Let `vecb=4hati+3hatj and vecc` be two vectors perpendicular to each other in the xy-plane. Find all vetors in te same plane having projection 1 and 2 along `vecb and vecc` respectively.

Explanation

To solve the problem, we need to find all vectors in the xy-plane that have specific projections along two given vectors, \(\vec{b}\) and \(\vec{c}\), which are perpendicular to each other. Here are the steps to arrive at the solution:

### Step 1: Identify the Given Vectors
We have:
\[ \vec{b} = 4\hat{i} + 3\hat{j} \]
Since \(\vec{c}\) is perpendicular to \(\vec{b}\), we can find \(\vec{c}\) by ensuring that the dot product \(\vec{b} \cdot \vec{c} = 0\).

### Step 2: Find the Perpendicular Vector \(\vec{c}\)
Let \(\vec{c} = a\hat{i} + b\hat{j}\). The condition for perpendicularity gives us:
\[ \vec{b} \cdot \vec{c} = (4\hat{i} + 3\hat{j}) \cdot (a\hat{i} + b\hat{j}) = 4a + 3b = 0 \]
From this equation, we can express \(b\) in terms of \(a\):
\[ 3b = -4a \implies b = -\frac{4}{3}a \]
Choosing \(a = 3\) gives \(b = -4\). Thus, we can take:
\[ \vec{c} = 3\hat{i} - 4\hat{j} \]

### Step 3: Set Up the Vector \(\vec{a}\)
Let \(\vec{a} = \alpha \vec{b} + \beta \vec{c}\):
\[ \vec{a} = \alpha(4\hat{i} + 3\hat{j}) + \beta(3\hat{i} - 4\hat{j}) = (4\alpha + 3\beta)\hat{i} + (3\alpha - 4\beta)\hat{j} \]

### Step 4: Find Projections
The projection of \(\vec{a}\) along \(\vec{b}\) is given as 1:
\[ \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = 1 \]
Calculating \(|\vec{b}|\):
\[ |\vec{b}| = \sqrt{4^2 + 3^2} = 5 \]
Thus:
\[ \vec{a} \cdot \vec{b} = 1 \cdot 5 = 5 \]
Calculating \(\vec{a} \cdot \vec{b}\):
\[ (4\alpha + 3\beta)(4) + (3\alpha - 4\beta)(3) = 16\alpha + 12\beta + 9\alpha - 12\beta = 25\alpha = 5 \]
This gives:
\[ \alpha = \frac{1}{5} \]

The projection of \(\vec{a}\) along \(\vec{c}\) is given as 2:
\[ \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} = 2 \]
Calculating \(|\vec{c}|\):
\[ |\vec{c}| = \sqrt{3^2 + (-4)^2} = 5 \]
Thus:
\[ \vec{a} \cdot \vec{c} = 2 \cdot 5 = 10 \]
Calculating \(\vec{a} \cdot \vec{c}\):
\[ (4\alpha + 3\beta)(3) + (3\alpha - 4\beta)(-4) = 12\alpha + 9\beta - 12\alpha + 16\beta = 25\beta = 10 \]
This gives:
\[ \beta = \frac{2}{5} \]

### Step 5: Substitute \(\alpha\) and \(\beta\) Back into \(\vec{a}\)
Now substituting \(\alpha\) and \(\beta\) back into \(\vec{a}\):
\[ \vec{a} = \frac{1}{5}(4\hat{i} + 3\hat{j}) + \frac{2}{5}(3\hat{i} - 4\hat{j}) \]
Calculating:
\[ \vec{a} = \frac{1}{5}(4\hat{i} + 3\hat{j}) + \frac{6}{5}\hat{i} - \frac{8}{5}\hat{j} = \frac{4 + 6}{5}\hat{i} + \frac{3 - 8}{5}\hat{j} = \frac{10}{5}\hat{i} - \frac{5}{5}\hat{j} = 2\hat{i} - \hat{j} \]

### Step 6: Final Result
Thus, the vector \(\vec{a}\) is:
\[ \vec{a} = 2\hat{i} - \hat{j} \]