Stuck on a vector question.

Question

Particle A is stationary at the point $2\vec{i} +3\vec{j}$, particle B is stationary at the point $3\vec{i} -\vec{j}$ and particle C is stationary at the point $4\vec{i} +13\vec{j}$. Particle B undergoes a displacement of $k(\vec{i} +2\vec{j})$ and particle C undergoes a displacement of $k(4\vec{i} - \vec{j})$ so that all three particles are aligned in a straight line. Determine the possible values of k.

Answer 1

After the displacements, the vector AB and AC become

$$AB = (3i-j)+k(i +2j)-(2i +3)= (1+k)i +2(k-2)j$$ $$AC =( 4i +13j )+ k(4i-j) -(2i+3j)= 2(1+2k)i + (10-k)j$$

Since AB || AC, we have

$$\frac {1+k}{2+4k}=\frac{2k-4}{10-k}$$

which yields $k=-\frac23$ and $k=3$.

Answer 2

For a given vector $\vec{v}=\alpha \hat{i}+\beta\hat{j}$, with $\alpha \neq 0$, we can calculate the angle $\theta$ between $\vec{v}$ and the $Ox$ axis by: $$\tan \theta = \frac{\beta}{\alpha}$$ Now, you know how to find this $\theta$ for $\vec{u}_{A} = 2\hat{i}+3\hat{j}$. Besides, you know that points $B$ and $C$ are now represented, respectivelly, by the vectors $\vec{u}_{B}=(3+k)\hat{i}+(-1+2k)\hat{j}$ and $\vec{u}_{C}=(4+4k)\hat{i}+(13-k)\hat{j}$. Now, if they are all aligned in the same straight line, what you can conclude using these facts?