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Project (4, 1, 1) onto (1, 1, 0)

Step by step, with the reason for every step · Projection onto a vector · Linear Algebra

Step by step

The question

Question: find the projection of (4, 1, 1) onto (1, 1, 0).

v = [4; 1; 1], s = [1; 1; 0]

The projection of v onto s is the part of v that goes along s.

Think of the shadow of v on the line through s, with the light straight above.

projection: the part of v along s
like the shadow of v on the line of s

1 Write the rule

The rule: v dot s, over s dot s, times s.

The fraction is one number: how much of s we need.

proj = v • ss • s × s

2 Work out the number

The top: v • s, matching entries multiplied and added.

v • s = 4 × 1 + 1 × 1 + 1 × 0 = 4 + 1 + 0 = 5

The bottom: s • s.

s • s = 1 × 1 + 1 × 1 + 0 × 0 = 1 + 1 + 0 = 2

So the number is 5/2.

number = 52

Careful!

The bottom is s dot s, the vector we project onto.

Not v dot v.

Careful: the bottom is s • s, not v • v

3 Times s

Times s: every entry times 5/2.

So the projection is (5/2, 5/2, 0).

proj = 52 × [1; 1; 0] = [5 over 2; 5 over 2; 0]

4 Check

Let's check: the leftover, v minus the projection, must be at a right angle to s.

v − proj = [4; 1; 1] − [5 over 2; 5 over 2; 0]
v − proj = [4 − 5 over 2; 1 − 5 over 2; 1 − 0]
v − proj = [3 over 2; −3 over 2; 1]

Its dot product with s is 0.

So the leftover is at a right angle to s: the answer is right!

(v − proj) • s = 32 × 1 + (−32) × 1 + 1 × 0 = 32 − 32 + 0 = 0

Let's look back.

First: write the rule.

Then: work out the number.

Then: times s.

Last: check.

Answer
proj = 52 × [1; 1; 0] = [5 over 2; 5 over 2; 0]

Your turn

Try this one yourself, then check:

Project [6; 1; 1] onto [1; 1; 0]
Show the answer
[7 over 2; 7 over 2; 0]

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