If `vec(beta)` is perpendicular to both `vec(alpha) and vec(gamma)` where `vec(alpha)= hat(k) and vec(gamma) = 2hat(i) + 3hat(j) + 4hat(k)`, then what is `vec(beta)` equal to ?

Options

A. `3hat(i) + 2hat(j)`
B. `-3hat(i) + 2hat(j)`
C. `2hat(i) - 3hat(j)`
D. `-2hat(i) + 3hat(j)`

Answer

To find the vector \(\vec{\beta}\) that is perpendicular to both \(\vec{\alpha}\) and \(\vec{\gamma}\), we can follow these steps:

### Step 1: Define the vectors
We know:
- \(\vec{\alpha} = \hat{k}\)
- \(\vec{\gamma} = 2\hat{i} + 3\hat{j} + 4\hat{k}\)

Let \(\vec{\beta} = a\hat{i} + b\hat{j} + c\hat{k}\).

### Step 2: Use the condition of perpendicularity
For two vectors to be perpendicular, their dot product must equal zero. Therefore, we need to set up the following equations based on the dot products:

1. \(\vec{\alpha} \cdot \vec{\beta} = 0\)
2. \(\vec{\gamma} \cdot \vec{\beta} = 0\)

### Step 3: Calculate the dot product with \(\vec{\alpha}\)
\[ \vec{\alpha} \cdot \vec{\beta} = \hat{k} \cdot (a\hat{i} + b\hat{j} + c\hat{k}) = c \]
Setting this equal to zero gives:
\[ c = 0 \]

### Step 4: Substitute \(c\) into \(\vec{\beta}\)
Now we can rewrite \(\vec{\beta}\) as:
\[ \vec{\beta} = a\hat{i} + b\hat{j} \]

### Step 5: Calculate the dot product with \(\vec{\gamma}\)
Now we calculate the dot product with \(\vec{\gamma}\):
\[ \vec{\gamma} \cdot \vec{\beta} = (2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (a\hat{i} + b\hat{j}) = 2a + 3b + 4c \]
Since we already know \(c = 0\), we have:
\[ 2a + 3b = 0 \]

### Step 6: Solve for \(a\) in terms of \(b\)
Rearranging the equation gives:
\[ 2a = -3b \implies a = -\frac{3}{2}b \]

### Step 7: Write the final form of \(\vec{\beta}\)
Substituting \(a\) back into \(\vec{\beta}\):
\[ \vec{\beta} = -\frac{3}{2}b \hat{i} + b \hat{j} \]
Factoring out \(b\):
\[ \vec{\beta} = b\left(-\frac{3}{2} \hat{i} + \hat{j}\right) \]

### Conclusion
Thus, \(\vec{\beta}\) can be expressed as:
\[ \vec{\beta} = k\left(-\frac{3}{2} \hat{i} + \hat{j}\right) \quad \text{for any scalar } k \]