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  1. For 𝐮,𝐯 ∈ ℝ𝑛, we have ‖𝐮−𝐯‖≤‖𝐮+𝐯‖.

  2. The dot product of two vectors is a vector.

  3. For 𝐮,𝐯∈ℝ𝑛, we have ‖𝐮−𝐯‖≤‖𝐮‖+‖𝐯‖.

  4. A homogeneous system of linear equations with more equations than variables will always have at least one parameter in its solution.

  5. Given a non-zero vector 𝐯, there exist exactly two unit vectors that are parallel to 𝐯.

My answers were

  1. FALSE because if we assumed that a= (-1,-2) and b= (3,4) it would make the statement false
  2. FALSE because the dot product of 2 vectors is a scalar
  3. FALSE this would have the same assumption as for question 1
  4. FALSE I am not sure
  5. TRUE I am not sure

I am not sure which one of my answers is/are wrong

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  • $\begingroup$ Look at 3, call $-{\bf v} = {\bf w}$ and look up the triangle inequality. $\endgroup$ – David G. Stork May 19 at 20:53
  • $\begingroup$ wouldn't it be the same for question 1 since the triangle inequality states that ‖𝐮+𝐯‖ ≤ ‖𝐮‖+‖𝐯‖? $\endgroup$ – Drake May 19 at 21:09
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$1$ and $2$ are both right.

$3$ is wrong. The triangle inequality actually implies $3$:

$$||u-v||\leq ||u||+||-v||=||u||+||v||$$

$4$ is right. Just consider

$$\left\{ \begin{array}{ll} x=0 \\ x=0 \end{array} \right.$$ The only solution is $x=0$. This statement would be true the other way around: a homogeneous system of linear equations with more variables than equations will always have at least one parameter in its solution.

And $5$ is also right: $\bf u=\frac{v}{||v||}$ is a unit vector. Any other vector parallel to $\bf v$ (and thus also parallel to $\bf u$) is of the form $k\bf u$ for some real number $k$. And $k\bf u$ is a unit vector if and only if $k=\pm 1$. Hence the two unit vectors are $\pm \bf u$.

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  • $\begingroup$ thanks for the explanation $\endgroup$ – Drake May 19 at 22:32
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Number 3 is incorrect. Why? Because of the well-known fact that $|\bf{x} + \bf{y}| \le |\bf{x}| + |\bf{y}|$ (the Triangle Inequality).

In particular, $|\bf{u} - \bf{v}| = |\bf{u} + (- \bf{v})| \le |\bf{u}| + |- \bf{v}| = |\bf{u}| + |\bf{v}|$.

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  • $\begingroup$ thanks for the explanation $\endgroup$ – Drake May 19 at 22:32

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