Question

How do you find the vector $\mathrm{v}$ with the given magnitude of $9$ and in the same direction as $u=[2,5]$ ?

Answer 1

Step 1: Defining vector:

Vector:

1. If a quantity has both magnitude and direction, it is known as a vector.
2. Velocity, force, and electromagnetic field are examples of a vector.

Vector with magnitude and the components of another vector:

Consider two vectors $\overrightarrow{u}$and $\overrightarrow{v}$ are in same direction.

The magnitude of $\overrightarrow{u}$ is $\Vert \overrightarrow{u}\Vert$ and $\overrightarrow{v}$ is $\Vert \overrightarrow{v}\Vert$.

If the components of $\overrightarrow{\mathrm{u}}$ and the magnitude of $\overrightarrow{\mathrm{v}}$ are known, the components of $\overrightarrow{v}$ can be obtained by the formula:

$\overrightarrow{v}=\frac{\Vert \overrightarrow{v}\Vert ·\overrightarrow{u}}{\Vert \overrightarrow{u}\Vert }$.

Step 2: Calculate the magnitude of $\overrightarrow{\mathrm{u}}:$

$\begin{array}{rcl}\Vert \overrightarrow{\mathrm{u}}\Vert & =& \sqrt{2^2+5^2}\\ & =& \sqrt{4+25}\\ & =& \sqrt{29}\end{array}$

Step 3: Find the components of $\overrightarrow{v}:$

Magnitude,$\parallel \overrightarrow{\mathrm{v}}\parallel =9,\overrightarrow{\mathrm{u}}=<2,5>\text{and}\parallel \overrightarrow{\mathrm{u}}\parallel =\sqrt{29}$
Substitute all the values in the formula,
$\begin{array}{rcl}\overrightarrow{v}& =& \frac{9\cdot ⟨2,5⟩}{\sqrt{29}}\\ & =& \frac{⟨9\cdot 2\cdot 9\cdot 5⟩}{\sqrt{29}}\\ & =& \frac{⟨18\cdot 45⟩}{\sqrt{29}}\end{array}$
Hence, the vector $\mathrm{v}$ with the given magnitude of $9$ and in the same direction as $u=[2,5]$ is $\overrightarrow{v}=\frac{18}{\sqrt{29}}\hat{i}+\frac{45}{\sqrt{29}}\hat{j}$.