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Q. 53

Expert-verified
Found in: Page 603

### Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

# Find a vector $v$ whose magnitude is $4$ and whose componentin the $i$direction is twice the component in the $j$ direction.

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

See the 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$\sqrt{4{x}^{2}+{x}^{2}}=4$

$\sqrt{5{x}^{2}}=4\phantom{\rule{0ex}{0ex}}5{x}^{2}=16\phantom{\rule{0ex}{0ex}}{x}^{2}=\frac{16}{5}\phantom{\rule{0ex}{0ex}}x=\frac{4}{\sqrt{5}}$

## Step 3. Required vector

Now the vector will be

$v=2xi+xj\phantom{\rule{0ex}{0ex}}v=2\left(\frac{4}{\sqrt{5}}\right)i+\frac{4}{\sqrt{5}}j\phantom{\rule{0ex}{0ex}}v=\frac{8}{\sqrt{5}}i+\frac{4}{\sqrt{5}}j$