Let `P M` be the perpendicular from the point `P(1,2,3)` to the `x-y` plane. If ` vec O P` makes an angle `theta` with the positive direction of the `z-` axis and` vec O M` makes an angle `varphi` with the positive direction of `x-` axis, `w h e r eO` is the origin and `thetaa n dvarphi` are acute angels, then a. `costhetacosvarphi=1//sqrt(14)` b. `sinthetasinvarphi=2//sqrt(14)` c. `""tanvarphi=2` d. `tantheta=sqrt(5)//3`

Options

A. `costhetacosphi=1//sqrt14`
B. `sinthetasinphi=2//sqrt14`
C. `tanphi=2`
D. `tantheta=sqrt5//3`

Explanation

To solve the problem, we need to analyze the given point \( P(1, 2, 3) \) and its relationship with the angles \( \theta \) and \( \varphi \) with respect to the coordinate axes.

### Step 1: Understanding the Geometry
The point \( P(1, 2, 3) \) is located in three-dimensional space. The perpendicular \( PM \) from point \( P \) to the \( x-y \) plane means that the coordinates of point \( M \) will be \( (1, 2, 0) \).

### Step 2: Finding the Length of OP
The vector \( \vec{OP} \) can be represented as:
\[ \vec{OP} = (1, 2, 3) \]
The length \( r \) of \( \vec{OP} \) is given by:
\[ r = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

### Step 3: Finding \( \cos \theta \)
The angle \( \theta \) is the angle between the vector \( \vec{OP} \) and the positive direction of the \( z \)-axis. The cosine of this angle is given by:
\[ \cos \theta = \frac{z}{r} = \frac{3}{\sqrt{14}} \]

### Step 4: Finding \( \sin \theta \)
Using the Pythagorean identity, we can find \( \sin \theta \):
\[ \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{3}{\sqrt{14}}\right)^2} = \sqrt{1 - \frac{9}{14}} = \sqrt{\frac{5}{14}} = \frac{\sqrt{5}}{\sqrt{14}} \]

### Step 5: Finding \( \cos \varphi \) and \( \sin \varphi \)
The angle \( \varphi \) is the angle between the vector \( \vec{OM} \) and the positive direction of the \( x \)-axis. The coordinates of point \( M \) are \( (1, 2, 0) \), so:
\[ \vec{OM} = (1, 2, 0) \]
The length of \( \vec{OM} \) is:
\[ |\vec{OM}| = \sqrt{1^2 + 2^2} = \sqrt{5} \]
Thus:
\[ \cos \varphi = \frac{x}{|\vec{OM}|} = \frac{1}{\sqrt{5}}, \quad \sin \varphi = \frac{y}{|\vec{OM}|} = \frac{2}{\sqrt{5}} \]

### Step 6: Finding \( \tan \theta \) and \( \tan \varphi \)
Now we can find \( \tan \theta \) and \( \tan \varphi \):
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{\sqrt{5}}{\sqrt{14}}}{\frac{3}{\sqrt{14}}} = \frac{\sqrt{5}}{3} \]
\[ \tan \varphi = \frac{\sin \varphi}{\cos \varphi} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = 2 \]

### Conclusion
Now we can summarize the results:
- \( \cos \theta \cos \varphi = \frac{3}{\sqrt{14}} \cdot \frac{1}{\sqrt{5}} = \frac{3}{\sqrt{70}} \) (not an option)
- \( \sin \theta \sin \varphi = \frac{\sqrt{5}}{\sqrt{14}} \cdot \frac{2}{\sqrt{5}} = \frac{2}{\sqrt{14}} \) (option b)
- \( \tan \varphi = 2 \) (option c)
- \( \tan \theta = \frac{\sqrt{5}}{3} \) (option d)

### Final Answers
- a. False
- b. True
- c. True
- d. True