Question

A vector perpendicular to the vector ($\mathrm{i}+2\mathrm{j}$) and having magnitude $3\sqrt{5}$ units is _____

Answer 1

Step 1: Given and assume

The magnitude of r is $3\sqrt{5}$.

Let,

A vector, $\mathrm{r}=\mathrm{xi}+\mathrm{yj}$ perpendicular to $\mathrm{A}=(\mathrm{i}+2\mathrm{j})$.

Step 2: To find

We have to find a vector perpendicular to the vector ($\mathrm{i}+2\mathrm{j}$) and have magnitude $3\sqrt{5}$ units.

Step 3: Calculation

The dot product of r and A must be zero because $\mathrm{Cos}90°=0$.

Example., $\mathrm{r}.\mathrm{A}=(\mathrm{xi}+\mathrm{yj}).(\mathrm{i}+2\mathrm{j})$

then,

$\mathrm{x}+2\mathrm{y}=0...\left(1\right)$

According to the given information,

$$\begin{gathered} \left|\mathrm{r}\right|=\sqrt{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)} \\ =3\sqrt{5} \end{gathered}$$

By taking square both sides, we get

$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2=9\times 5 \\ {\mathrm{x}}^2+{\mathrm{y}}^2=45 \end{gathered}$$

From equation $\left(1\right)$,

$$\begin{gathered} {(-2\mathrm{y})}^2+{\mathrm{y}}^2=45 \\ 4{\mathrm{y}}^2+{\mathrm{y}}^2=45 \\ 5{\mathrm{y}}^2=45 \\ {\mathrm{y}}^2=9 \\ \mathrm{y}=3 \end{gathered}$$

By substituting the value of “y” in the equation $\left(1\right)$, we get

$$\begin{gathered} \mathrm{x}+2\left(3\right)=0 \\ \mathrm{x}+6=0 \\ \mathrm{x}=-6 \end{gathered}$$

Therefore, the vector perpendicular is $\mathrm{r}=(6\mathrm{i}-3\mathrm{j})$.