Chapter 9.4, Problem 53AYU

To determine

To calculate: The vector $v$ with magnitude 4 and whose $i$ component is twice the $j$ component.

Answer to Problem 53AYU

The vector $v$ is $\frac{8\sqrt{5}}{5}i+\frac{4\sqrt{5}}{5}j\text{ or }-\frac{8\sqrt{5}}{5}i-\frac{4\sqrt{5}}{5}j$ .

Explanation of Solution

Given information:

A vector should have magnitude 4 and whose $i$ component is twice the $j$ component.

Formula used:

The magnitude of a vector $v=ai+bj$ is $\Vert v\Vert =\sqrt{a^2+b^2}$ .

Calculation:

Consider the vector should have magnitude 4 and whose $i$ component is twice the $j$ component.

Assume the vector to be $v=ai+bj$ .

Since, $i$ component is twice the $j$ component therefore $a=2b$ .

So, the vector is $v=2bi+bj$ .

The magnitude of vector is 4. That is $\Vert v\Vert =4$ .

The magnitude of a vector $v=ai+bj$ is $\Vert v\Vert =\sqrt{a^2+b^2}$ .

Therefore,

  $\begin{array}{c}4=\sqrt{{\left(2b\right)}^2+b^2}\\ 4=\sqrt{5b^2}\\ 5b^2=16\\ b^2=\frac{16}{5}\end{array}$

Simplify it further as,

  $b=\pm \frac{4}{\sqrt{5}}$

Therefore,

  $\begin{array}{c}a=2b\\ a=\pm \frac{8}{\sqrt{5}}\end{array}$

Therefore, vector when $a=\frac{8}{\sqrt{5}},b=\frac{4}{\sqrt{5}}$ is $\frac{8\sqrt{5}}{5}i+\frac{4\sqrt{5}}{5}j$

And vector when $a=-\frac{8}{\sqrt{5}},b=-\frac{4}{\sqrt{5}}$ is $-\frac{8\sqrt{5}}{5}i-\frac{4\sqrt{5}}{5}j$

Thus, the vector $v$ is $\frac{8\sqrt{5}}{5}i+\frac{4\sqrt{5}}{5}j\text{ or }-\frac{8\sqrt{5}}{5}i-\frac{4\sqrt{5}}{5}j$ .