Question

Magnitude of Vectors

Find $\left|\mathbf{u}\right|,\left|\mathbf{v}\right|,\left|2\mathbf{u}\right|,\left|\frac{1}{2}\mathbf{v}\right|,|\mathbf{u}+\mathbf{v}|,|\mathbf{u}-\mathbf{v}|\&\left|\mathbf{u}\right|-\left|\mathbf{v}\right|$

$\mathbf{u}=-2\mathbf{i}+3\mathbf{j},\mathbf{v}=\mathbf{i}-2\mathbf{j}$

Short answer

The required values are-

$\begin{array}{l}\left|u\right|=\sqrt{13}\\ \left|v\right|=\sqrt{5}\\ \left|2u\right|=2\sqrt{13}\\ \left|\frac{1}{2}v\right|=\frac{\sqrt{5}}{2}\\ |u+v|=\sqrt{2}\\ |u-v|=\sqrt{34}\\ \left|u\right|-\left|v\right|=\sqrt{13}-\sqrt{5}\end{array}$

Step-by-step solution

Step 1. Given information

Two points are given as $\mathbf{u}=-2\mathbf{i}+3\mathbf{j},\mathbf{v}=\mathbf{i}-2\mathbf{j}$

Step 2. Concept used

Modulus represents the magnitude of vector which can be written as-

$\left|v\right|=\sqrt{a_1^2+a_2^2}$ for a point $u=a_{1}i+a_{2}j$.

We can also add vectors as-

$u+v=\left(a_1+b_1\right)i+\left(a_2+b_2\right)j$

Use the vector formulas to get the results.

Step 3. Calculation

Substitute the values of u and v in the required terms,

$\begin{array}{l}\left|u\right|=\sqrt{{(-2)}^2+3^2}\\ =\sqrt{4+9}\\ =\sqrt{13}\\ \left|v\right|=\sqrt{1^2+{(-2)}^2}\\ =\sqrt{1+4}\\ =\sqrt{5}\end{array}$

Also in,

$\begin{array}{l}\left|2u\right|=2\left|u\right|\\ =2\sqrt{3}\\ \left|\frac{1}{2}v\right|=\frac{1}{2}\left|v\right|\\ =\frac{1}{2}\sqrt{5}\\ =\frac{\sqrt{5}}{2}\end{array}$

And,

$\begin{array}{l}|u+v|=\sqrt{{(-2+1)}^2+{(3-2)}^2}\\ =\sqrt{1^2+1^2}\\ =\sqrt{2}\\ |u-v|=\sqrt{{(-2-1)}^2+{(3-(-2\left)\right)}^2}\\ =\sqrt{{(-3)}^2+5^2}\\ =\sqrt{9+25}\\ =\sqrt{34}\\ \left|u\right|-\left|v\right|=\sqrt{13}-\sqrt{5}\end{array}$

$\begin{array}{l}|u-v|=\sqrt{{(2-3)}^2+{(1-(-2\left)\right)}^2}\\ =\sqrt{{(-1)}^2+3^2}\\ =\sqrt{1+9}\\ =\sqrt{10}\\ \left|u\right|-\left|v\right|=\sqrt{5}-\sqrt{13}\end{array}$