Question-24
Let \(v = (3, 1, 2)\) be a vector in \(\mathbb{R}^{3}\). If \((a, b, c)\) is the vector obtained from \(v\) after an anti-clockwise rotation of the \(XZ\) plane with angle \(45^{\circ}\) about the \(Y\)-axis, then find the value of \(\sqrt{2}(a + b + c - 1)\).
\(6\)
We can form the rotation matrix as follows:
\[ Q=\begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0\\ \sin \theta & 0 & \cos \theta \end{bmatrix} \]
Setting \(\displaystyle \theta =45^{\circ }\), we get:
\[ Q=\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 0 & -1\\ 0 & \sqrt{2} & 0\\ 1 & 0 & 1 \end{bmatrix} \]
\(\displaystyle Qv\) is the result of rotating \(\displaystyle v\) by \(\displaystyle 45^{\circ }\):
\[ \begin{aligned} \begin{bmatrix} a\\ b\\ c \end{bmatrix} & =Qv\\ & =\frac{1}{\sqrt{2}}\begin{bmatrix} 1 & 0 & -1\\ 0 & \sqrt{2} & 0\\ 1 & 0 & 1 \end{bmatrix}\begin{bmatrix} 3\\ 1\\ 2 \end{bmatrix}\\ & =\frac{1}{\sqrt{2}}\begin{bmatrix} 1\\ \sqrt{2}\\ 5 \end{bmatrix} \end{aligned} \]
Thus, \(\displaystyle a=\frac{1}{\sqrt{2}} ,b=1,c=\frac{5}{\sqrt{2}}\). Now:
\[ \sqrt{2}( a+b+c-1) =6 \]