Find the position vector of M.

Step 1: Find the position vector of M. M is the midpoint of XY. The position vector of M, $\vec{OM}$, is given by: $$ \vec{OM} = \frac{1}{2}(\vec{OX} + \vec{OY}) $$ Substitute the given vectors $\vec{OX} = -4\vec{i} + 6\vec{j}$ and $\vec{OY} = 12\vec{i} - 6\vec{j}$: $$ \vec{OM} = \frac{1}{2}((-4\vec{i} + 6\vec{j}) + (12\vec{i} - 6\vec{j})) $$ $$ \vec{OM} = \frac{1}{2}((-4+12)\vec{i} + (6-6)\vec{j}) $$ $$ \vec{OM} = \frac{1}{2}(8\vec{i} + 0\vec{j}) $$ $$ \vec{OM} = 4\vec{i} + 0\vec{j} $$ Step 2: Find the position vector of N. Given that ON:NY = 1:2, N divides the line segment OY in the ratio 1:2. Therefore, the position vector of N, $\vec{ON}$, is: $$ \vec{ON} = \frac{1}{1+2}\vec{OY} = \frac{1}{3}\vec{OY} $$ Substitute $\vec{OY} = 12\vec{i} - 6\vec{j}$: $$ \vec{ON} = \frac{1}{3}(12\vec{i} - 6\vec{j}) $$ $$ \vec{ON} = (\frac{1}{3} \times 12)\vec{i} - (\frac{1}{3} \times 6)\vec{j} $$ $$ \vec{ON} = 4\vec{i} - 2\vec{j} $$ Step 3: Find the vector $\vec{NM}$. The vector $\vec{NM}$ is found by subtracting the position vector of N from the position vector of M: $$ \vec{NM} = \vec{OM} - \vec{ON} $$ Substitute the calculated $\vec{OM}$ and $\vec{ON}$: $$ \vec{NM} = (4\vec{i} + 0\vec{j}) - (4\vec{i} - 2\vec{j}) $$ $$ \vec{NM} = (4-4)\vec{i} + (0 - (-2))\vec{j} $$ $$ \vec{NM} = 0\vec{i} + 2\vec{j} $$ Step 4: Calculate the magnitude of $\vec{NM}$. The magnitude of a vector $a\vec{i} + b\vec{j}$ is given by $\sqrt{a^2 + b^2}$. $$ |\vec{NM}| = \sqrt{0^2 + 2^2} $$ $$ |\vec{NM}| = \sqrt{0 + 4} $$ $$ |\vec{NM}| = \sqrt{4} $$ $$ |\vec{NM}| = 2 $$ Correct to 1 decimal place, the magnitude is 2.0. The magnitude of NM is $\boxed{\text{2.0 units}}$. Send me the next one 📸