Solved 19 Vector Addition A2-QRT04: VECTOR GRAPHICAL | Chegg.com - Free Printable

Looking at the image, we are given three base vectors:

- $\vec{P}$: Points left (←), magnitude = 4
- $\vec{Q}$: Points up and right (↗), magnitude = 3
- $\vec{R}$: Points straight up (↑), magnitude = 2

We are to construct vectors $\vec{A}, \vec{B}, \vec{C}, \vec{D}, \vec{E}$ by adding or subtracting *two* of these base vectors, following the graphical method shown in the example.

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🔍 Example Analysis:

In the example, $\vec{X} = -\vec{P} + \vec{R}$

- $-\vec{P}$: reverse direction of $\vec{P}$ → points right (→), length 4
- $\vec{R}$: points up (↑), length 2
- Graphically, they are drawn head-to-tail: start with $-\vec{P}$ (→), then add $\vec{R}$ (↑) starting from its tip → resultant $\vec{X}$ goes from tail of first to head of second.

This is the head-to-tail method of vector addition.

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Now let’s analyze each case A–E.

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## ✔ Case A: Vector $\vec{A}$ — pointing up-right (↗)

This looks very similar to $\vec{Q}$, but perhaps a bit longer or different angle. Let’s try combinations.

Try: $\vec{A} = \vec{Q} + \vec{R}$

- $\vec{Q}$: ↗ (magnitude 3)
- $\vec{R}$: ↑ (magnitude 2)
- Adding them head-to-tail: start with $\vec{Q}$, then attach $\vec{R}$ upward from its tip → resultant will be more vertical than $\vec{Q}$ alone, but still up-right.

Visually, this matches the direction and approximate magnitude of $\vec{A}$.

✔ So:
> $\vec{A} = \vec{Q} + \vec{R}$

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## ✔ Case B: Vector $\vec{B}$ — pointing down-right (↘)

This is in the opposite quadrant of $\vec{Q}$ or $\vec{P}$.

Try: $\vec{B} = \vec{Q} - \vec{P}$

Recall: $\vec{Q} - \vec{P} = \vec{Q} + (-\vec{P})$

- $-\vec{P}$: → (right), magnitude 4
- $\vec{Q}$: ↗, magnitude 3

Add them: start with $-\vec{P}$ (→), then add $\vec{Q}$ (↗) on top → resultant will point up-right? No — wait, that would be too high.

Alternatively, try: $\vec{B} = -\vec{Q} + \vec{P}$? That might point down-left.

Wait — what if we do: $\vec{B} = \vec{P} - \vec{Q}$?

That’s $\vec{P} + (-\vec{Q})$

- $\vec{P}$: ← (left), magnitude 4
- $-\vec{Q}$: ↙ (down-left), magnitude 3

Adding those → would give something even more left/down → not matching.

Try: $\vec{B} = -\vec{Q} + \vec{R}$?

- $-\vec{Q}$: ↙ (down-left), magnitude 3
- $\vec{R}$: ↑, magnitude 2

Resultant: starts at origin, go down-left, then up → net direction: down-right? Possibly.

But visually, $\vec{B}$ looks like it could be $\vec{P} + \vec{R}$? No — $\vec{P}$ is left, $\vec{R}$ is up → that gives northwest.

Wait — here's a better idea:

Look at the direction of $\vec{B}$: it’s down-right (southeast). The only way to get that is to combine a right-pointing vector and a down-pointing vector.

We don’t have a down vector directly, but we can use negative of up vector: $-\vec{R}$.

So try: $\vec{B} = -\vec{P} + (-\vec{R})$

- $-\vec{P}$: → (right), 4 units
- $-\vec{R}$: ↓ (down), 2 units

Head-to-tail: start with →, then ↓ → resultant is southeast (↘), which matches $\vec{B}$!

✔ So:
> $\vec{B} = -\vec{P} - \vec{R}$

*(or equivalently, $\vec{B} = -(\vec{P} + \vec{R})$)*

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## ✔ Case C: Vector $\vec{C}$ — pointing up-left (↖)

This is northwest direction.

Try: $\vec{C} = -\vec{Q} + \vec{P}$?

- $-\vec{Q}$: ↙ (down-left), 3 units
- $\vec{P}$: ← (left), 4 units

Resultant: mostly left, slightly down → not quite.

Try: $\vec{C} = \vec{P} + \vec{R}$?

- $\vec{P}$: ←, 4
- $\vec{R}$: ↑, 2

Head-to-tail: start with ←, then ↑ → resultant is northwest (↖) — YES! Matches $\vec{C}$.

✔ So:
> $\vec{C} = \vec{P} + \vec{R}$

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## ✔ Case D: Vector $\vec{D}$ — pointing right (→)

This is horizontal right.

Only vector that points right naturally is $-\vec{P}$ (since $\vec{P}$ is left).

Could we get pure right by combining others?

Try: $\vec{D} = -\vec{P} + \vec{Q}$?

- $-\vec{P}$: →, 4
- $\vec{Q}$: ↗, 3 → adds an upward component → resultant would be up-right, not pure right.

What if we cancel the vertical component?

$\vec{Q}$ has both horizontal and vertical components. If we subtract $\vec{R}$ (which is purely vertical), maybe?

Try: $\vec{D} = \vec{Q} - \vec{R}$

- $\vec{Q}$: ↗ (has right and up components)
- $-\vec{R}$: ↓ (down), 2

If $\vec{Q}$’s vertical component is exactly 2, then subtracting $\vec{R}$ would cancel vertical → leaving pure horizontal right.

Is that plausible?

Magnitude of $\vec{Q}$ is 3. If its vertical component were 2, then horizontal component would be $\sqrt{3^2 - 2^2} = \sqrt{5} ≈ 2.24$, so yes, possible.

Graphically, if you draw $\vec{Q}$, then subtract $\vec{R}$ (i.e., add $-\vec{R}$ downward), the resultant would be horizontal right.

And that’s exactly what $\vec{D}$ is: pure right.

✔ So:
> $\vec{D} = \vec{Q} - \vec{R}$

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## ✔ Case E: Vector $\vec{E}$ — pointing down-right (↘), steeper than $\vec{B}$

Similar to $\vec{B}$, but longer and more downward.

Try: $\vec{E} = -\vec{P} + (-\vec{Q})$

- $-\vec{P}$: →, 4
- $-\vec{Q}$: ↙, 3 → has left and down components

Resultant: right + (left + down) → net: (4 - left-component-of-Q) right, plus down.

Since $\vec{Q}$ has magnitude 3, and assuming it’s at 45° for simplicity (though not specified), its components would be ~2.12 right and 2.12 up → so $-\vec{Q}$ is 2.12 left and 2.12 down.

Then $-\vec{P} + (-\vec{Q})$ = (4 right - 2.12 left) = 1.88 right, and 2.12 down → resultant is down-right, which matches $\vec{E}$.

Alternatively, we could write: $\vec{E} = -\vec{P} - \vec{Q}$

✔ So:
> $\vec{E} = -\vec{P} - \vec{Q}$

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## 📝 Final Answers:

| Case | Vector Equation |
|------|------------------------------|
| A | $\vec{A} = \vec{Q} + \vec{R}$ |
| B | $\vec{B} = -\vec{P} - \vec{R}$ |
| C | $\vec{C} = \vec{P} + \vec{R}$ |
| D | $\vec{D} = \vec{Q} - \vec{R}$ |
| E | $\vec{E} = -\vec{P} - \vec{Q}$ |

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These equations represent the graphical addition/subtraction using the head-to-tail method, matching the directions and relative magnitudes shown in the diagram.