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

• So, what is the dot product of $b+tc$ and $a$? – Gerry Myerson Mar 30 '14 at 12:00
• Why was this downvoted?! – MPW Mar 30 '14 at 12:03
• Initially I was thinking that to find a vector that is perpendicular to $a$, I had to use cross product somehow and while typing the question I thought about using the definition of $cos\theta$. I still don't know how to do the second part though. – please delete me Mar 30 '14 at 12:06
• Note that $(b\cdot c)$ is a scalar, not a vector, so $a\times (b\cdot c)$ doesn't make sense. Similarly with $(a\cdot b)$ so that $(a\cdot b)\cdot c$ doesn't make sense. You can make a triple product e.g. $a\cdot (b \times c)$ which gives a scalar, or $a\times (b\times c)$ which gives a vector. – Mark Bennet Mar 30 '14 at 13:07

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 $t=\frac{5}{6}$.
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 $t=\frac56$.
• $(b+tc)\cdot a=0\\ (1+2t, 4+t, -4-3t)\cdot(-3, 3, 1)=0\\ (1+2t)\cdot(-3)+(4+t)\cdot 3+(-4-3t)\cdot 1=0\\ t=\frac{5}{6}$ – zaarcis Mar 30 '14 at 12:05