# Thread: vectors: sum, parallel and perpendicular

1. ## vectors: sum, parallel and perpendicular

4 vectors a, b, c and d.

a= 3i + 2j -6k
d= 2i - j + k

a is the sum of vectors b and c.
Find b and c so that b is PARALLEL to d and c is PERPENDICULAR to d.

2. Originally Posted by jenny
4 vectors a, b, c and d.

a= 3i + 2j -6k
d= 2i - j + k

a is the sum of vectors b and c.
Find b and c so that b is PARALLEL to d and c is PERPENDICULAR to d.
Let vectors b and c be
$b = xi + yj + zk$
and
$c = pi + qj + rk$

Then
$a = b + c$

$3i + 2j - 6k = (x + p)i + (y + q)j + (z + r)k$

Thus
$\begin{matrix} x + p = 3 \\ y + q = 2 \\ z + r = -6 \end{matrix}$

Now, b is parallel to d, so
$d = eb$

Thus
$2i - j + k = e(xi + yj + zk)$

Thus
$\begin{matrix}ex = 2 \\ ey = -1 \\ e = z \end{matrix}$

And finally c is perpendicular to d, so
$c \cdot d = 0$

$(pi + qj + rk) \cdot (2i - j + k) = 0$

$2p - q + r = 0$

So we have the conditions:

$\begin{matrix} x + p = 3 \\ y + q = 2 \\ z + r = -6 \end{matrix}$
and
$\begin{matrix}ex = 2 \\ ey = -1 \\ e = z \end{matrix}$
and
$2p - q + r = 0$

We have 7 variables in 7 unknowns. It'll take a while, but it should be solvable.

-Dan

3. Originally Posted by jenny
4 vectors a, b, c and d.

a= 3i + 2j -6k
d= 2i - j + k

a is the sum of vectors b and c.
Find b and c so that b is PARALLEL to d and c is PERPENDICULAR to d.
Vector $\bold{b}$ is the orthogonal projection of $\bold{a}$ on $\bold{d}$ calculated by

$\bold{b} = \frac{\bold{a}\cdot \bold{d}}{\bold{d}\cdot\bold{d}} \bold{d} = \frac{-2}{6} \bold{d} = \frac{-1}{3} \bold{d}.$

Then set $\bold{c} = \bold{a} - \bold{b}.$

Now check that $\bold{a} = \bold{b} + \bold{c}$ and $\bold{b}$ is parallel to $\bold{d}$ since it is a multiple of it.

Finally $\bold{c}$ is perpendicular to $\bold{d}$ since

\begin{aligned} \bold{c}\cdot\bold{d} &= (\bold{a} - \bold{b})\cdot\bold{d}.\\
&= \bold{a}\cdot\bold{d} - \bold{b}\cdot\bold{d} \\
&= \bold{a}\cdot\bold{d} - \frac{\bold{a}\cdot \bold{d}}{\bold{d}\cdot\bold{d}} \bold{d}\cdot\bold{d} \\
&= \bold{a}\cdot\bold{d} - \bold{a}\cdot\bold{d} \\
&= 0.
\end{aligned}

4. Originally Posted by topsquark
why should we automatically exclude the possibility that b might have a longer length than d?
That method does not exclude the possibility that b might be longer than d.
Moreover, the solution is unique.

5. Originally Posted by Plato
That method does not exclude the possibility that b might be longer than d.
Thanks for pointing that out. I wasn't looking at it right. :\

-Dan