Question

Find a vector $v$whose magnitude is $4$and whose component

in the $i$direction is twice the component in the $j$direction.

Short answer

The vector is$\frac{8}{\sqrt{5}}i+\frac{4}{\sqrt{5}}j$

Step-by-step solution

Step 1. Given information

According to the question the vector will be,

$v=2xi+xj$

Step 2. Magnitude of the vector

The magnitude of the vector $v=2xi+xj$is$\sqrt{4x^2+x^2}=4$

$$\begin{gathered} \sqrt{5x^2}=4 \\ 5x^2=16 \\ {x}^2=\frac{16}{5} \\ x=\frac{4}{\sqrt{5}} \end{gathered}$$

Step 3. Required vector

Now the vector will be

$$\begin{gathered} v=2xi+xj \\ v=2\left(\frac{4}{\sqrt{5}}\right)i+\frac{4}{\sqrt{5}}j \\ v=\frac{8}{\sqrt{5}}i+\frac{4}{\sqrt{5}}j \end{gathered}$$