Question

If $\sin \beta \ne \cos \beta$ and $x,y$ and $z$ satisfy the equations
$x\cos \alpha -y\sin \alpha +z=\cos \beta +1$
$x\sin \alpha +y\cos \alpha +z=-\sin \beta +1$
$x\cos (\alpha +\beta )-y\sin (\alpha +\beta )+z=2$,

then $x^2+y^2+z^2$ equals

• A. $1$
• B. $2$
• C. $3$
• D. $5$

Answer 1

The correct option is B $2$

The determinant of the coefficient of the given set of equation is
$\Delta =\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 1\\ \sin \alpha & \cos \alpha & 1\\ \cos (\alpha +\beta )& -\sin (\alpha +\beta )& 1\end{array}\right|$

Applying $R_3\to R_3-\cos \beta R_1+\sin \beta R_2$
$\Delta =\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & 1\\ \sin \alpha & \cos \alpha & 1\\ 0& 0& 1+\sin \beta -\cos \beta \end{array}\right|$

$\Rightarrow \Delta =(1+\sin \beta -\cos \beta )\left|\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right|$

$\Rightarrow \Delta =(1+\sin \beta -\cos \beta )({\cos }^2\alpha +{\sin }^2\alpha )$

$\Rightarrow \Delta =(1+\sin \beta -\cos \beta )$

To solve the equation by cramer's rule
Let
${\Delta }_1=\left|\begin{array}{ccc}\cos \beta +1& -\sin \alpha & 1\\ -\sin \beta +1& \cos \alpha & 1\\ 2& -\sin (\alpha +\beta )& 1\end{array}\right|$

Applying $C_1\to C_1-C_3$
$=\left|\begin{array}{ccc}\cos \beta & -\sin \alpha & 1\\ -\sin \beta & \cos \alpha & 1\\ 1& -\sin (\alpha +\beta )& 1\end{array}\right|$

Applying $R_3\to R_3-\cos \beta R_1+\sin \beta R_2$
$=\left|\begin{array}{ccc}\cos \beta & -\sin \alpha & 1\\ -\sin \beta & \cos \alpha & 1\\ 1-{\cos }^2\beta -{\sin }^2\beta & 0& 1-\cos \beta +\sin \beta \end{array}\right|{\Delta }_1=(1+\sin \beta -\cos \beta )\cos (\alpha +\beta )$

Similarly,

${\Delta }_2=\left|\begin{array}{ccc}\cos \alpha & \cos \beta +1& 1\\ \sin \alpha & -\sin \beta +1& 1\\ \cos (\alpha +\beta )& 2& 1\end{array}\right|$

${\Delta }_2=-(1+\sin \beta -\cos \beta )\sin (\alpha +\beta )$

And
${\Delta }_3=\left|\begin{array}{ccc}\cos \alpha & -\sin \alpha & \cos \beta +1\\ \sin \alpha & \cos \alpha & -\sin \beta +1\\ \cos (\alpha +\beta )& -\sin (\alpha +\beta )& 2\end{array}\right|{\Delta }_3=1+\sin \beta -\cos \beta$

Therefore from cramer's rule
$x=\frac{{\Delta }_1}{\Delta }=\cos (\alpha +\beta ),y=\frac{{\Delta }_2}{\Delta }=-\sin (\alpha +\beta ),z=\frac{{\Delta }_3}{\Delta }=1$

Hence,
$x^2+y^2+z^2={\cos }^2(\alpha +\beta )+{\sin }^2(\alpha +\beta )+1=2$

Hence, option B.