If begin{bmatrix} alpha beta gamma end{bmat | vedclass

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(A) Given the matrix equation: $\begin{bmatrix} \alpha \ \beta \ \gamma \end{bmatrix} = \begin{bmatrix} \cos 	heta & -\sin 	heta & 0 \ \sin 	het

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$\frac{\alpha^2+\beta^2+\gamma^2}{z}$

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(A) Given the matrix equation: $\begin{bmatrix} \alpha \ \beta \ \gamma \end{bmatrix} = \begin{bmatrix} \cos 	heta & -\sin 	heta & 0 \ \sin 	heta & \cos 	heta & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \ y \ z \end{bmatrix}$ Multiplying the matrices,we get: $\alpha = x \cos 	heta - y \sin 	heta$ $(i)$ $\beta = x \sin 	heta + y \cos 	heta$ $(ii)$ $\gamma = z$ $(iii)$ Squaring and adding equations $(i)$ and $(ii)$: $\alpha^2 + \beta^2 = (x \cos 	heta - y \sin 	heta)^2 + (x \sin 	heta + y \cos 	heta)^2$ $\alpha^2 + \beta^2 = x^2 \cos^2 	heta + y^2 \sin^2 	heta - 2xy \sin 	heta \cos 	heta + x^2 \sin^2 	heta + y^2 \cos^2 	heta + 2xy \sin 	heta \cos 	heta$ $\alpha^2 + \beta^2 = x^2(\cos^2 	heta + \sin^2 	heta) + y^2(\sin^2 	heta + \cos^2 	heta) = x^2 + y^2$ Now,consider the expression $\frac{x^2+y^2+z^2}{\gamma}$. Substituting $x^2+y^2 = \alpha^2+\beta^2$ and $z = \gamma$: $\frac{x^2+y^2+z^2}{\gamma} = \frac{\alpha^2+\beta^2+\gamma^2}{\gamma}$ Since $\gamma = z$,this is equal to $\frac{\alpha^2+\beta^2+\gamma^2}{z}$.

If $\begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}$,then $\frac{x^2+y^2+z^2}{\gamma} =$

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