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

We are given three vectors:
- $\vec{P}$: magnitude 4, pointing left (so it's in the negative x-direction)
- $\vec{Q}$: magnitude 3, pointing up and to the right (diagonal, northeast direction)
- $\vec{R}$: magnitude 2, pointing straight up (positive y-direction)

We are to determine vector equations for vectors $\vec{A}, \vec{B}, \vec{C}, \vec{D}, \vec{E}$ shown in cases A through E, using combinations of $\vec{P}, \vec{Q}, \vec{R}$, or their negatives.

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Step-by-step Analysis

We will analyze each case based on the direction and relative magnitude of the resulting vector, comparing it to the known vectors.

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#### Example Given:
$$
\vec{X} = -\vec{P} + \vec{R}
$$
This means:
- $-\vec{P}$ is a vector of length 4 pointing right (opposite of $\vec{P}$)
- Add $\vec{R}$ (upward, length 2)
- Resultant is a vector starting from the tail of $-\vec{P}$, ending at the tip of $\vec{R}$ — this matches the example diagram.

So we use vector addition/subtraction via head-to-tail method.

Now let’s go through each case.

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Case A: Vector $\vec{A}$

- Direction: Diagonal, up and to the right, similar to $\vec{Q}$
- Length: Appears to be about 3 units → same as $\vec{Q}$

So $\vec{A}$ looks just like $\vec{Q}$.

But is it exactly $\vec{Q}$? Let's check if it could be a combination.

But since it's identical in direction and magnitude to $\vec{Q}$, we can write:

$$
\boxed{\vec{A} = \vec{Q}}
$$

---

Case B: Vector $\vec{B}$

- Direction: Down and to the right (southeast), opposite to $\vec{Q}$
- Magnitude: Seems shorter than $\vec{Q}$, but comparable

Wait — notice that $\vec{Q}$ is up and to the right. So $-\vec{Q}$ would point down and to the left.

But $\vec{B}$ points down and to the right — not the same as $-\vec{Q}$.

Let’s think: Could this be $\vec{P} - \vec{R}$?

- $\vec{P}$ is left (negative x), so $-\vec{P}$ is right (positive x)
- $\vec{R}$ is up

Try: $\vec{B} = -\vec{P} - \vec{R}$?
- $-\vec{P}$: right, length 4
- $-\vec{R}$: down, length 2
- Adding them: right and down → southeast direction → matches $\vec{B}$

Yes! That seems correct.

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

Alternatively, $\vec{B} = -(\vec{P} + \vec{R})$

But simplest form is:
$$
\boxed{\vec{B} = -\vec{P} - \vec{R}}
$$

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Case C: Vector $\vec{C}$

- Direction: Down and to the left (southwest)
- Magnitude: Looks about 3–4 units

Compare with $\vec{P}$: left, length 4
$\vec{R}$: up, length 2 → so $-\vec{R}$: down, length 2

If we do: $\vec{P} - \vec{R}$

- $\vec{P}$: left
- $-\vec{R}$: down
- So $\vec{P} - \vec{R} = \vec{P} + (-\vec{R})$: left and down → southwest → matches $\vec{C}$

Magnitude: $\vec{P}$ has length 4, $-\vec{R}$ has length 2 → resultant should be roughly $\sqrt{4^2 + 2^2} = \sqrt{20} \approx 4.47$, which is reasonable.

So:
$$
\boxed{\vec{C} = \vec{P} - \vec{R}}
$$

---

Case D: Vector $\vec{D}$

- Direction: Right (positive x-direction)
- Magnitude: Small, appears to be about 1–2 units

We know:
- $\vec{P}$ is left (length 4), so $-\vec{P}$ is right, length 4

But $\vec{D}$ is much shorter.

Could it be $\vec{Q} - \vec{R}$?

- $\vec{Q}$: up-right
- $\vec{R}$: up
- So $\vec{Q} - \vec{R}$: subtract upward component → results in a vector that is still right, but less upward → net: slightly right and down?

But $\vec{D}$ is purely horizontal.

Wait — what if $\vec{D}$ is just the horizontal component of $\vec{Q}$?

But we’re limited to only adding/subtracting $\vec{P}, \vec{Q}, \vec{R}$ directly.

Another idea: Is $\vec{D}$ equal to $-\vec{P}$? No — $-\vec{P}$ has length 4, but $\vec{D}$ is smaller.

Wait — maybe it's $\vec{Q} + \vec{R}$? But that would be more upward.

Alternatively, consider: Can we get a small rightward vector?

Let’s suppose $\vec{D} = \vec{Q} - \vec{R}$?

- $\vec{Q}$ has components: assume it's at 45° (common in such problems), then:
- $Q_x = 3\cos(45^\circ) = 3 \cdot \frac{\sqrt{2}}{2} \approx 2.12$
- $Q_y = 3\sin(45^\circ) \approx 2.12$
- $\vec{R} = (0, 2)$
- So $\vec{Q} - \vec{R} = (2.12, 0.12)$ → almost horizontal right → matches $\vec{D}$

So yes! This gives a vector slightly right and barely up — but in the diagram, $\vec{D}$ is drawn purely right, so perhaps it's intended to represent the horizontal component.

But since we can't break into components, we must express it as a combination.

Wait — is there another way?

What if $\vec{D} = -\vec{P} + \text{something}$?

No — we need two of the vectors.

Wait — perhaps $\vec{D} = \vec{Q} - \vec{R}$?

As above, this gives approximately $(2.12, 0.12)$ → mostly right.

But the diagram shows a purely horizontal arrow.

Alternatively, maybe it's $\vec{D} = -\vec{P} + \vec{Q}$?

That would be:
- $-\vec{P}$: right 4
- $\vec{Q}$: up-right 3
- Sum: right and up → northeast → no, doesn’t match

Wait — what if $\vec{D} = \vec{Q} - \vec{R}$?

Yes — that gives a vector with small upward component, but maybe the diagram simplifies it.

But look again: the problem says "construct the vectors shown" by adding or subtracting two of the vectors.

So we must find a combination of two vectors that gives the direction and magnitude of $\vec{D}$.

But $\vec{D}$ is short and horizontal.

Only possible horizontal vectors:
- $\vec{P}$: left
- $-\vec{P}$: right

But $-\vec{P}$ is length 4, too long.

Unless it's scaled — but we can't scale.

Wait — maybe $\vec{D}$ is not made from full vectors, but the diagram shows the resultant of a combination.

But the instructions say: “Construct the vectors shown... by adding or subtracting two of the vectors”

So likely, $\vec{D}$ is the result of combining two of the given vectors.

But none of the combinations give a purely horizontal vector.

Wait — unless $\vec{D} = \vec{Q} - \vec{R}$, and the slight upward component is ignored in the drawing.

But that seems unlikely.

Alternative idea: Could $\vec{D}$ be $-\vec{P} + \vec{Q}$? No — that’s up and right.

Wait — let’s reconsider.

Is it possible that $\vec{D}$ is simply $-\vec{P}$? But no — too long.

Wait — perhaps the vector $\vec{D}$ is meant to be the horizontal component of $\vec{Q}$, but we are to express it as a combination.

But we can’t do that unless we have a vector that cancels the vertical part.

But we don’t have a horizontal vector.

Wait — here's a better idea: Maybe $\vec{D} = \vec{Q} - \vec{R}$?

Let’s test:

- $\vec{Q}$: up and right
- $\vec{R}$: up
- So $\vec{Q} - \vec{R}$: subtract the upward component → leaves a vector that is only the horizontal part of $\vec{Q}$ plus zero vertical? No — because $\vec{Q}$ has both components.

Actually, $\vec{Q} - \vec{R}$ has:
- $x$-component: $Q_x$
- $y$-component: $Q_y - R_y = 2.12 - 2 = 0.12$

So it’s almost horizontal, but slightly up.

But $\vec{D}$ is drawn purely right.

So maybe not.

Wait — what if $\vec{D} = \vec{Q} + (-\vec{R})$? Same thing.

Still not purely horizontal.

But perhaps the diagram is schematic.

Alternatively, is there a possibility that $\vec{D} = \vec{Q} - \vec{R}$, and the small upward component is negligible in the sketch?

But let’s look at the other options.

Wait — what if $\vec{D} = -\vec{P} + \vec{R}$?

- $-\vec{P}$: right 4
- $\vec{R}$: up 2
- Sum: up and right → no

Not matching.

Wait — what if $\vec{D} = \vec{Q} + \vec{R}$? Even more up.

No.

Wait — could $\vec{D}$ be $-\vec{P} + \vec{Q}$?

- $-\vec{P}$: right 4
- $\vec{Q}$: up-right 3
- Sum: right and up → northeast → no

Wait — maybe I’m missing something.

Let’s re-express all possibilities.

Perhaps $\vec{D}$ is just the horizontal component of $\vec{Q}$, but since we can't decompose, maybe it's not expressible?

But the problem says we can construct it using two of the vectors.

Wait — what if $\vec{D} = \vec{Q} - \vec{R}$, and the diagram shows it as horizontal for simplicity?

But that seems forced.

Alternatively, maybe $\vec{D} = -\vec{P} + \vec{Q}$, but again, not horizontal.

Wait — another idea: Could $\vec{D}$ be $\vec{R} - \vec{Q}$?

- $\vec{R}$: up 2
- $\vec{Q}$: up-right 3
- So $\vec{R} - \vec{Q}$: down-left → no

Not matching.

Wait — let’s go back to Case A.

In Case A, $\vec{A}$ is clearly $\vec{Q}$.

In Case B, $\vec{B} = -\vec{P} - \vec{R}$

In Case C, $\vec{C} = \vec{P} - \vec{R}$

Now, $\vec{D}$ is small and right.

Wait — what if $\vec{D} = \vec{Q} - \vec{R}$?

Let’s compute its magnitude:

Assume $\vec{Q}$ is at 45°: $Q_x = Q_y = 3/\sqrt{2} \approx 2.12$

$\vec{R} = (0, 2)$

So $\vec{Q} - \vec{R} = (2.12, 0.12)$

Magnitude: $\sqrt{2.12^2 + 0.12^2} \approx \sqrt{4.49 + 0.014} \approx \sqrt{4.5} \approx 2.12$

Direction: arctan(0.12/2.12) ≈ 3.2° above horizontal → very close to horizontal.

So in the diagram, it might be drawn as horizontal for simplicity.

Thus, it’s plausible that:
$$
\boxed{\vec{D} = \vec{Q} - \vec{R}}
$$

Even though it has a small upward component, it’s approximately horizontal.

Alternatively, is there a better fit?

Wait — what if $\vec{D} = -\vec{P} + \vec{Q}$?

No — that’s longer and up.

No.

Another possibility: Could $\vec{D}$ be $\vec{Q} + \vec{P}$? No — left and up.

No.

So most likely:
$$
\boxed{\vec{D} = \vec{Q} - \vec{R}}
$$

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Case E: Vector $\vec{E}$

- Direction: Down and to the right (southeast)
- Magnitude: About 3–4 units

Compare to $\vec{Q}$: up-right → $-\vec{Q}$: down-left → not matching

But $\vec{E}$ is down and right.

So try: $-\vec{P} - \vec{Q}$?

- $-\vec{P}$: right 4
- $-\vec{Q}$: down-left 3 → so sum: right and down-left → net: right and down, but with left component?

Wait — $-\vec{P}$ is right, $-\vec{Q}$ is down-left → so adding them: right minus left → net right, and down.

So:
- $x$: $+4 - 3\cos(45^\circ) \approx 4 - 2.12 = 1.88$
- $y$: $-3\sin(45^\circ) \approx -2.12$

So vector: (1.88, -2.12) → down and right → matches $\vec{E}$

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

Alternatively, $\vec{E} = -(\vec{P} + \vec{Q})$

But:
$$
\boxed{\vec{E} = -\vec{P} - \vec{Q}}
$$

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Final Answers:

| Case | Vector Equation |
|------|-----------------|
| A | $\vec{A} = \vec{Q}$ |
| 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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Verification:

- A: $\vec{Q}$ → up-right, length 3 → matches
- B: $-\vec{P}$ (right 4) + $-\vec{R}$ (down 2) → southeast → matches
- C: $\vec{P}$ (left 4) + $-\vec{R}$ (down 2) → southwest → matches
- D: $\vec{Q}$ (up-right 3) + $-\vec{R}$ (down 2) → slightly up-right → approximated as horizontal → acceptable
- E: $-\vec{P}$ (right 4) + $-\vec{Q}$ (down-left 3) → net: right and down → southeast → matches

✔ All consistent.

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✔ Final Answer:

$$
\begin{aligned}
\vec{A} &= \vec{Q} \\
\vec{B} &= -\vec{P} - \vec{R} \\
\vec{C} &= \vec{P} - \vec{R} \\
\vec{D} &= \vec{Q} - \vec{R} \\
\vec{E} &= -\vec{P} - \vec{Q}
\end{aligned}
$$