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Encyclopedia>Numbers and Quantities>Vectors>Two Dimensions>Introduction to Vectors>What Is the Length of a Vector?

A vector a with its x and y component

The length of a vector $\vec{a} = (x, y)$ is denoted by an absolute value sign $| \vec{a} |$ , and can be found by using this formula (notice the similarity to the Pythagorean theorem):

Formula

The Length of a Vector

| \vec{a} | = | (x, y) | = \sqrt{x^{2} + y^{2}}

Example 1

Find the length of the vector $\vec{v} = (3, 4)$ .

$\begin{array}{l} | \vec{v} | & = | (3, 4) | = \sqrt{3^{2} + 4^{2}} \\ = \sqrt{9 + 16} = \sqrt{25} = 5 \end{array}$

$| \vec{v} | = | (3, 4) | = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$

Example 2

For what values of $t$ does the vector $(t, t^{2})$ have length $\sqrt{2}$ ?

In this case, you use the formula as an equation and solve it for $t$ . This is how you do it:

\begin{array}{l} \sqrt{2} & = \sqrt{{(t)}^{2} + {(t^{2})}^{2}} \\ = \sqrt{t^{2} (1 + t^{2})} \\ = t \sqrt{1 + t^{2}} . \end{array}

You divide each side of the equation by

t

and get

\begin{array}{l} \frac{\sqrt{2}}{t} & = \sqrt{1 + t^{2}} & Squaring \\ \frac{2}{t^{2}} & = 1 + t^{2} & | \cdot t^{2} \\ 2 & = t^{2} + t^{4} & Move over \\ 0 & = t^{4} + t^{2} - 2 & u = t^{2} \\ 0 & = u^{2} + u - 2 & Factorizing \\ 0 & = (u + 2) (u - 1) . \end{array}

\begin{array}{l} \frac{\sqrt{2}}{t} & = \sqrt{1 + t^{2}} & Squaring \\ \frac{2}{t^{2}} & = 1 + t^{2} & | \cdot t^{2} \\ 2 & = t^{2} + t^{4} & Move over \\ 0 & = t^{4} + t^{2} - 2 & Substitute u = t^{2} \\ 0 & = u^{2} + u - 2 & Factorizing \\ 0 & = (u + 2) (u - 1) . \end{array}

You can now substitute back. Then you get this from the first factor:

\begin{align} u + 2 & = 0 & (1) \\ t^{2} + 2 & = 0 \\ t^{2} & = - 2 \end{align}

From the second factor, you get

\begin{align} u - 1 & = 0 & (2) \\ t^{2} - 1 & = 0 \\ t^{2} & = 1 \\ t & = \pm 1 . \end{align}

Since $t^{2} = - 2$ from (1) has no solution, $t = \pm 1$ from (2).

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