How do you find the unit vector in the direction of the given vector of $w = 7 j - 3 i$ $w=7j-3i$ ?

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Answer 1 · 2017-02-08T11:41:12

The answer is $= \frac{1}{\sqrt{58}} \cdot < 7, - 3 >$ $=1/sqrt58*<7,-3>$

The unit vector in direction of $\to a$ $veca$ is $= \frac{\to a}{∣ ∣ ∣ ∣ \to a ∣ ∣ ∣ ∣}$ $=veca/(||veca||)$

Here $\to w = < 7, - 3 >$ $vecw= <7,-3>$

The modulus is $= ∣ ∣ ∣ ∣ \to w ∣ ∣ ∣ ∣ = | | < 7, - 3 > | |$ $=||vecw||=||<7,-3>||$

$= \sqrt{49 + 9} = \sqrt{58}$ $=sqrt(49+9)=sqrt(58)$

So,

the unit vector is

$= \frac{\to w}{∣ ∣ ∣ ∣ \to w ∣ ∣ ∣ ∣} = \frac{1}{\sqrt{58}} \cdot < 7, - 3 >$ $=vecw/(||vecw||)=1/sqrt58*<7,-3>$

How do you find the unit vector in the direction of the given vector of w=7j−3iw=7j-3i?