A vector `vecr` is inclined to x-axis at `45^(@)` and to y-axis at `60^(@)`. If `|vecr|=8` units find `vecr`.

Answer

To find the vector \(\vec{r}\) that is inclined to the x-axis at \(45^\circ\) and to the y-axis at \(60^\circ\) with a magnitude of \(8\) units, we can follow these steps:

### Step 1: Determine the direction cosines
The direction cosines \(l\), \(m\), and \(n\) of the vector \(\vec{r}\) can be determined using the angles given:

- The direction cosine \(l\) with respect to the x-axis is given by:
\[ l = \cos(45^\circ) = \frac{1}{\sqrt{2}} \]

- The direction cosine \(m\) with respect to the y-axis is given by:
\[ m = \cos(60^\circ) = \frac{1}{2} \]

- The direction cosine \(n\) with respect to the z-axis can be calculated using the property that the sum of the squares of the direction cosines equals \(1\):
\[ l^2 + m^2 + n^2 = 1 \]

### Step 2: Substitute the values of \(l\) and \(m\)
Substituting the values of \(l\) and \(m\):
\[ \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + n^2 = 1 \]
Calculating the squares:
\[ \frac{1}{2} + \frac{1}{4} + n^2 = 1 \]

### Step 3: Solve for \(n^2\)
To solve for \(n^2\), we first find a common denominator:
\[ \frac{2}{4} + \frac{1}{4} + n^2 = 1 \]
This simplifies to:
\[ \frac{3}{4} + n^2 = 1 \]
Subtracting \(\frac{3}{4}\) from both sides:
\[ n^2 = 1 - \frac{3}{4} = \frac{1}{4} \]
Taking the square root:
\[ n = \pm \frac{1}{2} \]

### Step 4: Write the vector in component form
Now that we have the direction cosines:
- \(l = \frac{1}{\sqrt{2}}\)
- \(m = \frac{1}{2}\)
- \(n = \pm \frac{1}{2}\)

The vector \(\vec{r}\) can be expressed as:
\[ \vec{r} = |r| (l \hat{i} + m \hat{j} + n \hat{k}) \]
Given that \(|r| = 8\):
\[ \vec{r} = 8 \left(\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{2} \hat{j} + \left(\pm \frac{1}{2}\right) \hat{k}\right) \]

### Step 5: Simplify the vector
Distributing \(8\):
\[ \vec{r} = 8 \cdot \frac{1}{\sqrt{2}} \hat{i} + 8 \cdot \frac{1}{2} \hat{j} + 8 \cdot \left(\pm \frac{1}{2}\right) \hat{k} \]
This simplifies to:
\[ \vec{r} = 4\sqrt{2} \hat{i} + 4 \hat{j} + 4 \left(\pm 1\right) \hat{k} \]
Thus, the vector \(\vec{r}\) can be expressed as:
\[ \vec{r} = 4\sqrt{2} \hat{i} + 4 \hat{j} + 4 \hat{k} \quad \text{or} \quad \vec{r} = 4\sqrt{2} \hat{i} + 4 \hat{j} - 4 \hat{k} \]

### Final Result
The vector \(\vec{r}\) is:
\[ \vec{r} = 4\sqrt{2} \hat{i} + 4 \hat{j} + 4 \hat{k} \quad \text{or} \quad \vec{r} = 4\sqrt{2} \hat{i} + 4 \hat{j} - 4 \hat{k} \]