For , ∈ ℝ, we have ‖−‖≤‖+‖.
The dot product of two vectors is a vector.
For ,∈ℝ, we have ‖−‖≤‖‖+‖‖.
A homogeneous system of linear equations with more equations than variables will always have at least one parameter in its solution.
Given a non-zero vector , there exist exactly two unit vectors that are parallel to .
My answers were
- FALSE because if we assumed that a= (-1,-2) and b= (3,4) it would make the statement false
- FALSE because the dot product of 2 vectors is a scalar
- FALSE this would have the same assumption as for question 1
- FALSE I am not sure
- TRUE I am not sure
I am not sure which one of my answers is/are wrong
$1$ 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 $\pm \bf u$.