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I need to find the components of vectors $\vec{c}_{i}$that are perpendicular to $\vec{a} = [3, -1, 5]$ and $\vec{b} = [2, 2, 5]$, knowing that the length of obtained vectors $\vec{c}_{i}$ has to be exactly $\sqrt{42}$.

I know two things:

  1. Vector $\vec{c} = [c_{1}, c_{2}, c_{3}]$ is perpendicular to $\vec{a} = [a_{1}, a_{2}, a_{3}]$ if their dot product equals zero. Thus the following condition has to be satisfied: $c_{1}a_{1} + c_{2}a_{2} + c_{3}a_{3} = 0$.
  2. Cross product of vectors $\vec{a}$ and $\vec{b}$ would give me as a result vector $\vec{c}$ that is indeed perpendicular to both vectors. However, is this information useless for this task?

Anyway how should I solve this task?

Perhaps solving the following system of equations would yield the vectors I am looking for?

$$\begin{cases} a_{1}c_{1} + a_{2}c_{2} + a_{3}c_{3} = 0 \\ b_{1}c_{1} + b_{2}c_{2} + b_{3}c_{3} = 0 \\ \sqrt{c^2_{1} + c^2_{2} + c^2_{3}} = \sqrt{42} \end{cases}$$

$$\begin{cases} 3c_{1} -c_{2} + 5c_{3} = 0 \\ 2c_{1} + 2c_{2} + 5c_{3} = 0 \\ \sqrt{c^2_{1} + c^2_{2} + c^2_{3}} = \sqrt{42} \end{cases}$$

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  • $\begingroup$ Yep, you'll have to solve these 3 equations and 3 unknowns. $\endgroup$
    – Vizag
    May 25 '19 at 16:34
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Get the cross product $\vec{c}$ The answer you want is $\vec{c'}=\frac{\sqrt{42}}{|\vec{c}|}\vec{c}$

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Take the cross product of $\vec{a}$ and $\vec{b}$ (as well as its negative). This will give you vectors perpendicular to both. Calculate the length of the obtained vectors, and multiply by an appropriate constant to make the length $\sqrt {42}$.

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