1. **Problem:** Compute the dot product $\vec{a} \cdot \vec{b}$ for given vectors. **Formula:** The dot product of two vectors $\vec{a} = \langle a_1, a_2, a_3 \rangle$ and $\vec{b} = \langle b_1, b_2, b_3 \rangle$ is given by: $$\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$$ **Step 1:** For (i) $\vec{a} = \langle 2,4,-3 \rangle$ and $\vec{b} = \langle -1,5,2 \rangle$: $$\vec{a} \cdot \vec{b} = 2 \times (-1) + 4 \times 5 + (-3) \times 2 = -2 + 20 - 6 = 12$$ **Step 2:** For (ii) $\vec{a} = 3\vec{i} - 2\vec{j} + \vec{k}$ and $\vec{b} = 4\vec{i} + 5\vec{j} - 2\vec{k}$, rewrite as components: $$\vec{a} = \langle 3, -2, 1 \rangle, \quad \vec{b} = \langle 4, 5, -2 \rangle$$ Calculate dot product: $$3 \times 4 + (-2) \times 5 + 1 \times (-2) = 12 - 10 - 2 = 0$$ --- 2. **Problem:** Find the angle $\theta$ between $\vec{a} = \langle 4, -3, 1 \rangle$ and $\vec{b} = \langle -1, -2, 2 \rangle$. **Formula:** $$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$$ where $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$ is the magnitude. **Step 1:** Compute dot product: $$4 \times (-1) + (-3) \times (-2) + 1 \times 2 = -4 + 6 + 2 = 4$$ **Step 2:** Compute magnitudes: $$|\vec{a}| = \sqrt{4^2 + (-3)^2 + 1^2} = \sqrt{16 + 9 + 1} = \sqrt{26}$$ $$|\vec{b}| = \sqrt{(-1)^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$$ **Step 3:** Calculate cosine of angle: $$\cos \theta = \frac{4}{\sqrt{26} \times 3} = \frac{4}{3\sqrt{26}}$$ **Step 4:** Find angle: $$\theta = \cos^{-1} \left( \frac{4}{3\sqrt{26}} \right)$$ --- 3. **Problem:** Show vectors are orthogonal if their dot product is zero. **Step 1:** For (i) $\vec{i} = \langle 1,0,0 \rangle$ and $\vec{j} = \langle 0,1,0 \rangle$: $$\vec{i} \cdot \vec{j} = 1 \times 0 + 0 \times 1 + 0 \times 0 = 0$$ Orthogonal. **Step 2:** For (ii) $\vec{i} = \langle 1,0,0 \rangle$ and $\vec{k} = \langle 0,0,1 \rangle$: $$\vec{i} \cdot \vec{k} = 1 \times 0 + 0 \times 0 + 0 \times 1 = 0$$ Orthogonal. **Step 3:** For (iii) $\vec{j} = \langle 0,1,0 \rangle$ and $\vec{k} = \langle 0,0,1 \rangle$: $$\vec{j} \cdot \vec{k} = 0 \times 0 + 1 \times 0 + 0 \times 1 = 0$$ Orthogonal. **Step 4:** For (iv) $\vec{c} = 3\vec{i} - 7\vec{j} + 2\vec{k} = \langle 3, -7, 2 \rangle$ and $\vec{d} = 10\vec{i} + 4\vec{j} - \vec{k} = \langle 10, 4, -1 \rangle$: $$\vec{c} \cdot \vec{d} = 3 \times 10 + (-7) \times 4 + 2 \times (-1) = 30 - 28 - 2 = 0$$ Orthogonal. --- 4. **Problem:** Given $\vec{a} = \langle -2, 3, 1 \rangle$, $\vec{b} = \langle 7, 4, 5 \rangle$, $\vec{c} = \langle 1, -5, 2 \rangle$, evaluate: (i) $\vec{a} \cdot \vec{b}$ (ii) $\vec{a} \cdot (\vec{b} + \vec{c})$ (iii) $(2\vec{a} + \vec{b}) \cdot 3\vec{c}$ **Step 1:** Calculate $\vec{a} \cdot \vec{b}$: $$(-2) \times 7 + 3 \times 4 + 1 \times 5 = -14 + 12 + 5 = 3$$ **Step 2:** Calculate $\vec{b} + \vec{c}$: $$\langle 7 + 1, 4 + (-5), 5 + 2 \rangle = \langle 8, -1, 7 \rangle$$ Calculate $\vec{a} \cdot (\vec{b} + \vec{c})$: $$(-2) \times 8 + 3 \times (-1) + 1 \times 7 = -16 - 3 + 7 = -12$$ **Step 3:** Calculate $2\vec{a} + \vec{b}$: $$2 \times \langle -2, 3, 1 \rangle + \langle 7, 4, 5 \rangle = \langle -4, 6, 2 \rangle + \langle 7, 4, 5 \rangle = \langle 3, 10, 7 \rangle$$ Calculate $3\vec{c}$: $$3 \times \langle 1, -5, 2 \rangle = \langle 3, -15, 6 \rangle$$ Calculate dot product: $$\langle 3, 10, 7 \rangle \cdot \langle 3, -15, 6 \rangle = 3 \times 3 + 10 \times (-15) + 7 \times 6 = 9 - 150 + 42 = -99$$ --- 5. **Problem:** Find the angle between vectors $\vec{a}$ and $\vec{b}$. **Formula:** $$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$$ **(i)** $\vec{a} = \vec{i} - 7\vec{j} + 4\vec{k} = \langle 1, -7, 4 \rangle$, $\vec{b} = 5\vec{i} - \vec{k} = \langle 5, 0, -1 \rangle$ Calculate dot product: $$1 \times 5 + (-7) \times 0 + 4 \times (-1) = 5 + 0 - 4 = 1$$ Calculate magnitudes: $$|\vec{a}| = \sqrt{1^2 + (-7)^2 + 4^2} = \sqrt{1 + 49 + 16} = \sqrt{66}$$ $$|\vec{b}| = \sqrt{5^2 + 0^2 + (-1)^2} = \sqrt{25 + 0 + 1} = \sqrt{26}$$ Calculate angle: $$\cos \theta = \frac{1}{\sqrt{66} \times \sqrt{26}} = \frac{1}{\sqrt{1716}}$$ $$\theta = \cos^{-1} \left( \frac{1}{\sqrt{1716}} \right)$$ **(ii)** $\vec{a} = \langle 3, -5, -1 \rangle$, $\vec{b} = \langle 2, 1, -3 \rangle$ Calculate dot product: $$3 \times 2 + (-5) \times 1 + (-1) \times (-3) = 6 - 5 + 3 = 4$$ Calculate magnitudes: $$|\vec{a}| = \sqrt{3^2 + (-5)^2 + (-1)^2} = \sqrt{9 + 25 + 1} = \sqrt{35}$$ $$|\vec{b}| = \sqrt{2^2 + 1^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14}$$ Calculate angle: $$\cos \theta = \frac{4}{\sqrt{35} \times \sqrt{14}} = \frac{4}{\sqrt{490}}$$ $$\theta = \cos^{-1} \left( \frac{4}{\sqrt{490}} \right)$$