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

To solve this problem, we need to use graphical vector addition — specifically, the head-to-tail method. In this method:

- To add two vectors, place the tail of the second vector at the head of the first.
- The resultant vector goes from the tail of the first to the head of the last.
- To subtract a vector, reverse its direction (i.e., add its negative).

We are given three base vectors:
- P: points right (→)
- Q: points up-right (↗)
- R: points down (↓)

Now let’s analyze each diagram (A–E) one by one.

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🔹 A: Vector A

The diagram shows:
- First vector: P (→)
- Second vector: R (↓), placed head-to-tail with P
- Resultant A goes from start of P to end of R

✔ So:
A = P + R

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🔹 B: Vector B

This is a triangle showing:
- One side: Q (↗)
- Another side: R (↓), but pointing *upward* in the diagram → that means it’s –R
- Resultant B goes from start of Q to end of –R

Wait — actually, looking closely: the vector labeled B is the *resultant*, and the two component vectors are drawn head-to-tail forming a triangle.

From the diagram:
- Start with Q (↗)
- Then add R (↓) — but since the arrow for R points *downward* and is placed after Q, it's just +R

But wait — if you follow the path: start at origin → go along Q → then go along R → final point gives vector B.

So again: B = Q + R

✔ B = Q + R

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🔹 C: Vector C

Diagram shows:
- First vector: R (↓) — but drawn pointing *upward*? Wait — no, look carefully.

Actually, the bottom vector is P (→), and the left vector is Q (↗), but pointing *left-down*? No — let’s be precise.

In diagram C:
- The resultant C is the top vector (↖)
- The two component vectors form a triangle with C as the closing side.

If you trace:
- Start at tail of Q (which is pointing ↗), go to its head.
- Then from there, go along P (→) — but in the diagram, P is drawn going *left*? No — wait, P is drawn going to the right, but the arrowhead is at the left? That suggests it’s –P

Actually — here’s the key: in vector diagrams, when a vector is drawn opposite to its defined direction, it represents the negative.

In diagram C:
- One leg is Q (↗)
- The other leg is P, but drawn pointing *left* (←), which is –P
- The resultant C goes from the tail of Q to the head of –P

So: C = Q + (–P) = Q – P

✔ C = Q – P

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🔹 D: Vector D

Diagram shows:
- First vector: R (↓) — but drawn pointing *upward* → so it’s –R
- Second vector: Q (↗), placed head-to-tail with –R
- Resultant D goes from start of –R to end of Q

So: D = –R + Q = Q – R

✔ D = Q – R

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🔹 E: Vector E

Diagram shows:
- First vector: Q (↗)
- Second vector: P (→), placed head-to-tail with Q
- But the resultant E is drawn from the *start of Q* to the *end of P* — that’s standard addition: Q + P

However — wait! Look at the direction of E: it’s pointing to the right, almost horizontal, while Q + P should be diagonal up-right.

Actually — let’s re-express:

In diagram E:
- The vector P is drawn going to the right (→), and Q is drawn going up-left (↖)? No — Q is still ↗, but placed such that its tail is at the head of P? Or vice versa?

Actually — the diagram shows:
- Start at origin → draw P (→)
- Then from end of P, draw Q (↗)
- Resultant E goes from start of P to end of Q → so E = P + Q

Which is the same as Q + P (addition is commutative).

✔ E = P + Q

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## ✔ Final Answers:

| Diagram | Expression |
|---------|-------------------|
| A | A = P + R |
| B | B = Q + R |
| C | C = Q – P |
| D | D = Q – R |
| E | E = P + Q |

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## 📌 Summary of Reasoning:

- Always apply head-to-tail rule.
- If a vector is drawn in the *opposite direction* of its defined orientation, it’s the negative of that vector.
- Order doesn’t matter for addition (commutative), but direction matters for subtraction.

Let me know if you’d like diagrams or animations to visualize these!