Chapter 9.4, Problem 54AYU

To determine

To find: a vector vwhosethe component in the i direction is equal to the component in the j direction.

Answer to Problem 54AYU

  $\text{v}=\frac{3}{\sqrt{2}}\text{i}+\frac{3}{\sqrt{2}}\text{j}\text{or }-\frac{3}{\sqrt{2}}\text{i}-\frac{3}{\sqrt{2}}\text{j}$ .

Explanation of Solution

Given:

The given condition is “the component in the i direction is equal to the component in the j direction”.

And the magnitude of the vector is 3.

Calculation:

As the magnitude of the vector $\text{v}=a\text{i}+b\text{j}=\langle a,b\rangle \text{is }\Vert \text{v}\Vert =\sqrt{a^2+b^2}$ .

Since the component in the i direction is equal to the component in the j direction.

Therefore,

  $\begin{array}{l}\Vert 3\Vert =\sqrt{a^2+b^2}\{\because \text{v}=3\}\\ \Vert 3\Vert =\sqrt{a^2+a^2}\{a=b\}\\ \therefore \sqrt{2a^2}=3\\ \Rightarrow 2a^2=9\left\{squaringbothsides\right\}\\ a^2=\frac{9}{2}\\ a=\sqrt{\frac{9}{2}}=\pm \frac{3}{\sqrt{2}}\end{array}$

  $\text{as}a=\pm \frac{3}{\sqrt{2}},\text{then}b=\pm \frac{3}{\sqrt{2}}$

Hence, the vectoris $\text{v}=\frac{3}{\sqrt{2}}\text{i}+\frac{3}{\sqrt{2}}\text{j}\text{or }-\frac{3}{\sqrt{2}}\text{i}-\frac{3}{\sqrt{2}}\text{j}$ .