Let `vec(a) = hat(i) + hat(j), vec(b) = 3 hat(i) + 4 hat(k) and vec (b) = vec(c) + vec(d)`, where `vec(c)` is parallel to `vec(a) and vec(d) ` is perpendicular to `vec(a)`.
What is `vec(c)` equal to ?

Options

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

Explanation

To solve the problem step by step, we need to find the vector \(\vec{c}\) given the conditions that \(\vec{c}\) is parallel to \(\vec{a}\) and \(\vec{d}\) is perpendicular to \(\vec{a}\).

### Step 1: Define the vectors
We have:
\[ \vec{a} = \hat{i} + \hat{j} \]
\[ \vec{b} = 3\hat{i} + 4\hat{k} \]
We also know that:
\[ \vec{b} = \vec{c} + \vec{d} \]

### Step 2: Use the dot product
Since \(\vec{c}\) is parallel to \(\vec{a}\), we can express \(\vec{c}\) as:
\[ \vec{c} = k\vec{a} = k(\hat{i} + \hat{j}) \quad \text{for some scalar } k \]

### Step 3: Find \(\vec{d}\)
From the equation \(\vec{b} = \vec{c} + \vec{d}\), we can rearrange it to find \(\vec{d}\):
\[ \vec{d} = \vec{b} - \vec{c} = (3\hat{i} + 4\hat{k}) - k(\hat{i} + \hat{j}) = (3 - k)\hat{i} - k\hat{j} + 4\hat{k} \]

### Step 4: Use the perpendicular condition
Since \(\vec{d}\) is perpendicular to \(\vec{a}\), we can use the dot product:
\[ \vec{a} \cdot \vec{d} = 0 \]
Calculating the dot product:
\[ (\hat{i} + \hat{j}) \cdot ((3 - k)\hat{i} - k\hat{j} + 4\hat{k}) = (1)(3 - k) + (1)(-k) + (0)(4) = 3 - k - k = 3 - 2k \]
Setting this equal to zero gives:
\[ 3 - 2k = 0 \implies 2k = 3 \implies k = \frac{3}{2} \]

### Step 5: Calculate \(\vec{c}\)
Now substituting \(k\) back into the equation for \(\vec{c}\):
\[ \vec{c} = k(\hat{i} + \hat{j}) = \frac{3}{2}(\hat{i} + \hat{j}) = \frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} \]

### Final Answer
Thus, the vector \(\vec{c}\) is:
\[ \vec{c} = \frac{3}{2}\hat{i} + \frac{3}{2}\hat{j} \]