In a model of a frame, three bars point along the vectors a = 2 i + j + 3 k, b = i + j + k and c =…

Step 1. The two dot products. $\vec a\cdot\vec c = 4 + 1 + 3 = 8$ and $\vec b\cdot\vec c = 2 + 1 + 1 = 4$.

Step 2. The condition. $\vec a\cdot\vec c + \lambda\,\vec b\cdot\vec c = 8 + 4\lambda = 0$, so $\lambda = -2$.

Step 3. The check. $\vec a - 2\vec b = -\hat j + \hat k$, and $(-\hat j + \hat k)\cdot\vec c = -1 + 1 = 0$.

Step 4. The angle. $\vec a\cdot\vec b = 2 + 1 + 3 = 6$, $\lvert\vec a\rvert = \sqrt{14}$ and $\lvert\vec b\rvert = \sqrt{3}$. So $\cos\theta = \dfrac{6}{\sqrt{42}} \approx 0.9258$ and $\theta \approx 22.21^\circ$.