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

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  • $\begingroup$ So, what is the dot product of $b+tc$ and $a$? $\endgroup$ – Gerry Myerson Mar 30 '14 at 12:00
  • $\begingroup$ Why was this downvoted?! $\endgroup$ – MPW Mar 30 '14 at 12:03
  • $\begingroup$ 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. $\endgroup$ – please delete me Mar 30 '14 at 12:06
  • $\begingroup$ 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. $\endgroup$ – Mark Bennet Mar 30 '14 at 13:07
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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 $t=\frac{5}{6}$.

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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.

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

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