The position vector of a point `P` is `vecr=xhati+yhatj+zhatk` where `x,y,zepsilonN` and `veca=hati+hatj+hatk`. If `vecr.veca=10`, then the number of possible position of `P` is

Options

A. 36
B. 72
C. 66
D. none of these

Explanation

To solve the problem step by step, we need to find the number of possible positions of point \( P \) given its position vector \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) and the vector \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) such that \( \vec{r} \cdot \vec{a} = 10 \), where \( x, y, z \in \mathbb{N} \) (the set of natural numbers).

### Step 1: Set up the dot product equation
The dot product of the vectors \( \vec{r} \) and \( \vec{a} \) is given by:
\[ \vec{r} \cdot \vec{a} = (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = x + y + z \]
According to the problem, this dot product equals 10:
\[ x + y + z = 10 \]

### Step 2: Understand the constraints on \( x, y, z \)
Since \( x, y, z \) are natural numbers, they must be at least 1. This means:
\[ x \geq 1, \quad y \geq 1, \quad z \geq 1 \]

### Step 3: Substitute to simplify the equation
To simplify the equation, we can substitute \( x' = x - 1 \), \( y' = y - 1 \), and \( z' = z - 1 \). This gives us:
\[ x' + 1 + y' + 1 + z' + 1 = 10 \]
which simplifies to:
\[ x' + y' + z' = 7 \]
where \( x', y', z' \geq 0 \) (since \( x, y, z \) are at least 1).

### Step 4: Use the stars and bars combinatorial method
We need to find the number of non-negative integer solutions to the equation \( x' + y' + z' = 7 \). This can be solved using the "stars and bars" theorem, which states that the number of ways to distribute \( n \) indistinguishable objects (stars) into \( r \) distinguishable boxes (variables) is given by:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \( n = 7 \) (the total we need to distribute) and \( r = 3 \) (the three variables \( x', y', z' \)):
\[ \text{Number of solutions} = \binom{7 + 3 - 1}{3 - 1} = \binom{9}{2} \]

### Step 5: Calculate the binomial coefficient
Now we compute \( \binom{9}{2} \):
\[ \binom{9}{2} = \frac{9 \times 8}{2 \times 1} = \frac{72}{2} = 36 \]

### Conclusion
Thus, the number of possible positions of point \( P \) is \( 36 \).

---