Scalar Multiplication

Scalar multiplication. For a vector $\bar a$ and a real number (scalar) $k$, the scalar multiple $k\bar a$

is the vector with magnitude $|k\bar a|=|k||\bar a|$, pointing in the same direction as $\bar a$ when $k>0$

and in the opposite direction when $k<0$; when $k=0$, $k\bar a=\bar 0$. So $2\bar a$ has the same direction as

$\bar a$ but is twice as long.

Some direct consequences: $\bar a$ and $k\bar a$ are always collinear (parallel) vectors; two non-zero vectors

$\bar a$ and $\bar b$ are collinear exactly when $\bar a=m\bar b$ for some non-zero scalar $m$; if $\hat a$ is

the unit vector along a non-zero vector $\bar a$, then $\bar a=|\bar a|\hat a$; and a vector of length $k$

along $\bar a$'s direction is $k\hat a=\dfrac{k}{|\bar a|}\bar a$.

Worked examples (using scalar multiples to test collinearity/parallelism, and to solve for unknown scalars).

• To show $2\hat i-3\hat j+4\hat k$ and $-4\hat i+6\hat j-8\hat k$ are parallel: writing the second vector as $-2(2\hat i-3\hat j+4\hat k)$ shows it is a scalar multiple of the first, so the two are parallel.
• If $\bar a$ and $\bar b$ are given to be non-collinear, an equation like $\bar a+3\bar b=2\lambda\bar a-\mu \bar b$ is solved by matching the coefficients of $\bar a$ and of $\bar b$ separately on both sides (since a non-zero combination of two non-collinear vectors can only be zero when both coefficients are zero) -- here $2\lambda=1$ and $-\mu=3$ give $\lambda=\tfrac12,\ \mu=-3$.
• To decide whether vectors such as $\bar a=-2\hat i+3\hat j-\hat k$, $\bar b=3\hat i-6\hat j+9\hat k$ are …