4.5.4 · D5Linear Algebra (Full)
Question bank — Projection of vectors
Above: the right triangle behind (left) and the "already-on-the-line" picture behind idempotence (right). Refer back to these as you work the drills.
True or false — justify
Each line: decide true/false, then give the reason — bare yes/no scores nothing.
The vector projection always points in the same direction as .
False. When the angle is obtuse () the scalar factor is negative, so the projection points opposite to . It stays on the line of , but not always the same way.
If you double the length of , the vector projection stays the same.
True. Replace by : the numerator , the denominator , and the vector factor becomes . So — unchanged. Only the direction of matters, not its length.
If you double the length of , the scalar projection stays the same.
True. depends only on the unit direction , and doubling leaves (its direction) unchanged.
whenever .
False. Equal lengths don't fix the direction: one result lives along , the other along . They only agree if and already point the same way.
The scalar projection can be larger than .
False. and , so its magnitude never exceeds . A shadow is never longer than the stick.
If , then .
False. The projection is zero whenever (their dot product is zero), no matter how long is. A vertical stick under an overhead sun casts no floor-shadow.
The perpendicular part is always perpendicular to , by construction.
True. algebraically, so orthogonality is guaranteed, not lucky. This is the basis of Orthogonal decomposition.
Projecting onto , then projecting that result onto again, changes nothing the second time.
True. Call . Projecting gives ; but , so the second projection returns exactly. Doing it twice equals doing it once — that is what idempotent means (see the right-hand figure).
For a unit vector (so ), the scalar projection and the magnitude of the vector projection are equal.
True. The scalar is the number ; the vector is , whose magnitude is . So the vector's length equals the absolute value of the scalar — same size, different type of object.
Spot the error
Each states a "result." Say what's wrong and give the correct reasoning.
"."
Wrong denominator. This divides by one but multiplies by the full , making the answer too long by a factor . Correct: — one length for the shadow, one to unitise the direction.
"Since the angle is , the scalar projection is ."
, not . Dropping the sign hides that the shadow falls on the back side of ; the scalar projection is negative here.
"To find how much of points along , I compute so I always get a clean positive answer."
Taking the absolute value destroys direction information. The sign is the answer to "which way along " — throwing it away merges obtuse and acute cases.
"The projection of onto is because that's the vector we started with."
The projection onto the -axis keeps only the -component, which is here. Correct answer is : a purely vertical vector has no horizontal shadow.
" — the dot product is the scalar projection."
Only true when . In general you must divide by : . The raw dot product bundles in 's length.
"Projection is linear, so by symmetry of the dot product."
The dot product is symmetric, but projection also carries a direction factor (or ). Swapping the roles changes the line the shadow lands on; the operation is not symmetric.
" only holds in 2D."
It holds in any dimension. The identity is pure algebra: is defined as , so adding them back always returns , regardless of how many coordinates the vectors have.
Why questions
Why does the scalar projection use and not ?
We want the component along (adjacent side of the right triangle in the figure), and gives exactly that. would give the perpendicular (opposite) part, i.e. .
Why does the vector projection divide by rather than ?
One factor of converts the dot product into the true shadow length; a second factor turns into the unit direction . Together that's .
Why is the dot product the "engine" of projection instead of measuring the angle directly?
The dot product secretly contains via , so it lets us extract the shadow from coordinates with no protractor. See Dot product.
Why must be a unit vector when we write ?
The formula "(length)×(direction)" only works if the direction has length exactly 1; otherwise the direction factor would sneak an extra length into the answer. See Unit vectors.
Why does subtracting the projection give something perpendicular, in Gram-Schmidt process?
Removing the part of that lies along leaves only what is orthogonal to . Repeating this against each earlier vector is exactly how Gram–Schmidt builds an orthonormal set.
Why does work in Work done by a force secretly use projection?
Only the component of force along the displacement does work; that component is the scalar projection , and multiplying by recovers .
Why can the scalar projection be negative while a physical "length" cannot?
The scalar projection is a signed length that also records direction along the -line. The minus sign says "the shadow points backward," which a plain length would hide.
Edge cases
What is when is perpendicular to ()?
The zero vector. Since , the numerator vanishes — no part of lies along .
What is when is already parallel to (same direction, )?
It equals itself. All of lies along , so the shadow captures the whole vector and .
What happens to the projection formulas when ?
Both are undefined: the scalar and the vector each divide by zero (since and ), and there is no direction to project onto.
What is when ?
The zero vector. A zero vector has no shadow — the numerator kills it, and too.
At exactly , is the projection a "small" vector or exactly zero?
Exactly zero, since . It's the clean boundary between positive (acute) and negative (obtuse) shadows.
As , what does the scalar projection approach?
It approaches , its most negative value, because . The shadow is longest when points straight back along .
If and point the exact opposite way (), what is ?
The zero vector. Anti-parallel vectors are still on the same line, so the projection captures all of and nothing is left over perpendicular.
Recall One-line self-test before you leave
Cover these and answer in your head. Sign of scalar projection at ? ::: Negative — just past , . Denominator of the vector projection? ::: . Does lengthening change ? ::: No — only 's direction matters. Projection onto ? ::: Undefined — both scalar and vector forms divide by zero. What does "idempotent" mean here? ::: Projecting a second time changes nothing: .