FURTHER MATH. The position vectors of the points P and Q are P = -i + 3j and q = 2i + 2j. Find (i) 12PQ (ii) angle between p and q (iii) projection of p on q (iv) the scalar h, if p+h = 5i+7j.

$$3\vec{p} = 3(-i + 3j) = -3i + 9j$$ $$4\vec{q} = 4(2i + 2j) = 8i + 8j$$

$$3\vec{p} - 4\vec{q} = (-3i + 9j) - (8i + 8j)$$ $$3\vec{p} - 4\vec{q} = (-3 - 8)i + (9 - 8)j$$ $$3\vec{p} - 4\vec{q} = -11i + j$$

The result is $\boxed{-11i + j}$.

$$\vec{p} \cdot \vec{q} = (-1)(2) + (3)(2)$$ $$\vec{p} \cdot \vec{q} = -2 + 6$$ $$\vec{p} \cdot \vec{q} = 4$$

$$12\vec{p} \cdot \vec{q} = 12 \times 4$$ $$12\vec{p} \cdot \vec{q} = 48$$

The result is $\boxed{48}$.

$$|\vec{p}| = \sqrt{(-1)^2 + (3)^2} = \sqrt{1 + 9} = \sqrt{10}$$ $$|\vec{q}| = \sqrt{(2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$$

We already found $\vec{p} \cdot \vec{q} = 4$ from part b.

$$\cos \theta = \frac{\vec{p} \cdot \vec{q}}{|\vec{p}| |\vec{q}|}$$ $$\cos \theta = \frac{4}{\sqrt{10} \cdot 2\sqrt{2}}$$ $$\cos \theta = \frac{4}{2\sqrt{20}}$$ $$\cos \theta = \frac{4}{2 \cdot 2\sqrt{5}}$$ $$\cos \theta = \frac{4}{4\sqrt{5}}$$ $$\cos \theta = \frac{1}{\sqrt{5}}$$

$$\theta = \arccos\left(\frac{1}{\sqrt{5}}\right)$$ $$\theta \approx 63.43^\circ$$

The angle between $\vec{p}$ and $\vec{q}$ is approximately $\boxed{63.43^\circ}$.

$\text{proj}_{\vec{q}} \vec{p} = \frac{\vec{p} \cdot \vec{q}}{|\vec{q}|^2} \vec{q}$.
We know $\vec{p} \cdot \vec{q} = 4$ and $|\vec{q}|^2 = (\sqrt{8})^2 = 8$.

$$\text{proj}_{\vec{q}} \vec{p} = \frac{4}{8} (2i + 2j)$$

$$\text{proj}_{\vec{q}} \vec{p} = \frac{1}{2} (2i + 2j)$$ $$\text{proj}_{\vec{q}} \vec{p} = i + j$$

The projection of $\vec{p}$ on $\vec{q}$ is $\boxed{i + j}$.