Chapter 9.5, Problem 39AYU

a.

To determine

To prove: the given vector $\text{u}+\text{v and u}-\text{v}$ are orthogonal.

Explanation of Solution

Given:

The givenvectoris $\text{u and v}$ have same magnitude.

Calculation:

As per the question,the vector is $\text{u and v}$ have same magnitude

  $\Vert \text{u}\Vert =\Vert \text{v}\Vert$ .

Now, let $u=a_{1}i+b_{1}j\text{ and }v=a_{2}i+b_{2}j$ .

Therefore,

  $\begin{array}{l}\Vert \text{u}\Vert =\Vert \text{v}\Vert \\ \Rightarrow \sqrt{a_1^2+b_1^2}=\sqrt{a_2^2+b_2^2}\\ \therefore \text{ squaring both sides,}\\ a_1^2+b_1^2=a_2^2+b_2^2\end{array}$

Two vectors are orthogonal if and only if dot product of two vectors is zero.

  $\Rightarrow \text{u}\cdot v=0$

  $\begin{array}{l}\text{u}\cdot \text{v}=\left(a_{1}i+b_{1}j+a_{2}i+b_{2}j\right)\cdot \left(a_{1}i+b_{1}j-a_{2}i-b_{2}j\right)\\ =\left[\left(a_1+a_2\right)i+\left(b_1+b_2\right)j\right]\left[\left(a_1-a_2\right)i+\left(b_1-b_2\right)j\right]\\ =\left[\left(a_1+a_2\right)\left(a_1-a_2\right)+\left(b_1+b_2\right)\left(b_1-b_2\right)\right]\\ =a_1^2+b_1^2-a_2^2-b_2^2\\ =\left(a_1^2-a_2^2\right)+\left(b_1^2-b_2^2\right)\{\because u=v\}\\ =0\end{array}$

Hence, the given vector $\text{u}+\text{v and u}-\text{v}$ are orthogonal

b.

To determine

To prove: that an angle inscribed in a semicircle is a right angle given vector $\text{u}+\text{v and u}-\text{v}$ are orthogonal.

Explanation of Solution

Given:

The given vector is $\text{u and v}$ have same magnitude.

Calculation:

As per the question, the vector is $\text{u and v}$ have same magnitude

  $\Vert \text{u}\Vert =\Vert \text{v}\Vert$ .

  

From the figure,

  $\begin{array}{l}AC=OA+OC\\ AC=-\left(-v\right)+u\\ AC=v+u\\ CB=CO+OB\\ CB=-u+v\end{array}$ .

Therefore, $v+u,v-u$ are orthogonal.

  $\begin{array}{l}\Rightarrow AC\text{ is perpendicular on }CB\\ \Rightarrow \angle ACB={90}^{\circ }\end{array}$

Hence, the given vector $\text{u}+\text{v and u}-\text{v}$ are orthogonal