Question

Find scalar $\mathit{a}\mathbf{,}\mathit{b}\mathbf{,}\mathit{c}\mathbf{,}\mathit{d}\mathbf{,}\mathit{e}\mathbf{,}\mathit{f}\mathbf{,}\mathit{g}$ such that vectors$\left[\begin{array}{c}a\\ d\\ f\end{array}\right]\mathbf{,}\left[\begin{array}{c}b\\ 1\\ g\end{array}\right]\mathbf{,}\left[\begin{array}{c}c\\ \begin{array}{c}e\\ \frac{1}{2}\end{array}\end{array}\right]$are orthonormal.

Short answer

The value of the coefficient are $a=\frac{1}{2},b=0,c=\pm \frac{\sqrt{3}}{2},d=0,e=0,f=-\frac{\sqrt{3}}{2},g=0$.

Step-by-step solution

Step by Step Solution Step 1: Orthonormal

Consider a vectors ${\overrightarrow{v}}_1=\left[\begin{array}{c}a\\ d\\ f\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}b\\ 1\\ g\end{array}\right],{\overrightarrow{v}}_3=\left[\begin{array}{c}c\\ e\\ \frac{1}{2}\end{array}\right]$such that ${\overrightarrow{v}}_1,{\overrightarrow{v}}_2\text{and}{\overrightarrow{v}}_3$are orthonormal.

If the vector $\overrightarrow{x}$is orthonormal to $\overrightarrow{y}$then, $\overrightarrow{x}·\overrightarrow{y}=0$and $||\overrightarrow{x}||=1$.

By the definition, the value of the vectors $||{\overrightarrow{v}}_1||=1$,$||{\overrightarrow{v}}_2||=1$and $||{\overrightarrow{v}}_3||=1$

Determine the coefficients

Simplify the equation $||{\overrightarrow{v}}_1||=1$as follows,

role="math" localid="1659532723265" $\begin{array}{rcl}||{\overrightarrow{v}}_1||& =& 1\\ \sqrt{a^2+d^2+f^2}& =& 1\\ a^2+d^2+f^2& =& 1\end{array}$

Simplify the equation $||{\overrightarrow{v}}_2||=1$as follows.

$\begin{array}{rcl}||{\overrightarrow{v}}_2||& =& 1\\ \sqrt{b^2+1^2+g^2}& =& 1\\ b^2+1^2+g^2& =& 1\\ b& =& g=0\end{array}$

Simplify the equation $||{\overrightarrow{v}}_3||=1$as follows,

$\begin{array}{rcl}||{\overrightarrow{v}}_3||& =& 1\\ \sqrt{c^2+e^2+{\left(\frac{1}{2}\right)}^2}& =& 1\\ c^2+e^2+{\left(\frac{1}{2}\right)}^2& =& 1\\ c^2+e^2+\frac{1}{4}& =& 1\end{array}$

Further, simplify the equation as follows.

$\begin{array}{rcl}{c}^2+e^2+\frac{1}{4}& =& 1\\ c^2+e^2& =& \frac{3}{4}\end{array}$

As ${\overrightarrow{v}}_1\perp {\overrightarrow{v}}_2$, simplify the equation ${\overrightarrow{v}}_1·{\overrightarrow{v}}_2=0$as follows.

$\begin{array}{rcl}{\overrightarrow{v}}_1·{\overrightarrow{v}}_2& =& 0\\ \left[\begin{array}{ccc}a& d& f\end{array}\right]·\left[\begin{array}{c}b\\ 1\\ g\end{array}\right]& =& 0\\ a·b+d·1+f·g& =& 0\end{array}$

Substitute the values $0$for $b$and $0$for $g$in the equation $a·b+d·1+f·g=0$as follows.

$\begin{array}{rcl}a·b+d·1+f·g& =& 0\\ a·\left(0\right)+d·1+f·\left(0\right)& =& 0\\ d& =& 0\end{array}$

Determine the coefficients

As ${\overrightarrow{v}}_2\perp {\overrightarrow{v}}_3$, simplify the equation ${\overrightarrow{v}}_2·{\overrightarrow{v}}_3=0$as follows.

$\begin{array}{rcl}{\overrightarrow{v}}_2·{\overrightarrow{v}}_3& =& 0\\ \left[\begin{array}{ccc}b& 1& g\end{array}\right]·\left[\begin{array}{c}c\\ e\\ \frac{1}{2}\end{array}\right]& =& 0\\ b·c+1·e+g·\frac{1}{2}& =& 0\end{array}$

Substitute the values $0$for $b$and $0$for $g$in the equation $b·c+1·e+g·\frac{1}{2}=0$as follows

$\begin{array}{rcl}b·c+1·e+g·\frac{1}{2}& =& 0\\ \left(0\right)·c+1·e+\left(0\right)·\frac{1}{2}& =& 0\\ e& =& 0\end{array}$

Determine the coefficients

Substitute the values $0$for $e$in the equation $c^2+e^2=\frac{3}{4}$as follows,

$\begin{array}{rcl}{c}^2+e^2& =& \frac{3}{4}\\ c^2+{\left(0\right)}^2& =& \frac{3}{4}\\ c& =& \pm \frac{\sqrt{3}}{2}\end{array}$

As ${\overrightarrow{v}}_1\perp {\overrightarrow{v}}_3$, simplify the equation ${\overrightarrow{v}}_1·{\overrightarrow{v}}_3=0$as follows.

$\begin{array}{rcl}{\overrightarrow{v}}_1·{\overrightarrow{v}}_3& =& 0\\ \left[\begin{array}{ccc}a& d& f\end{array}\right]·\left[\begin{array}{c}c\\ e\\ \frac{1}{2}\end{array}\right]& =& 0\\ a·c+d·e+f·\frac{1}{2}& =& 0\end{array}$

Substitute the values $0$for $e$and role="math" localid="1659534199281" $\frac{\sqrt{3}}{2}$for $c$in the equation$a·c+d·e+f·\frac{1}{2}=0$ as follows.

$\begin{array}{rcl}a·c+d·e+f·\frac{1}{2}& =& 0\\ a·\frac{\sqrt{3}}{2}+d·\left(0\right)+f·\frac{1}{2}& =& 0\\ a·\frac{\sqrt{3}}{2}+d·\left(0\right)+f·\frac{1}{2}& =& 0\\ f& =& -\sqrt{3}a\end{array}$

Determine the coefficients

Substitute the values $0$for $d$and $-\sqrt{3}a$for $f$in the equation $a^2+d^2+f^2=1$as follows.

$\begin{array}{rcl}{a}^2+d^2+f^2& =& 1\\ a^2+{\left(0\right)}^2+{\left(-\sqrt{3}a\right)}^2& =& 1\\ a^2+3a^2& =& 1\\ a& =& \pm \frac{1}{2}\end{array}$

As $a=\pm \frac{1}{2}$implies $f=-\frac{\sqrt{3}}{2}$

Substitute the values $0$for $d$and $-\sqrt{3}a$for $f$in ${\overrightarrow{v}}_1=\left[\begin{array}{c}a\\ d\\ f\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}b\\ 1\\ g\end{array}\right],{\overrightarrow{v}}_3=\left[\begin{array}{c}c\\ e\\ \frac{1}{2}\end{array}\right]$as follows.

$$\begin{gathered} {\overrightarrow{v}}_1=\left[\begin{array}{c}1/2\\ 0\\ \sqrt{3}/2\end{array}\right] \\ {\overrightarrow{v}}_2=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right] \\ {\overrightarrow{v}}_3=\left[\begin{array}{c}\sqrt{3}/2\\ 0\\ 1/2\end{array}\right] \end{gathered}$$

Hence, the value of the coefficient are $a=\frac{1}{2},b=0,c=\pm \frac{\sqrt{3}}{2},d=0,e=0,f=-\frac{\sqrt{3}}{2},g=0$.