Through the point `P(h, k, l)` a plane is drawn at right angles to OP to meet co-ordinate axes at A, B and C. If OP=p, `A_xy` is area of projetion of `triangle(ABC)` on xy-plane. `A_zy` is area of projection of `triangle(ABC)` on yz-plane, then

Options

A. (a)`triangle=|(p^5)/(hkl)|`
B. (b)`triangle=|(p^5)/(2hkl)|`
C. (c)`(A_xy)/(A_yz)=|(1)/(h)|`
D. (d)`(A_xy)/(A_yz)=|(h)/(l)|`

Answer

To solve the problem, we need to find the areas of the projections of triangle ABC on the XY-plane and YZ-plane, given the point P(h, k, l) and the distance OP = p. Let's go through the solution step by step.

### Step 1: Determine the coordinates of points A, B, and C
The plane is drawn through point P(h, k, l) and is perpendicular to the line OP. The coordinates of the points where this plane meets the coordinate axes can be determined as follows:

- Point A (intersection with x-axis): When y = 0 and z = 0, we have:
\[ \frac{h}{p} x + \frac{k}{p} y + \frac{l}{p} z = 1 \implies A = \left(\frac{p^2}{h}, 0, 0\right) \]

- Point B (intersection with y-axis): When x = 0 and z = 0, we have:
\[ \frac{h}{p} x + \frac{k}{p} y + \frac{l}{p} z = 1 \implies B = \left(0, \frac{p^2}{k}, 0\right) \]

- Point C (intersection with z-axis): When x = 0 and y = 0, we have:
\[ \frac{h}{p} x + \frac{k}{p} y + \frac{l}{p} z = 1 \implies C = \left(0, 0, \frac{p^2}{l}\right) \]

### Step 2: Calculate the area of triangle ABC
The area of triangle ABC can be calculated using the formula for the area of a triangle formed by three points in 3D space. The area can be given by:
\[ \text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| \]
Where:
- \(\vec{AB} = B - A = \left(0 - \frac{p^2}{h}, \frac{p^2}{k} - 0, 0 - 0\right) = \left(-\frac{p^2}{h}, \frac{p^2}{k}, 0\right)\)
- \(\vec{AC} = C - A = \left(0 - \frac{p^2}{h}, 0 - 0, \frac{p^2}{l} - 0\right) = \left(-\frac{p^2}{h}, 0, \frac{p^2}{l}\right)\)

### Step 3: Calculate the cross product \(\vec{AB} \times \vec{AC}\)
The cross product can be calculated as follows:
\[ \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -\frac{p^2}{h} & \frac{p^2}{k} & 0 \\ -\frac{p^2}{h} & 0 & \frac{p^2}{l} \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \left(\frac{p^2}{k} \cdot \frac{p^2}{l} - 0\right) - \hat{j} \left(-\frac{p^2}{h} \cdot \frac{p^2}{l} - 0\right) + \hat{k} \left(-\frac{p^2}{h} \cdot 0 - \left(-\frac{p^2}{h} \cdot \frac{p^2}{k}\right)\right) \]
\[ = \hat{i} \frac{p^4}{kl} + \hat{j} \frac{p^4}{hl} + \hat{k} \frac{p^4}{hk} \]

### Step 4: Calculate the magnitude of the cross product
The magnitude of the cross product is:
\[ \left| \vec{AB} \times \vec{AC} \right| = \sqrt{\left(\frac{p^4}{kl}\right)^2 + \left(\frac{p^4}{hl}\right)^2 + \left(\frac{p^4}{hk}\right)^2} \]
\[ = p^4 \sqrt{\frac{1}{k^2l^2} + \frac{1}{h^2l^2} + \frac{1}{h^2k^2}} \]

### Step 5: Area of triangle ABC
Thus, the area of triangle ABC is:
\[ \text{Area} = \frac{1}{2} \left| \vec{AB} \times \vec{AC} \right| = \frac{p^4}{2} \sqrt{\frac{1}{k^2l^2} + \frac{1}{h^2l^2} + \frac{1}{h^2k^2}} \]

### Step 6: Projection areas
The areas of projection on the XY-plane and YZ-plane can be calculated as follows:
- Area of projection on XY-plane:
\[ A_{xy} = \frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot \frac{p^2}{h} \cdot \frac{p^2}{k} = \frac{p^4}{2hk} \]

- Area of projection on YZ-plane:
\[ A_{yz} = \frac{1}{2} \cdot \frac{p^2}{k} \cdot \frac{p^2}{l} = \frac{p^4}{2kl} \]

### Final Result
Thus, the areas of projections are:
- \(A_{xy} = \frac{p^4}{2hk}\)
- \(A_{yz} = \frac{p^4}{2kl}\)