Review question

# If $a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$ is perpendicular to these vectors, what is $a:b:c$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6493

## Solution

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 $-\dfrac{2}{3}\mathbf{i}+\dfrac{2}{3}\mathbf{j}-\dfrac{1}{3}\mathbf{k}$.