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

Let's solve this step-by-step based on the principles of vector addition and subtraction using graphical methods.

We are given three vectors:

- $\vec{P}$: a horizontal vector pointing to the right.
- $\vec{Q}$: a vector pointing upward and to the right (diagonal).
- $\vec{R}$: a vertical vector pointing downward.

We need to express each resultant vector ($\vec{A}, \vec{B}, \vec{C}, \vec{D}, \vec{E}$) in terms of $\vec{P}, \vec{Q},$ and $\vec{R}$, using vector addition/subtraction rules.

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🔹 General Rules:

- Vector Addition: When adding vectors graphically, place the tail of the second vector at the head of the first. The resultant goes from the tail of the first to the head of the last.
- Vector Subtraction: $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$, so reverse the direction of the subtracted vector.

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Now let’s analyze each case:

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✔ Example: $\vec{X} = \vec{P} + \vec{Q}$

This is already shown as an example: start with $\vec{P}$, then add $\vec{Q}$ from its tip — the resultant is $\vec{X}$.

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🟨 A. Vector $\vec{A}$

- $\vec{A}$ starts from the origin (tail), goes right (like $\vec{P}$), then up (like $\vec{Q}$), but ends at a point that matches $\vec{P} + \vec{Q}$.
- But wait — look carefully: it's not exactly $\vec{P} + \vec{Q}$ because the direction seems different.

Wait — actually, comparing with the example, we see that in the example, $\vec{X} = \vec{P} + \vec{Q}$ forms a triangle with $\vec{P}$ and $\vec{Q}$ added head-to-tail.

But here, in A, the vector $\vec{A}$ is formed by:
- Starting from the origin,
- Going along $\vec{P}$ (right),
- Then going vertically upward — but that's not $\vec{Q}$, since $\vec{Q}$ has a diagonal direction.

Wait — no! Let's re-express.

Actually, in figure A, the two vectors being added are:
- First: $\vec{P}$ (horizontal right)
- Second: a vector upward — which looks like the vertical component of $\vec{Q}$? But no — we can only use $\vec{P}, \vec{Q}, \vec{R}$.

But notice: $\vec{R}$ points downward. So $-\vec{R}$ would be upward.

So if we have:
- $\vec{P}$ → right
- $-\vec{R}$ → upward (since $\vec{R}$ is down)

Then $\vec{A} = \vec{P} + (-\vec{R}) = \vec{P} - \vec{R}$

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

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🟨 B. Vector $\vec{B}$

In this diagram, we see:
- Two vectors forming a triangle: one labeled $\vec{B}$, and two others.
- The triangle shows a vector starting from the origin, going to the right (like $\vec{P}$), then turning down-left to close the triangle.

But more clearly: the two input vectors are:
- One going right: $\vec{P}$
- One going down: $\vec{R}$

And the resultant $\vec{B}$ is the third side of the triangle, from the end of $\vec{R}$ back to the start.

Wait — actually, in vector addition, the resultant is from tail of first to head of last.

Here, the diagram shows:
- Start at origin → go along $\vec{P}$ → then go along $\vec{R}$ (but $\vec{R}$ is down), but the arrow is drawn from the head of $\vec{P}$ to the head of $\vec{B}$?

Wait — better: this is a closed triangle, so it represents:
- $\vec{P} + \vec{R} + \vec{B} = 0$ → so $\vec{B} = -(\vec{P} + \vec{R})$

But looking at the arrows:
- From origin → right → down → back to origin?
- No: the vector $\vec{B}$ is drawn as the closing side of the triangle.

The triangle shows:
- First vector: $\vec{P}$ (right)
- Second vector: $\vec{R}$ (down)
- Third vector: $\vec{B}$ (from end of $\vec{R}$ back to origin)

So: $\vec{P} + \vec{R} + \vec{B} = 0$ ⇒ $\vec{B} = -(\vec{P} + \vec{R})$

But wait — the direction of $\vec{B}$ is up and left, which is opposite to $\vec{P} + \vec{R}$.

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

Alternatively, if you're adding vectors in sequence: $\vec{P} + \vec{R}$ gives a vector down-right; then $\vec{B}$ closes the triangle, so it must be the negative of that.

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

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🟨 C. Vector $\vec{C}$

This is a triangle with:
- One vector: $\vec{Q}$ (diagonal up-right)
- Another vector: $\vec{R}$ (down)
- And $\vec{C}$ closing the triangle.

From the diagram:
- Starts at origin → goes along $\vec{Q}$ → then goes along $\vec{R}$ (down) → back to origin via $\vec{C}$

So: $\vec{Q} + \vec{R} + \vec{C} = 0$ ⇒ $\vec{C} = -(\vec{Q} + \vec{R})$

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

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🟨 D. Vector $\vec{D}$

This one shows:
- A vector starting from origin, going up-left (opposite to $\vec{Q}$), then another vector from there to the final point.

Wait — let's see the path:
- First vector: appears to be $-\vec{Q}$ (because $\vec{Q}$ is up-right, so this is down-left)
- Then a vector from there to the head of $\vec{D}$ — looks like $\vec{P}$ (right)

But the order: tail to head.

From the diagram:
- Start at origin → go along $-\vec{Q}$ → then go along $\vec{P}$ → result is $\vec{D}$

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

✔ So:
$$
\boxed{\vec{D} = \vec{P} - \vec{Q}}
$$

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🟨 E. Vector $\vec{E}$

This one is a bit trickier.

We see:
- A vector $\vec{E}$ that starts from the head of $\vec{Q}$ and ends at the tail of $\vec{P}$?

Wait — no.

Looking at the diagram:
- There is a vector $\vec{Q}$ drawn from origin.
- Then a vector $\vec{P}$ drawn from the head of $\vec{Q}$.
- Then $\vec{E}$ connects the tail of $\vec{Q}$ to the head of $\vec{P}$?

No — wait: $\vec{E}$ is drawn from the head of $\vec{Q}$ to the head of $\vec{P}$? That doesn't make sense.

Wait — let’s interpret carefully.

Actually, the diagram shows:
- $\vec{Q}$: from origin to some point
- $\vec{P}$: from origin to right
- $\vec{E}$: from the head of $\vec{Q}$ to the head of $\vec{P}$

But that means $\vec{E} = \vec{P} - \vec{Q}$, because it's the vector from the tip of $\vec{Q}$ to the tip of $\vec{P}$, assuming both start from origin.

Yes! This is a common trick: when two vectors originate from the same point, the vector from the tip of $\vec{Q}$ to the tip of $\vec{P}$ is $\vec{P} - \vec{Q}$.

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

Wait — but that's the same as $\vec{D}$?

No — check again.

Wait — in D, we had $\vec{D} = \vec{P} - \vec{Q}$, but the configuration was different.

But in E, it's geometrically equivalent to $\vec{P} - \vec{Q}$, since both vectors start from origin.

But wait — in E, $\vec{E}$ is drawn from the tip of $\vec{Q}$ to the tip of $\vec{P}$, so yes:
$$
\vec{E} = \vec{P} - \vec{Q}
$$

But let’s double-check: is that consistent?

Yes — because:
- $\vec{P}$ and $\vec{Q}$ both start at origin.
- Then the vector from tip of $\vec{Q}$ to tip of $\vec{P}$ is $\vec{P} - \vec{Q}$.

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

But wait — is this correct?

Let’s think: if you have $\vec{P}$ and $\vec{Q}$ from the same origin, then:
- $\vec{P} - \vec{Q}$ is a vector from tip of $\vec{Q}$ to tip of $\vec{P}$ — yes.

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

But now we have D and E both equal to $\vec{P} - \vec{Q}$? That might be possible.

But let’s recheck D.

In D, the vector $\vec{D}$ is drawn as:
- Start at origin → go along $-\vec{Q}$ → then go along $\vec{P}$ → so total: $-\vec{Q} + \vec{P} = \vec{P} - \vec{Q}$

Yes — so both D and E represent $\vec{P} - \vec{Q}$, just drawn differently.

So they are equal.

But is there a difference?

Wait — in E, the vector $\vec{E}$ is not starting from origin. It starts from the head of $\vec{Q}$ and ends at the head of $\vec{P}$, so it's $\vec{P} - \vec{Q}$, but positioned differently.

But as a vector, it's the same.

So yes, $\vec{E} = \vec{P} - \vec{Q}$

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Wait — but let’s check if any of these involve $\vec{R}$?

Only A, B, and C involve $\vec{R}$.

But in D and E, no $\vec{R}$ appears — so we can’t include it unless necessary.

So final answers:

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✔ Final Answers:

- A: $\vec{A} = \vec{P} - \vec{R}$
- B: $\vec{B} = -\vec{P} - \vec{R}$
- C: $\vec{C} = -\vec{Q} - \vec{R}$
- D: $\vec{D} = \vec{P} - \vec{Q}$
- E: $\vec{E} = \vec{P} - \vec{Q}$

Wait — but E is drawn differently than D. Is E really $\vec{P} - \vec{Q}$?

Let me re-analyze E.

In E, we see:
- A vector $\vec{Q}$ from origin.
- A vector $\vec{P}$ from origin.
- Then $\vec{E}$ is drawn from the tip of $\vec{Q}$ to the tip of $\vec{P}$.

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

But is there a possibility it’s something else?

Wait — could it be $\vec{Q} + \vec{R}$ or something?

No — $\vec{R}$ is not involved.

But let’s consider: is there any way $\vec{E}$ involves $\vec{R}$?

No — the diagram doesn’t show $\vec{R}$ in E.

So it must be $\vec{P} - \vec{Q}$

Similarly, D is also $\vec{P} - \vec{Q}$

So both D and E are equal to $\vec{P} - \vec{Q}$

That’s acceptable.

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But wait — let’s double-check C.

In C, the triangle has:
- One side: $\vec{Q}$ (up-right)
- One side: $\vec{R}$ (down)
- And $\vec{C}$ closing the triangle.

So: $\vec{Q} + \vec{R} + \vec{C} = 0$ → $\vec{C} = -(\vec{Q} + \vec{R})$

Yes.

So:
$$
\vec{C} = -\vec{Q} - \vec{R}
$$

Correct.

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✔ Final Answer Summary:

| Letter | Expression |
|--------|------------|
| A | $\vec{A} = \vec{P} - \vec{R}$ |
| B | $\vec{B} = -\vec{P} - \vec{R}$ |
| C | $\vec{C} = -\vec{Q} - \vec{R}$ |
| D | $\vec{D} = \vec{P} - \vec{Q}$ |
| E | $\vec{E} = \vec{P} - \vec{Q}$ |

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✔ Final Boxed Answers:

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

Let me know if you want explanations for each!