Q:

Adding two vectors such that the resulting vector is perpendicular to a third vector

Let $$a = (-3, 3, 1)$$
    $$b = (1, 4, -4)$$
    $$c = (2, 1, -3)$$
For which values of $t \in \Re$ is $b + tc$ perpendicular to a?
For a vector to be perpendicular to $a$, the dot product of that vector and $a$ must equal to 0, right? I don't know where to go from here.
Also, can I do the following and in which case is there a solution and which case is impossible:

$a \times  (b.c)$
$(a.b).c$

In the first case I am doing the cross product of a vector and a scalar and in the second case I am doing the dot product of a scalar and a vector, so I am not sure how that works.
For the first one, is this correct of $(-3, 3, 1)\times (18, 0, 0)$?

A:

Let ${\bf v}=(v_1,v_2,v_3)$ and ${\bf u}=(u_1,u_2,u_3)$ be two vectors in $\mathbb{R}^3$. Denote by $\langle {\bf v},{\bf u}\rangle$ the usual inner product on $\mathbb{R}^3$, that is:
$$\langle {\bf v},{\bf u}\rangle=v_1\cdot u_1 + v_2\cdot u_2 + v_3\cdot u_3.$$
You want to find all $t\in \mathbb{R}$ such that $\langle b+tc,a\rangle=0$, so you have:
$$0=\langle b+tc,a\rangle=\langle b,a\rangle+\langle tc,a\rangle=\langle b,a\rangle+t\langle c,a\rangle=5+t\cdot(-6).$$
Thus you get the unique solution $t=\frac{5}{6}$.

