Find the vector $\vec{c}$ which is orthogonal to $\vec{a}$ and $\vec{b}$ and whose first component is 1

$\vec{a}$ = (0 / 4 / -2), start point: $P$ (2 / -3 / 5), end point: $A$ (2 / 1 / 3)

$\vec{b}$ = (-5 / 5 / -1), start point: $P$ (2 / -3 / 5), end point: $B$ (-3 / 2 / 4)

I can only assume that $\vec{c}$ is supposed to be an unit vector. What I don't understand, is, how one vector is supposed to be orthogonal to both vectors $\vec{a}$ and $\vec{b}$ at once?

Maybe the Point $P$ provides the place for the new vector to be orthogonal to both vectors, but how to find $\vec{c}$ respectively determine its properties (direction, x-, y-, and z-values)?

Edit - As suggested by rschwieb here's the solution I worked out:

The first component of $\vec{c}$ is $x = 1$, the other ones are unknown, thus:

$\vec{c}$ = (1, y, z)

Because the vector $\vec{c}$ is orthogonal to the other vectors $\vec{a}$ and $\vec{b}$, the scalar products are: $\vec{c} * \vec{a} = 0$ and $\vec{c} * \vec{b} = 0$. This results in the following system of equations:

$\vec{a} * \vec{c} => (0, 4, -2) * (1, y, z) = (0*1) + (4*y) + (-2*z) = 0$

$\vec{b} * \vec{c} => (-5, 5, -1) * (1, y, z) = (-5*1) + (5*y) + (-1*z) = 0$

Which is then simplified to:

$4y - 2z = 0$

$-5 + 5y - z = 0$

When calculating this system of equations, the solutions then are

y = $\dfrac{5}{3}$ and z = $\dfrac{10}{3}$

Thus, the end result is:

$\vec{c} = (1, \dfrac{5}{3}, \dfrac{10}{3})$

  • $\begingroup$ Good! Be sure to try the cross-product method, too. $\endgroup$
    – rschwieb
    May 26 '12 at 22:11

Of course a vector can be perpendicular to two vectors at once: think of the x-y-z coordinate axes!

If you simply let $\vec{c} =(1,a,b)$ be your vector, and find the resulting system of equations that arises if you assume $\vec{c}$ has dot product 0 the other two vectors, you will find that $\vec{c}=(1,5/3,10/3)$.

I would like to encourage you to show this work in a solution of your own.

Edit: Somehow made a mistake in my haste. $\vec{c}=(1,5/3,10/3)$. Corrected above.

  • $\begingroup$ Allright, so the first component (1) is the x-coordinate of $\vec{c}$. What do you mean by $\vec{c}$ having "dot product o the other two vectors"? $\endgroup$ May 26 '12 at 18:57
  • $\begingroup$ @MiroslavCetojevic: Do you know what dot product is? Calculate $a \cdot c$ and $b \cdot c$. $\endgroup$
    – Gigili
    May 26 '12 at 19:01
  • $\begingroup$ Oh right, I'm german-speaking and didn't know that dot procuct is the same as scalar product. $\endgroup$ May 26 '12 at 19:04
  • $\begingroup$ So this means (ac) + (bc)? $\endgroup$ May 26 '12 at 19:07
  • $\begingroup$ You have a system of equations to find $y$ and $z$, as rschwieb explained. $a \cdot c=0$ and $b \cdot c=0$. Note that the zero is due to orthogonality. $\endgroup$
    – Gigili
    May 26 '12 at 19:17

If you take the cross product of two non-parallel vectors, you automatically get a vector that is perpendicular to the original two. Now all you have to do is divide all three components of the result by the value of the first component. The result should be the same as rschwieb's solution. You'd skip the equation system part, but instead have to calculate a cross product, so I'm not sure which solution is easier.

  • $\begingroup$ Also a good alternative! @MiroslavCetojevic be sure to try it both ways! $\endgroup$
    – rschwieb
    May 26 '12 at 22:08

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