1
$\begingroup$

Find a vector $$\vec{w}$$ such that it is perpendicular to the following vectors:

$$ \vec{v_1} = [-1,2,-1,0]^{\,T} $$ $$ \vec{v_2} = [-3,9,-3,2]^{\,T} $$ $$ \vec{v_3} = [-1,-10,2,-8]^{\,T} $$

Now I know that in order for two vectors to be perpendicular, the dot product must equal zero.

Can I take the three vectors, augment them into a 4X3 matrix and then some how get it into the form Ax = b?

$\endgroup$
  • $\begingroup$ That is exactly what you do, but simply write $\vec w$ instead of $x$ in the equation $Ax=b$. $\endgroup$ – Lee Mosher Feb 17 '16 at 17:49
  • $\begingroup$ @Redshift? Ever heard of Null space? Because after applying dot products, this is what it comes down to...Get the RREF mode on your graphing calculator ready! You may work it out and we can confirm if you did it right $\endgroup$ – imranfat Feb 17 '16 at 17:52
0
$\begingroup$

Or take their cross product :

$$v_1\times v_2\times v_3=\begin{vmatrix}i&j&k&w\\-1&2&-1&0\\-3&9&-3&2\\-1&-10&2&-8\end{vmatrix}=$$$${}$$

$$=\begin{vmatrix}i&j&k&w\\-1&2&-1&0\\0&3&0&2\\0&-12&3&-8\end{vmatrix}=\begin{vmatrix}i&j&k&w\\-1&2&-1&0\\0&3&0&2\\0&0&3&0\end{vmatrix}=-3\begin{vmatrix}i&j&w\\-1&2&0\\0&3&2\end{vmatrix}=\left(-12,\,-6,\,0,\,9\right)=:u$$$${}$$

Now check $\;u^t\;$ indeed perpendicular to all your three vectors.

$\endgroup$
0
$\begingroup$

Note that a vector $\vec v$ is orthogonal to your given vectors if and only if $A\vec v=\vec 0$ where $$ A= \left[\begin{array}{rrrr} -1 & 2 & -1 & 0 \\ -3 & 9 & -3 & 2 \\ -1 & -10 & 2 & -8 \end{array}\right] $$ Thus we wish to compute a basis for the null space of $A$. Row-reducing gives $$ \DeclareMathOperator{rref}{rref}\rref A= \left[\begin{array}{rrrr} 1 & 0 & 0 & \frac{4}{3} \\ 0 & 1 & 0 & \frac{2}{3} \\ 0 & 0 & 1 & 0 \end{array}\right] $$ Hence a $\vec v$ is orthogonal to your given vectors if and only if $$ \vec v = \begin{bmatrix}v_1\\ v_1\\ v_3\\ v_4\end{bmatrix}= \begin{bmatrix}-\frac{4}{3}\,v_4\\ -\frac{2}{3}\,v_4\\0\\ v_4\end{bmatrix}= \begin{bmatrix}-\frac{4}{3}\\ -\frac{2}{3}\\0\\ 1\end{bmatrix}v_4 $$ That is, the vectors orthogonal to your given vectors are exactly the scalar multiples of $\left\langle -\frac{4}{3}, -\frac{2}{3}, 0, 1\right\rangle$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.