To find the vector \(\vec{A}\) given the direction cosines, we can follow these steps:
### Step 1: Understand the Direction Cosines
The direction cosines of a vector \(\vec{A}\) are given as:
- \(\cos \alpha = \frac{4}{5\sqrt{2}}\)
- \(\cos \beta = \frac{1}{\sqrt{2}}\)
- \(\cos \gamma = \frac{3}{5\sqrt{2}}\)
### Step 2: Relate Direction Cosines to Vector Components
The direction cosines are defined as:
\[
\cos \alpha = \frac{A}{\|\vec{A}\|}, \quad \cos \beta = \frac{B}{\|\vec{A}\|}, \quad \cos \gamma = \frac{C}{\|\vec{A}\|}
\]
where \(A\), \(B\), and \(C\) are the components of the vector \(\vec{A} = A\hat{i} + B\hat{j} + C\hat{k}\) and \(\|\vec{A}\|\) is the magnitude of the vector.
### Step 3: Set Up the Equations
From the definitions, we can express the components \(A\), \(B\), and \(C\) in terms of the magnitude \(\|\vec{A}\|\):
\[
A = \cos \alpha \cdot \|\vec{A}\|, \quad B = \cos \beta \cdot \|\vec{A}\|, \quad C = \cos \gamma \cdot \|\vec{A}\|
\]
### Step 4: Find the Magnitude
To find the magnitude \(\|\vec{A}\|\), we can use the relationship:
\[
\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1
\]
Calculating each term:
\[
\cos^2 \alpha = \left(\frac{4}{5\sqrt{2}}\right)^2 = \frac{16}{50} = \frac{8}{25}
\]
\[
\cos^2 \beta = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} = \frac{25}{50}
\]
\[
\cos^2 \gamma = \left(\frac{3}{5\sqrt{2}}\right)^2 = \frac{9}{50}
\]
Now, summing these:
\[
\frac{8}{25} + \frac{25}{50} + \frac{9}{50} = \frac{16}{50} + \frac{25}{50} + \frac{9}{50} = \frac{50}{50} = 1
\]
This confirms that the direction cosines are valid.
### Step 5: Calculate the Components
Assuming \(\|\vec{A}\| = k\), we can express the components:
\[
A = \frac{4}{5\sqrt{2}} \cdot k, \quad B = \frac{1}{\sqrt{2}} \cdot k, \quad C = \frac{3}{5\sqrt{2}} \cdot k
\]
### Step 6: Choose a Suitable Magnitude
To simplify calculations, we can choose \(k = 5\sqrt{2}\):
\[
A = 4, \quad B = 5, \quad C = 3
\]
### Step 7: Write the Vector
Thus, the vector \(\vec{A}\) can be expressed as:
\[
\vec{A} = 4\hat{i} + 5\hat{j} + 3\hat{k}
\]
### Conclusion
The vector \(\vec{A}\) is \(4\hat{i} + 5\hat{j} + 3\hat{k}\).
