Question

prove that magnitude of unit vector is one​

Answer 1

Explanation:

A unit vector is a vector with magnitude of 1. ... In some situations it is helpful to find a unit vector that has the same direction as a given vector. A unit vector of v, in the same direction as v, can be found by dividing v by its magnitude ∥ v ∥ .

Answer 2

We have to prove that the magnitude of a unit vector is one.

Concepts :

• When a vector is divided by its magnitude, we get the unit vector along the given vector. for example, if a is a vector than, unit vector along a is a/|a|.
• unit vector is represented by symbol '^' which is called a cap or hat.

• $\text{unit vector along}\:\:\vec a=\hat a=\frac{\vec a}{|\vec a|}$

A vector , A is given by,

$\vec A=a_x\hat i+a_y\hat j+a_z\hat k$

∴ The magnitude of vector A is given by,

$|\vec A|=\sqrt{a_x^2+a_y^2+a_z^2}$

Now, unit vector along vector A is given by,

$\hat A=\frac{\vec A}{|\vec A|}=\frac{a_x\hat i+a_y\hat j+a_z\hat k}{\sqrt{a_x^2+a_y^2+a_z^2}}\\\\\\\implies\hat A=\left(\frac{a_x}{\sqrt{a_x^2+a_y^2+a_z^2}}\right)\hat i+\left(\frac{a_y}{\sqrt{a_x^2+a_y^2+a_z^2}}\right)\hat j+\left(\frac{a_z}{\sqrt{a_x^2+a_y^2+a_z^2}}\right)\hat k$

now if we find the magnitude of unit vector along A , we will get,

$|\hat A|=\sqrt{\left(\frac{a_x}{\sqrt{a_x^2+a_y^2+a_z^2}}\right)^2+\left(\frac{a_y}{\sqrt{a_x^2+a_y^2+a_z^2}}\right)^2+\left(\frac{a_z}{\sqrt{a_x^2+a_y^2+a_z^2}}\right)^2}\\\\\\=\sqrt{\left(\frac{a_x^2}{a_x^2+a_y^2+a_z^2}\right)+\left(\frac{a_y^2}{a_x^2+a_y^2+a_z^2}\right)+\left(\frac{a_z^2}{a_x^2+a_y^2+a_z^2}\right)}\\\\\\=\sqrt{\frac{a_x^2+a_y^2+a_z^2}{a_x^2+a_y^2+a_z^2}}\\\\=1$

Therefore the magnitude of a unit vector is always 1.

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