1. **Problem:** Determine if $P \lor (P \leftrightarrow Q) \lor Q$ is a tautology or contradiction using a truth table. **Step 1:** Recall that a tautology is always true, and a contradiction is always false. **Step 2:** Construct the truth table for $P$, $Q$, $P \leftrightarrow Q$, and the expression. | $P$ | $Q$ | $P \leftrightarrow Q$ | $P \lor (P \leftrightarrow Q) \lor Q$ | |-----|-----|-----------------------|---------------------------------------| | T | T | T | T | | T | F | F | T | | F | T | F | T | | F | F | T | T | **Step 3:** Since the expression is true for all truth values, it is a tautology. 2. **Problem:** Calculate the cross product magnitude of vectors $\mathbf{a}$ and $\mathbf{b}$ with $|\mathbf{a}|=2\sqrt{3}$, $|\mathbf{b}|=3$, and angle $53^\circ$ between them. **Step 1:** Use formula $|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$. **Step 2:** Calculate $\sin 53^\circ \approx 0.7986$. **Step 3:** Compute $|\mathbf{a} \times \mathbf{b}| = 2\sqrt{3} \times 3 \times 0.7986 = 6\sqrt{3} \times 0.7986$. **Step 4:** Since $\sqrt{3} \approx 1.732$, then $6 \times 1.732 \times 0.7986 \approx 8.3$. 3. **Problem:** Find $\mathbf{a} + 2\mathbf{b} - \mathbf{c}$ where $\mathbf{a} = 2\mathbf{i} + 5\mathbf{j}$, $\mathbf{b} = 12\mathbf{i} - 10\mathbf{j}$, $\mathbf{c} = -3\mathbf{i} + 9\mathbf{j}$ using column matrix notation. **Step 1:** Write vectors as matrices: $$\mathbf{a} = \begin{bmatrix}2 \\ 5\end{bmatrix}, \mathbf{b} = \begin{bmatrix}12 \\ -10\end{bmatrix}, \mathbf{c} = \begin{bmatrix}-3 \\ 9\end{bmatrix}$$ **Step 2:** Calculate $2\mathbf{b} = \begin{bmatrix}24 \\ -20\end{bmatrix}$. **Step 3:** Compute $\mathbf{a} + 2\mathbf{b} = \begin{bmatrix}2+24 \\ 5-20\end{bmatrix} = \begin{bmatrix}26 \\ -15\end{bmatrix}$. **Step 4:** Subtract $\mathbf{c}$: $$\begin{bmatrix}26 \\ -15\end{bmatrix} - \begin{bmatrix}-3 \\ 9\end{bmatrix} = \begin{bmatrix}26 + 3 \\ -15 - 9\end{bmatrix} = \begin{bmatrix}29 \\ -24\end{bmatrix}$$ 4. **Problem:** Given matrices $$A=\begin{bmatrix}a & 2 \\ 3 & 5\end{bmatrix}, B=\begin{bmatrix}2 & 4 \\ -1 & 2\end{bmatrix}, C=\begin{bmatrix}-1 & 2 \\ 3 & c\end{bmatrix}$$ and $2A - 3B = 4C$, find $a$, $b$, and $c$. **Step 1:** Compute $2A = \begin{bmatrix}2a & 4 \\ 6 & 10\end{bmatrix}$. **Step 2:** Compute $3B = \begin{bmatrix}6 & 12 \\ -3 & 6\end{bmatrix}$. **Step 3:** Calculate $2A - 3B = \begin{bmatrix}2a - 6 & 4 - 12 \\ 6 - (-3) & 10 - 6\end{bmatrix} = \begin{bmatrix}2a - 6 & -8 \\ 9 & 4\end{bmatrix}$. **Step 4:** Compute $4C = \begin{bmatrix}-4 & 8 \\ 12 & 4c\end{bmatrix}$. **Step 5:** Set $2A - 3B = 4C$ element-wise: - $2a - 6 = -4 \Rightarrow 2a = 2 \Rightarrow a = 1$ - $-8 = 8$ (contradiction, so check if $b$ is a variable or typo; $b$ not given in matrices, ignore) - $9 = 12$ (contradiction, so likely $b$ is a typo, ignore) - $4 = 4c \Rightarrow c = 1$ **Step 6:** Since $b$ is not in matrices, only $a=1$ and $c=1$ are found. 5. **Problem:** Find the angle between vectors $\mathbf{a}$ and $\mathbf{b}$ with magnitudes $\sqrt{7}$ and $6$, and dot product $\mathbf{a} \cdot \mathbf{b} = \frac{13}{2}$. **Step 1:** Use formula $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta$. **Step 2:** Substitute values: $$\frac{13}{2} = \sqrt{7} \times 6 \times \cos\theta$$ **Step 3:** Solve for $\cos\theta$: $$\cos\theta = \frac{13/2}{6\sqrt{7}} = \frac{13}{12\sqrt{7}}$$ **Step 4:** Approximate $\sqrt{7} \approx 2.6458$: $$\cos\theta \approx \frac{13}{12 \times 2.6458} = \frac{13}{31.75} \approx 0.409$$ **Step 5:** Find $\theta = \cos^{-1}(0.409) \approx 66.8^\circ$. 6. **Problem:** Given $n(U) = 169$, $n(A) = 81$, $n(B) = 96$, and $n(A \cup B) = 137$, find $n(A \cap B)$ and describe the Venn diagram. **Step 1:** Use formula: $$n(A \cup B) = n(A) + n(B) - n(A \cap B)$$ **Step 2:** Substitute values: $$137 = 81 + 96 - n(A \cap B)$$ **Step 3:** Solve for $n(A \cap B)$: $$n(A \cap B) = 81 + 96 - 137 = 177 - 137 = 40$$ **Step 4:** The Venn diagram has two overlapping circles $A$ and $B$ inside rectangle $U$ with: - $n(A) = 81$ - $n(B) = 96$ - Intersection $n(A \cap B) = 40$ - Union $n(A \cup B) = 137$ - Universal set $n(U) = 169$ This completes all problems.