Determine λ so that the vector vec a=2λvec i+vec j+(1-λ)vec…

Determine $\lambda$ so that the vector $\vec a=2\lambda\vec i+\vec j+(1-\lambda)\vec k$ makes equal angles with $\vec b=-\vec i+3\vec j$ and $\vec c=5\vec i-\vec j+8\vec k$.

Step 1: Condition

\[ \vec a=(2\lambda,1,1-\lambda),\qquad \vec b=(-1,3,0),\qquad \vec c=(5,-1,8). \]

Equal angles mean equal cosines. The common factor $|\vec a|\neq0$ cancels, so the projections of $\vec a$ onto $\vec b$ and onto $\vec c$ must be equal: $\frac{\vec a\cdot\vec b}{|\vec b|}=\frac{\vec a\cdot\vec c}{|\vec c|}$.

Step 2: Ingredients

\[ \vec a\cdot\vec b=-2\lambda+3,\qquad \vec a\cdot\vec c=10\lambda-1+8-8\lambda=2\lambda+7, \]

\[ |\vec b|=\sqrt{10},\qquad |\vec c|=\sqrt{90}=3\sqrt{10}. \]

Step 3: Equation

\[ \frac{-2\lambda+3}{\sqrt{10}}=\frac{2\lambda+7}{3\sqrt{10}}\ \Longrightarrow\ -6\lambda+9=2\lambda+7\ \Longrightarrow\ 8\lambda=2. \]

\[\lambda=\frac14\]

The other sign. The equation $\cos\angle(\vec a,\vec b)=-\cos\angle(\vec a,\vec c)$ gives $\lambda=4$. There the angles are supplementary, not equal: for $\vec a=(8,1,-3)$ the projections are $-\tfrac{5}{\sqrt{10}}$ and $+\tfrac{5}{\sqrt{10}}$. So $\lambda=4$ is rejected.

Step 4: Check and the common angle

For $\lambda=\tfrac14$: $\vec a=\left(\tfrac12,1,\tfrac34\right)$, $\vec a\cdot\vec b=\tfrac52$, $\vec a\cdot\vec c=\tfrac{15}{2}$, and

\[ \frac{5/2}{\sqrt{10}}=\frac{15/2}{3\sqrt{10}}=\frac{5}{2\sqrt{10}}.\ \checkmark \]

\[ |\vec a|^2=\tfrac14+1+\tfrac9{16}=\tfrac{29}{16},\qquad \cos\varphi=\frac{5/2}{\frac{\sqrt{29}}{4}\sqrt{10}}=\frac{10}{\sqrt{290}}=\sqrt{\frac{10}{29}}\approx0.5872. \]

\[\varphi=\arccos\sqrt{\tfrac{10}{29}}\approx54.0^\circ\]