If a plane cuts intercepts of lengths `8,4 and4` units on the coordinate axes respectively, then the length of perpendicular from origin to the plane is

Options

A. `3/8` units
B. `8/3` units
C. 8 units
D. `4/3` units

Answer

To find the length of the perpendicular from the origin to the plane that intercepts the coordinate axes at lengths of 8, 4, and 4 units, we can follow these steps:

### Step 1: Write the equation of the plane using intercept form
The intercept form of the equation of a plane is given by:

\[ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 \]

where \(a\), \(b\), and \(c\) are the x, y, and z intercepts of the plane respectively. Here, \(a = 8\), \(b = 4\), and \(c = 4\).

### Step 2: Substitute the intercepts into the equation
Substituting the values of the intercepts into the equation, we have:

\[ \frac{x}{8} + \frac{y}{4} + \frac{z}{4} = 1 \]

### Step 3: Clear the denominators
To eliminate the denominators, we can multiply the entire equation by the least common multiple (LCM) of the denominators, which is 8:

\[ 8 \left(\frac{x}{8}\right) + 8 \left(\frac{y}{4}\right) + 8 \left(\frac{z}{4}\right) = 8 \]

This simplifies to:

\[ x + 2y + 2z = 8 \]

### Step 4: Rearrange the equation into standard form
Rearranging the equation gives us:

\[ x + 2y + 2z - 8 = 0 \]

### Step 5: Use the distance formula 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 \((0, 0, 0)\), and the coefficients from the plane equation \(x + 2y + 2z - 8 = 0\) are \(A = 1\), \(B = 2\), \(C = 2\), and \(D = -8\).

### Step 6: Substitute into the distance formula
Substituting the values into the distance formula gives:

\[ D = \frac{|1(0) + 2(0) + 2(0) - 8|}{\sqrt{1^2 + 2^2 + 2^2}} \]

This simplifies to:

\[ D = \frac{|-8|}{\sqrt{1 + 4 + 4}} = \frac{8}{\sqrt{9}} = \frac{8}{3} \]

### Conclusion
Thus, the length of the perpendicular from the origin to the plane is:

\[ \frac{8}{3} \text{ units} \]

---