SOLUTION: Find two vectors, V1 and V2 whose sum is <-2,0>, where V1 is parallel to <0,2> while V2 is perpendicular to <0,2>.

Question 1166411: Find two vectors, V1 and V2 whose sum is <-2,0>, where V1 is parallel to <0,2> while V2 is perpendicular to <0,2>.
Answer by CPhill(2138) (Show Source): You can put this solution on YOUR website!
This problem can be solved by setting up a system of linear equations based on the properties of parallel and perpendicular vectors in the plane.
The two vectors are:
* $\mathbf{V}_1 = \langle 0, 2 \rangle$
* $\mathbf{V}_2 = \langle -2, 0 \rangle$
---
## 📐 Setting Up the Equations
Let the two unknown vectors be:
$$\mathbf{V}_1 = \langle a, b \rangle$$
$$\mathbf{V}_2 = \langle c, d \rangle$$
### 1. The Sum Condition
The sum of the two vectors is $\langle -2, 0 \rangle$:
$$\mathbf{V}_1 + \mathbf{V}_2 = \langle a + c, b + d \rangle = \langle -2, 0 \rangle$$
This gives us two scalar equations:
1. $a + c = -2$
2. $b + d = 0$
### 2. The Parallel Condition ($\mathbf{V}_1$ is parallel to $\langle 0, 2 \rangle$)
If $\mathbf{V}_1$ is parallel to $\langle 0, 2 \rangle$, then $\mathbf{V}_1$ must be a scalar multiple ($k$) of $\langle 0, 2 \rangle$:
$$\mathbf{V}_1 = k \langle 0, 2 \rangle = \langle 0, 2k \rangle$$
Comparing this to $\mathbf{V}_1 = \langle a, b \rangle$:
3. $a = 0$
4. $b = 2k$
### 3. The Perpendicular Condition ($\mathbf{V}_2$ is perpendicular to $\langle 0, 2 \rangle$)
Two vectors are perpendicular if their **dot product** is zero
[Image of two perpendicular vectors showing the dot product is zero]
.
$$\mathbf{V}_2 \cdot \langle 0, 2 \rangle = 0$$
$$\langle c, d \rangle \cdot \langle 0, 2 \rangle = (c)(0) + (d)(2) = 0$$
$$2d = 0$$
5. $d = 0$
---
## 🧠 Solving the System
Now we substitute the determined values back into the sum conditions:
* From Equation 3, we know **$a = 0$**.
* From Equation 5, we know **$d = 0$**.
1. **Use $a=0$ in the first sum equation:**
$$a + c = -2$$
$$0 + c = -2$$
$$\mathbf{c = -2}$$
2. **Use $d=0$ in the second sum equation:**
$$b + d = 0$$
$$b + 0 = 0$$
$$\mathbf{b = 0}$$
### Final Vectors
Substitute the values back into the vector definitions:
$$\mathbf{V}_1 = \langle a, b \rangle = \langle 0, 0 \rangle$$
$$\mathbf{V}_2 = \langle c, d \rangle = \langle -2, 0 \rangle$$
---
**Wait! Let's re-read the perpendicular condition.**
The vector $\mathbf{V}_1 = \langle 0, 0 \rangle$ is parallel to any vector. $\mathbf{V}_2 = \langle -2, 0 \rangle$ is perpendicular to $\langle 0, 2 \rangle$ because $\langle -2, 0 \rangle \cdot \langle 0, 2 \rangle = 0$.
However, there is a fundamental issue with $\mathbf{V}_1 = \langle 0, 0 \rangle$ being the unique solution for the parallel vector. A vector parallel to $\langle 0, 2 \rangle$ **must only have a vertical component**.
Let $\mathbf{P} = \langle 0, 2 \rangle$.
The space of vectors parallel to $\mathbf{P}$ is the set $S_{\parallel} = \{\langle 0, y \rangle\}$.
The space of vectors perpendicular to $\mathbf{P}$ is the set $S_{\perp} = \{\langle x, 0 \rangle\}$ (since $2y=0$ implies $y=0$ in the dot product).
Let $\mathbf{V}_1 = \langle 0, y_1 \rangle$ and $\mathbf{V}_2 = \langle x_2, 0 \rangle$.
Their sum must be $\langle -2, 0 \rangle$:
$$\mathbf{V}_1 + \mathbf{V}_2 = \langle 0 + x_2, y_1 + 0 \rangle = \langle -2, 0 \rangle$$
By comparison of components:
* $x_2 = -2$
* $y_1 = 0$
This confirms the initial result:
$$\mathbf{V}_1 = \langle 0, 0 \rangle$$
$$\mathbf{V}_2 = \langle -2, 0 \rangle$$
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