Let PN be the perpendicular from the point
`P(1, 2, 3)` to xy-plane. If OP makes an angle `alpha`
with positive direction of the z-axis and ON makes
an angle `beta` with the positive direction of x-axis,
where O is the origin (`alpha and beta ` are acute angles),
then

Options

A. `sin alpha sinbeta = 2/sqrt(14)`
B. `cos alphacos beta = 2/sqrt(14)`
C. `tan alpha = sqrt(5)/3`
D. `tan beta = 2`

Answer

To solve the problem step by step, we need to find the relationships between the angles and the coordinates of the point P(1, 2, 3) in relation to the origin O(0, 0, 0) and the xy-plane.

### Step 1: Identify the coordinates of point P and the origin O
- Given point \( P(1, 2, 3) \)
- The origin \( O(0, 0, 0) \)

### Step 2: Understand the geometry of the problem
- The point \( P \) projects perpendicularly onto the xy-plane at the point \( (1, 2, 0) \).
- The line \( OP \) makes an angle \( \alpha \) with the positive z-axis.
- The line \( ON \) (where \( N \) is the projection of \( P \) on the xy-plane) makes an angle \( \beta \) with the positive x-axis.

### Step 3: Calculate the length of \( OP \)
Using the distance formula:
\[ OP = \sqrt{(1-0)^2 + (2-0)^2 + (3-0)^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

### Step 4: Find \( \cos \alpha \)
From the definition of cosine in terms of the angle \( \alpha \):
\[ z = OP \cdot \cos \alpha \]
Substituting the known values:
\[ 3 = \sqrt{14} \cdot \cos \alpha \]
Thus,
\[ \cos \alpha = \frac{3}{\sqrt{14}} \]

### Step 5: Find \( OP \) in terms of \( \sin \alpha \)
Using the relationship:
\[ OM = OP \cdot \sin \alpha \]
Where \( OM \) is the projection of \( OP \) onto the xy-plane:
\[ OM = \sqrt{14} \cdot \sin \alpha \]

### Step 6: Find coordinates of point N
The coordinates of point \( N \) can be expressed as:
\[ N = (x, y, 0) = (OM \cdot \cos \beta, OM \cdot \sin \beta) \]
Substituting \( OM \):
\[ N = (\sqrt{14} \cdot \sin \alpha \cdot \cos \beta, \sqrt{14} \cdot \sin \alpha \cdot \sin \beta, 0) \]

### Step 7: Set up equations for x and y coordinates
From the coordinates of point \( P \):
1. \( x = 1 \)
2. \( y = 2 \)

Thus, we have:
\[ 1 = \sqrt{14} \cdot \sin \alpha \cdot \cos \beta \quad (1) \]
\[ 2 = \sqrt{14} \cdot \sin \alpha \cdot \sin \beta \quad (2) \]

### Step 8: Divide equations (2) by (1)
\[ \frac{2}{1} = \frac{\sqrt{14} \cdot \sin \alpha \cdot \sin \beta}{\sqrt{14} \cdot \sin \alpha \cdot \cos \beta} \]
This simplifies to:
\[ 2 = \frac{\sin \beta}{\cos \beta} = \tan \beta \]
Thus,
\[ \tan \beta = 2 \]

### Step 9: Find \( \sec^2 \beta \)
Using the identity:
\[ \tan^2 \beta + 1 = \sec^2 \beta \]
Substituting \( \tan \beta = 2 \):
\[ 4 + 1 = \sec^2 \beta \implies \sec^2 \beta = 5 \]
Thus,
\[ \sec \beta = \sqrt{5} \]

### Step 10: Find \( \cos \beta \)
Since \( \sec \beta = \frac{1}{\cos \beta} \):
\[ \cos \beta = \frac{1}{\sqrt{5}} \]

### Step 11: Find \( \sin \beta \)
Using \( \sin^2 \beta + \cos^2 \beta = 1 \):
\[ \sin^2 \beta + \left(\frac{1}{\sqrt{5}}\right)^2 = 1 \implies \sin^2 \beta + \frac{1}{5} = 1 \]
Thus,
\[ \sin^2 \beta = \frac{4}{5} \implies \sin \beta = \frac{2}{\sqrt{5}} \]

### Step 12: Find \( \sin \alpha \)
Using equation (1):
\[ 1 = \sqrt{14} \cdot \sin \alpha \cdot \frac{1}{\sqrt{5}} \implies \sin \alpha = \frac{\sqrt{5}}{\sqrt{14}} \]

### Step 13: Final results
We have:
- \( \tan \beta = 2 \)
- \( \sin \alpha = \frac{\sqrt{5}}{\sqrt{14}} \)
- \( \cos \alpha = \frac{3}{\sqrt{14}} \)