Let `veca = hati + hatj + hatk` and let `vecr` be a variable vector such that `vecr.hati, vecr.hatj` and `vecr.hatk` are posititve integers. If `vecr.veca le 12`, then the total number of such vectors is:

Options

A. `""_(12)C_(9)-1`
B. `""_(12)C_(3)`
C. `""_(12)C_(8)`
D. None of these

Answer

To solve the problem, we need to find the total number of positive integral solutions to the equation derived from the dot product of two vectors. Let's break down the solution step by step.

### Step 1: Define the vectors
Let:
- \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\)
- \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\)

Where \(x\), \(y\), and \(z\) are positive integers.

### Step 2: Set up the dot product equation
The dot product \(\vec{r} \cdot \vec{a}\) can be expressed as:
\[ \vec{r} \cdot \vec{a} = x + y + z \]
According to the problem, we have:
\[ x + y + z \leq 12 \]

### Step 3: Convert the inequality to an equation
To find the number of positive integral solutions, we can rewrite the inequality as an equation by introducing a new variable \(k\):
\[ x + y + z + k = 12 \]
Where \(k\) is a non-negative integer.

### Step 4: Count the number of solutions
Now, we need to find the number of positive integral solutions for the equation \(x + y + z + k = 12\). Since \(x\), \(y\), and \(z\) must be positive integers, we can substitute \(x' = x - 1\), \(y' = y - 1\), and \(z' = z - 1\) (where \(x'\), \(y'\), and \(z'\) are non-negative integers):
\[ (x' + 1) + (y' + 1) + (z' + 1) + k = 12 \]
This simplifies to:
\[ x' + y' + z' + k = 9 \]

### Step 5: Apply the stars and bars theorem
The number of non-negative integral solutions to the equation \(x' + y' + z' + k = 9\) can be found using the stars and bars theorem. The formula for the number of solutions is given by:
\[ \text{Number of solutions} = \binom{n + r - 1}{r - 1} \]
Where \(n\) is the total (9 in this case) and \(r\) is the number of variables (4 variables: \(x'\), \(y'\), \(z'\), and \(k\)).

Thus, we have:
\[ \text{Number of solutions} = \binom{9 + 4 - 1}{4 - 1} = \binom{12}{3} \]

### Step 6: Calculate the final answer
Now we calculate \(\binom{12}{3}\):
\[ \binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \]

### Final Answer
The total number of such vectors \(\vec{r}\) is \(220\).

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