5\hat{i} + 2\hat{j} - 7\hat{k}) + 9 = 0\), we can use the formula for the distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane given by the equation \(ax + by + cz + d = 0\):

\[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]

### Step 1: Identify the coefficients from the plane equation
From the plane equation \(\vec{r} \cdot (5\hat{i} + 2\hat{j} - 7\hat{k}) + 9 = 0\), we can identify:
- \(a = 5\)
- \(b = 2\)
- \(c = -7\)
- \(d = 9\)

### Step 2: Substitute the point coordinates into the formula
The coordinates of the point are \((x_1, y_1, z_1) = (3, 3, 3)\). We substitute these values into the distance formula:

\[ d = \frac{|5(3) + 2(3) - 7(3) + 9|}{\sqrt{5^2 + 2^2 + (-7)^2}} \]

### Step 3: Calculate the numerator
Calculating the numerator:

\[ 5(3) = 15 \]
\[ 2(3) = 6 \]
\[ -7(3) = -21 \]
Now, substituting these values into the numerator:

\[ 15 + 6 - 21 + 9 = 15 + 6 - 21 + 9 = 9 \]

So, the absolute value is:

\[ |9| = 9 \]

### Step 4: Calculate the denominator
Now, we calculate the denominator:

\[ \sqrt{5^2 + 2^2 + (-7)^2} = \sqrt{25 + 4 + 49} = \sqrt{78} \]

### Step 5: Combine the results to find the distance
Now we can combine the results to find the distance:

\[ d = \frac{9}{\sqrt{78}} \]

### Final Result
Thus, the distance of the point (3, 3, 3) from the plane is:

\[ d = \frac{9}{\sqrt{78}} \]