How do you normalize $(15 i - 3 j + 12 k)$ $(15i- 3j + 12k)$ ?

Question

How do you normalize $(15 i - 3 j + 12 k)$ $(15i- 3j + 12k)$ ?

1 Answer

Impact of this question

1361 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-04T05:30:19

Normalizing a vector involves dividing each of its elements by its length. In this case the normalized vector is $(\frac{15}{\sqrt{378}} i - \frac{3}{\sqrt{378}} j + \frac{12}{\sqrt{378}} k)$ $(15/sqrt378i-3/sqrt378j+12/sqrt378k)$ or $(\frac{15}{19.4} i - \frac{3}{19.4} j + \frac{12}{19.4} k)$ $(15/19.4i-3/19.4j+12/19.4k)$ .

Explanation:

Normalizing a vector is creating a vector of length $1$ $1$ unit in the same direction as the original vector.

To do that, we divide each of the elements by the length of the vector. The find the length of a vector $(a i + b j + c k)$ $(ai+bj+ck)$ the formula is:

$l = \sqrt{a^{2} + b^{2} + c^{2}}$ $l = sqrt(a^2+b^2+c^2)$

In this case:

$l = \sqrt{15^{2} + {(- 3)}^{2} + 12^{2}} = \sqrt{225 + 9 + 144} = \sqrt{378}$ $l = sqrt(15^2+(-3)^2+12^2)=sqrt(225+9+144) = sqrt378$ $(= 19.4)$ $(=19.4)$

The final vector can be expressed as $(\frac{15}{\sqrt{378}} i - \frac{3}{\sqrt{378}} j + \frac{12}{\sqrt{378}} k)$ $(15/sqrt378i-3/sqrt378j+12/sqrt378k)$ or $(\frac{15}{19.4} i - \frac{3}{19.4} j + \frac{12}{19.4} k)$ $(15/19.4i-3/19.4j+12/19.4k)$ .

Science

Math

Humanities

... and beyond

How do you normalize $(15 i - 3 j + 12 k)$ $(15i- 3j + 12k)$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you normalize (15i−3j+12k) (15i- 3j + 12k) ?

1 Answer

Explanation:

Related questions

Impact of this question

How do you normalize $(15 i - 3 j + 12 k)$ $(15i- 3j + 12k)$ ?