The vector `vecB = 3j + 4k` is to be written as the sum of a vector `vecB_(1)` parallel to `vecA = i+j` and a vector `vecB_(2)` perpendicular to `vecA`. Then `vecB_(1)` =

Answer

To solve the problem of expressing the vector \(\vec{B} = 3\vec{j} + 4\vec{k}\) as the sum of a vector \(\vec{B_1}\) parallel to \(\vec{A} = \vec{i} + \vec{j}\) and a vector \(\vec{B_2}\) perpendicular to \(\vec{A}\), we will follow these steps:

### Step 1: Write the equation for \(\vec{B}\)
We can express \(\vec{B}\) as the sum of \(\vec{B_1}\) and \(\vec{B_2}\):
\[ \vec{B} = \vec{B_1} + \vec{B_2} \]

### Step 2: Express \(\vec{B_1}\) in terms of a scalar multiple of \(\vec{A}\)
Since \(\vec{B_1}\) is parallel to \(\vec{A}\), we can write:
\[ \vec{B_1} = \lambda \vec{A} = \lambda (\vec{i} + \vec{j}) = \lambda \vec{i} + \lambda \vec{j} \]
where \(\lambda\) is a scalar.

### Step 3: Express \(\vec{B_2}\)
From the equation \(\vec{B} = \vec{B_1} + \vec{B_2}\), we can express \(\vec{B_2}\) as:
\[ \vec{B_2} = \vec{B} - \vec{B_1} = (3\vec{j} + 4\vec{k}) - (\lambda \vec{i} + \lambda \vec{j}) \]
This simplifies to:
\[ \vec{B_2} = -\lambda \vec{i} + (3 - \lambda) \vec{j} + 4\vec{k} \]

### Step 4: Set up the condition for perpendicularity
Since \(\vec{B_2}\) is perpendicular to \(\vec{A}\), their dot product must be zero:
\[ \vec{B_2} \cdot \vec{A} = 0 \]
Calculating the dot product:
\[ (-\lambda \vec{i} + (3 - \lambda) \vec{j} + 4\vec{k}) \cdot (\vec{i} + \vec{j}) = 0 \]
This expands to:
\[ -\lambda \cdot 1 + (3 - \lambda) \cdot 1 + 0 = 0 \]
Simplifying gives:
\[ -\lambda + 3 - \lambda = 0 \implies 3 - 2\lambda = 0 \]

### Step 5: Solve for \(\lambda\)
From the equation \(3 - 2\lambda = 0\), we can solve for \(\lambda\):
\[ 2\lambda = 3 \implies \lambda = \frac{3}{2} \]

### Step 6: Substitute \(\lambda\) back to find \(\vec{B_1}\)
Now we can substitute \(\lambda\) back into the expression for \(\vec{B_1}\):
\[ \vec{B_1} = \lambda \vec{A} = \frac{3}{2} (\vec{i} + \vec{j}) = \frac{3}{2} \vec{i} + \frac{3}{2} \vec{j} \]

### Final Answer
Thus, the vector \(\vec{B_1}\) is:
\[ \vec{B_1} = \frac{3}{2} \vec{i} + \frac{3}{2} \vec{j} \]

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