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

Question

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 $(-3, 3, 1)\times (18, 0, 0)$?

Answer 1

Your first intuition is correct. The dot product of $a$ with $b+tc$ must be zero. This should give you an equation to solve for $t$. I got $t=\frac56$.

The reason this works is that the dot product of two vectors measures the relative length of the "shadow" of one vector on the other (relative "projection" is the technical phrase). If the vectors are perpendicular, there is no shadow.

Answer 2

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 $t=\frac{5}{6}$.