The points `hat(i)-hat(j)+3hat(k) and 3hat(i)+3hat(j)+3hat(k)` are equidistant from the plane `rcdot(5hat(i)+2hat(j)-7hat(k))+9=0`, then they are

Options

A. on the same sides of the plane
B. parallel of the plane
C. on the opposite sides of the plane
D. None of these

Answer

To solve the problem, we need to show that the two given points are equidistant from the specified plane. Let's break down the solution step by step.

### Step 1: Identify the Points and the Plane
The points given are:
- Point A: \(\hat{i} - \hat{j} + 3\hat{k}\)
- Point B: \(3\hat{i} + 3\hat{j} + 3\hat{k}\)

The equation of the plane is given by:
\[ \mathbf{r} \cdot (5\hat{i} + 2\hat{j} - 7\hat{k}) + 9 = 0 \]

### Step 2: Find the Normal Vector of the Plane
From the equation of the plane, we can identify the normal vector \(\mathbf{n}\) as:
\[ \mathbf{n} = 5\hat{i} + 2\hat{j} - 7\hat{k} \]

### Step 3: Calculate the Distance from Each Point to the Plane
The distance \(d\) from a point \(\mathbf{P}(x_1, y_1, z_1)\) to the plane defined by \(\mathbf{n} \cdot \mathbf{r} + d = 0\) is given by the formula:
\[ d = \frac{|\mathbf{n} \cdot \mathbf{P} + d|}{|\mathbf{n}|} \]

#### Distance from Point A
For Point A \((1, -1, 3)\):
\[ \mathbf{n} \cdot \mathbf{P_A} = 5(1) + 2(-1) - 7(3) = 5 - 2 - 21 = -18 \]
Thus, the distance from Point A to the plane is:
\[ d_A = \frac{|-18 + 9|}{\sqrt{5^2 + 2^2 + (-7)^2}} = \frac{|-9|}{\sqrt{25 + 4 + 49}} = \frac{9}{\sqrt{78}} = \frac{9}{\sqrt{78}} \]

#### Distance from Point B
For Point B \((3, 3, 3)\):
\[ \mathbf{n} \cdot \mathbf{P_B} = 5(3) + 2(3) - 7(3) = 15 + 6 - 21 = 0 \]
Thus, the distance from Point B to the plane is:
\[ d_B = \frac{|0 + 9|}{\sqrt{78}} = \frac{9}{\sqrt{78}} \]

### Step 4: Compare the Distances
We have:
- Distance from Point A to the plane: \(d_A = \frac{9}{\sqrt{78}}\)
- Distance from Point B to the plane: \(d_B = \frac{9}{\sqrt{78}}\)

Since \(d_A = d_B\), we conclude that both points are equidistant from the plane.

### Conclusion
The points \(\hat{i} - \hat{j} + 3\hat{k}\) and \(3\hat{i} + 3\hat{j} + 3\hat{k}\) are equidistant from the plane given by the equation \(\mathbf{r} \cdot (5\hat{i} + 2\hat{j} - 7\hat{k}) + 9 = 0\).

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