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Magnitude of <6, -3>

Step by step, with the reason for every step · Vector from two points / magnitude · Precalculus

Step by step

The question

Question: find the magnitude of v = ⟨6, −3⟩.

v = ⟨6, −3⟩

1 Draw the arrow

Draw the grid.

The arrow starts at 0, at the origin.

It goes right 6 and down 3.

We call it v.

Across 6

and down 3.

They make a right angle.

The arrow is the longest side.

2 Find the length

Pythagoras: square the two parts, add them, then take the square root.

‖v‖ = √62 + (−3)2

6² is 6 times 6: 36.

62 = 6 × 6 = 36

(−3)² is 9.

Minus times minus is plus.

(−3)2 = (−3) × (−3) = 9

Write the squares: 36 + 9.

‖v‖ = √36 + 9

36 plus 9 is 45.

‖v‖ = √45

But 45 has a square number in it: 9.

45 is 9 times 5.

‖v‖ = √9 × 5

A root of a times splits into two roots.

‖v‖ = √9 × √5

√9 is 3.

3 × 3 = 9

So ‖v‖ = 3√5.

That is the exact answer, made simple.

‖v‖ = 3√5

Careful!

Not across plus up.

The arrow is the slanted side.

It is shorter than going across and then up.

Careful: not ‖v‖ = 6 + 3 = 9 ✗

3 Check

Let's check.

Square the length: we get 45.

The same as 36 plus 9.

Right!

(3√5)² = 9 × 5 = 45 and 36 + 9 = 45

Let's look back.

First: draw the arrow.

Then: find the length.

Last: check.

Answer
Careful: not ‖v‖ = 6 + 3 = 9 ✗

Your turn

Try this one yourself, then check:

Find the magnitude of v⟨5, −5⟩
Show the answer
‖v‖ = 5√2

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