Question bank — Dot product — formula, cosine formula, Cauchy-Schwarz inequality proof

False. The dot product is built to measure overlap (a single amount of "same-direction-ness"), so its output is a scalar. The Cross product is the operation that returns a vector.

False. It means they are perpendicular. Two nonzero vectors can be orthogonal, e.g. $(1,0)\cdot (0,1)=0$ — see Orthogonality and orthonormal bases.

False. $\mathbf{a}\cdot \mathbf{a}=a_1^2+\cdots +a_n^2$, a sum of squares, so it is $\ge 0$ and equals $0$ only for the zero vector. This positivity is the whole reason length $\Vert \mathbf{a}\Vert =\sqrt{\mathbf{a}\cdot \mathbf{a}}$ makes sense.

True. Each term $a_i{b}_i=b_i{a}_i$ since ordinary number multiplication commutes, so the whole sum is unchanged.

False. The discriminant proof uses only $\Vert \mathbf{a}-t\mathbf{b}{\Vert }^2\ge 0$, so it holds in any inner product space, even infinite-dimensional ones with no picture of an angle.

True. Equality forces $\cos \theta =+1$, i.e. $\theta =0^{\circ }$, so $\mathbf{b}$ is a positive scalar multiple of $\mathbf{a}$. (Recall $\theta$ is defined by the cosine formula in the box above.)

True. The absolute value ignores sign; opposite directions give $\cos \theta =-1$, so $|\cos \theta |=1$ and equality still holds. Both same and opposite directions are the "parallel" equality case.

False. It only says $(\mathbf{a}-\mathbf{b})\cdot \mathbf{c}=0$, i.e. $\mathbf{a}-\mathbf{b}$ is perpendicular to $\mathbf{c}$. Equality for all $\mathbf{c}$ would force $\mathbf{a}=\mathbf{b}$.

True. $(2\mathbf{a})\cdot (2\mathbf{b})=4(\mathbf{a}\cdot \mathbf{b})$ and both lengths double, so $\cos \theta =\frac{4(\mathbf{a}\cdot \mathbf{b})}{2\Vert \mathbf{a}\Vert \cdot 2\Vert \mathbf{b}\Vert }$ is unchanged. Scaling changes length, not direction.

The reasoning quietly assumes an angle $\theta$ already exists between $\mathbf{a}$ and $\mathbf{b}$. In abstract spaces the angle is only defined using Cauchy–Schwarz, so this is circular — the discriminant proof is needed first.

The middle term is dropped. Correctly $(\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}-\mathbf{b})=\Vert \mathbf{a}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2$; the $-2\mathbf{a}\cdot \mathbf{b}$ is exactly what carries the angle information.

Neither expression is even defined. $\mathbf{a}\cdot \mathbf{b}$ is a scalar, and dotting a scalar with a vector is meaningless. Dot product takes two vectors, so associativity cannot be asked of it.

False chain. In ${\mathbb{R}}^2$, $(1,0)\perp (0,1)$ and $(0,1)\perp (1,0)$, yet $(1,0)\cdot (1,0)=1\ne 0$. Perpendicularity is not transitive.

Expanding gives $f(t)=\Vert \mathbf{b}{\Vert }^2{t}^2-2(\mathbf{a}\cdot \mathbf{b})t+\Vert \mathbf{a}{\Vert }^2$, whose leading coefficient is $\Vert \mathbf{b}{\Vert }^2$. If $\mathbf{b}=0$ that leading coefficient is $0$, so $f$ is not quadratic and the discriminant step is invalid. But then both sides of C–S are $0$, so the inequality holds trivially — the $\mathbf{b}=0$ case must be handled separately.

The formula holds for every angle $0^{\circ }\le \theta \le 180^{\circ }$. Obtuse angles give $\cos \theta <0$, matching a negative dot product, and $\theta =90^{\circ }$ gives exactly $0$.

It gives $\sum a_i^2$, the sum of squared components, which is precisely $\Vert \mathbf{a}{\Vert }^2$. So the dot product was designed to contain squared distance.

A negative value means $\cos \theta <0$, i.e. the vectors point more than $90^{\circ }$ apart — they partly oppose each other. The sign encodes "same-ish direction" (+) versus "opposing" (−).

A squared length is a polynomial in $t$ (no square roots) and is guaranteed $\ge 0$. That non-negativity of a quadratic is the entire engine that forces the discriminant condition.

Discriminant $=0$ corresponds to $f(t_0)=0$ for some $t_0$, meaning $\mathbf{a}=t_0\mathbf{b}$ (parallel) — exactly the equality case. The "$\le$" keeps that boundary situation inside the theorem.

Perpendicular means $\cos \theta =0$, and since $\mathbf{a}\cdot \mathbf{b}=\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert \cos \theta$, the product is $0$ exactly when they are orthogonal (for nonzero vectors). One multiply-and-add answers it — see Orthogonality and orthonormal bases.

The three vectors $\mathbf{a},\mathbf{b},\mathbf{a}-\mathbf{b}$ form a triangle with $\theta$ between the first two, and $\mathbf{a}-\mathbf{b}$ opposite $\theta$. The Law of Cosines is exactly the tool relating a triangle's third side to two sides and the included angle.

Expanding $\Vert \mathbf{a}+\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2+2\mathbf{a}\cdot \mathbf{b}+\Vert \mathbf{b}{\Vert }^2$ and bounding $\mathbf{a}\cdot \mathbf{b}\le \Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert$ (which is C–S) gives $\Vert \mathbf{a}+\mathbf{b}\Vert \le \Vert \mathbf{a}\Vert +\Vert \mathbf{b}\Vert$.

It is $0$, giving $\Vert 0\Vert =0$. The angle is undefined because $\cos \theta =\frac{0}{0\cdot \Vert \mathbf{b}\Vert }$ divides by zero — the zero vector has no direction.

Yes, trivially. Both $|\mathbf{a}\cdot 0|=0$ and $\Vert \mathbf{a}\Vert \Vert 0\Vert =0$, so it reads $0\le 0$ — an equality, consistent with $0$ being a (trivial) scalar multiple of any vector.

$\cos \theta =\frac{\Vert \mathbf{a}{\Vert }^2}{\Vert \mathbf{a}{\Vert }^2}=1$, so $\theta =0^{\circ }$. Identical vectors point the same way, the maximum-overlap case.

$\cos \theta =-1$ so $\theta =180^{\circ }$, and $|\mathbf{a}\cdot \mathbf{b}|=\Vert \mathbf{a}\Vert \Vert \mathbf{b}\Vert$ — the equality case, since $\mathbf{b}$ is a scalar multiple of $\mathbf{a}$.

$\mathbf{b}$ must have enough negative "pull" in matching slots that the weighted overlap is net negative — the vectors sit more than $90^{\circ }$ apart, so $\mathbf{b}$ points broadly against $\mathbf{a}$.

It becomes ordinary multiplication $a_1{b}_1$, and C–S reads $|a_1{b}_1|\le |a_1||b_1|$, which is just $|xy|=|x||y|$ — always an equality, since any two 1-D vectors are parallel.