Examples

(- 3, 1)

$(- 3, 1)$ ,

(2, 14)

$(2, 14)$

Step 1

To find the position vector, subtract the initial point vector

P

$P$ from the terminal point vector

Q

$Q$ .

Q - P = (2 i + 14 j) - (- 3 i + 1 j)

$Q - P = (2 i + 14 j) - (- 3 i + 1 j)$

Step 2

Simplify each term.

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Step 2.1

Multiply

j

$j$ by

1

$1$ .

2 i + 14 j - (- 3 i + j)

$2 i + 14 j - (- 3 i + j)$

Step 2.2

Apply the distributive property.

2 i + 14 j - (- 3 i) - j

$2 i + 14 j - (- 3 i) - j$

Step 2.3

Multiply

- 3

$- 3$ by

- 1

$- 1$ .

2 i + 14 j + 3 i - j

$2 i + 14 j + 3 i - j$

2 i + 14 j + 3 i - j

$2 i + 14 j + 3 i - j$

Step 3

Simplify by adding terms.

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Step 3.1

Add

2 i

$2 i$ and

3 i

$3 i$ .

14 j + 5 i - j

$14 j + 5 i - j$

Step 3.2

Subtract

j

$j$ from

14 j

$14 j$ .

13 j + 5 i

$13 j + 5 i$

13 j + 5 i

$13 j + 5 i$

Step 4

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