Determine the parameter λ so that vectors a\2,λ,λ-1\ and…

Determine the parameter $\lambda$ so that vectors $\vec{a}\{2,\lambda,\lambda-1\}$ and $\vec{b}\{\lambda,\lambda-2,0\}$ have equal magnitudes, then find the angle between them.

Step 1: Condition $|\vec a|=|\vec b|$

\[ |\vec{a}|^2 = 4 + \lambda^2 + (\lambda-1)^2 = 2\lambda^2 - 2\lambda + 5, \qquad |\vec{b}|^2 = \lambda^2 + (\lambda-2)^2 = 2\lambda^2 - 4\lambda + 4 \]

\[ 2\lambda^2 - 2\lambda + 5 = 2\lambda^2 - 4\lambda + 4 \;\Longrightarrow\; 2\lambda = -1 \]

\[\lambda = -\frac{1}{2}\]

Step 2: The vectors

\[ \vec{a} = \left\{2,\;-\tfrac{1}{2},\;-\tfrac{3}{2}\right\}, \qquad \vec{b} = \left\{-\tfrac{1}{2},\;-\tfrac{5}{2},\;0\right\} \]

\[ |\vec{a}|^2 = 4+\tfrac14+\tfrac94 = \tfrac{13}{2}, \qquad |\vec{b}|^2 = \tfrac14+\tfrac{25}{4} = \tfrac{13}{2}, \qquad |\vec a|=|\vec b|=\frac{\sqrt{26}}{2} \;\checkmark \]

Step 3: Angle

\[ \vec{a}\cdot\vec{b} = (2)\left(-\tfrac{1}{2}\right) + \left(-\tfrac{1}{2}\right)\left(-\tfrac{5}{2}\right) + \left(-\tfrac{3}{2}\right)(0) = -1+\tfrac54 = \tfrac{1}{4} \]

\[ \cos\theta = \frac{\frac{1}{4}}{\frac{13}{2}} = \frac{1}{26} \]

\[\theta = \arccos\frac{1}{26} \approx 87.8^\circ\]

The vectors are almost perpendicular.