Find the length of the perpendicular from origin to the plane `vecr.(3i-4j-12k)+39=0`.

Answer

To find the length of the perpendicular from the origin to the plane given by the equation \(\vec{r} \cdot (3\hat{i} - 4\hat{j} - 12\hat{k}) + 39 = 0\), we can follow these steps:

### Step 1: Write the equation of the plane in standard form
The equation of the plane can be expressed as:
\[ \vec{r} \cdot (3\hat{i} - 4\hat{j} - 12\hat{k}) + 39 = 0 \]
This can be rewritten as:
\[ 3x - 4y - 12z + 39 = 0 \]
where \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\).

### Step 2: Identify the coefficients of the plane equation
From the equation \(3x - 4y - 12z + 39 = 0\), we can identify the coefficients:
- \(A = 3\)
- \(B = -4\)
- \(C = -12\)
- \(D = 39\)

### Step 3: Use the formula for the distance from a point to a plane
The formula for the distance \(d\) from a point \((x_0, y_0, z_0)\) to the plane \(Ax + By + Cz + D = 0\) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
In our case, the point is the origin \((0, 0, 0)\), so \(x_0 = 0\), \(y_0 = 0\), and \(z_0 = 0\).

### Step 4: Substitute the values into the distance formula
Substituting the values into the formula gives:
\[ d = \frac{|3(0) - 4(0) - 12(0) + 39|}{\sqrt{3^2 + (-4)^2 + (-12)^2}} \]
This simplifies to:
\[ d = \frac{|39|}{\sqrt{9 + 16 + 144}} \]

### Step 5: Calculate the denominator
Calculating the denominator:
\[ \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \]

### Step 6: Calculate the distance
Now substituting back into the distance formula:
\[ d = \frac{39}{13} = 3 \]

### Final Answer
Thus, the length of the perpendicular from the origin to the plane is \(3\) units.

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