Given that the vector \(a\mathbf{i}+b\mathbf{j}+c\mathbf{k}\) is perpendicular to each of the vectors \(5\mathbf{i}+3\mathbf{j}-4\mathbf{k}\) and \(2\mathbf{i}-\mathbf{j}-6\mathbf{k}\), find the ratios \(a:b:c\).
Two vectors are perpendicular exactly if their scalar product is equal to zero.
So we know that \((a\mathbf{i}+b\mathbf{j}+c\mathbf{k}).(5\mathbf{i}+3\mathbf{j}-4\mathbf{k})=0\) and \((a\mathbf{i}+b\mathbf{j}+c\mathbf{k}).(2\mathbf{i}-\mathbf{j}-6\mathbf{k})=0\).
Hence we can form the simultaneous equations
\[\begin{align*} 5a+3b-4c &= 0,\\ 2a-b-6c &=0. \end{align*}\]The second equation tells us \(b=2a-6c\), and substituting into the first equation, we find \(5a+3(2a-6c)-4c = 0\), which tells us \(a=2c\).
Substituting back into the second equation then reveals \(b=-2c\).
Hence \(a\mathbf{i}+b\mathbf{j}+c\mathbf{k} = 2c\mathbf{i}-2c\mathbf{j}+c\mathbf{k}\), and we conclude the desired ratio is \(2:-2:1\).
Hence find a unit vector which is perpendicular to each of the vectors \(5\mathbf{i}+3\mathbf{j}-4\mathbf{k}\) and \(2\mathbf{i}-\mathbf{j}-6\mathbf{k}\).
The vector \(2c\mathbf{i}-2c\mathbf{j}+c\mathbf{k}\) has length \(c\sqrt{2^2+(-2)^2+1^2} = 3c\).
Thus one unit vector in this direction is \(\dfrac{1}{3c}(2c\mathbf{i}-2c\mathbf{j}+c\mathbf{k})=\dfrac{2}{3}\mathbf{i}-\dfrac{2}{3}\mathbf{j}+\dfrac{1}{3}\mathbf{k}\).
The other is \(-\dfrac{2}{3}\mathbf{i}+\dfrac{2}{3}\mathbf{j}-\dfrac{1}{3}\mathbf{k}\).
