If the vectors vec{AB} = p hat{i} + q hat{j} | vedclass

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(B) We have, $\vec{AB} = p \hat{i} + q \hat{j} + r \hat{k}$, $\vec{AC} = s \hat{i} + 3 \hat{j} + 4 \hat{k}$, and $\vec{CB} = 3 \hat{i} + \hat{j} - 2 \

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$p=8, q=4, r=2, s=5$

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(B) We have, $\vec{AB} = p \hat{i} + q \hat{j} + r \hat{k}$, $\vec{AC} = s \hat{i} + 3 \hat{j} + 4 \hat{k}$, and $\vec{CB} = 3 \hat{i} + \hat{j} - 2 \hat{k}$.Since $\vec{CA} = -\vec{AC} = -s \hat{i} - 3 \hat{j} - 4 \hat{k}$, the area of $	riangle ABC$ is given by $\frac{1}{2} |\vec{CA} 	imes \vec{CB}| = 5 \sqrt{6}$.Calculating the cross product: $\vec{CA} 	imes \vec{CB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -s & -3 & -4 \ 3 & 1 & -2 \end{vmatrix} = \hat{i}(6+4) - \hat{j}(2s+12) + \hat{k}(-s+9) = 10 \hat{i} - (2s+12) \hat{j} + (9-s) \hat{k}$.Thus, $\frac{1}{2} \sqrt{100 + (2s+12)^2 + (9-s)^2} = 5 \sqrt{6} \implies \sqrt{100 + 4s^2 + 144 + 48s + 81 - 18s + s^2} = 10 \sqrt{6}$.Squaring both sides: $5s^2 + 30s + 325 = 600 \implies 5s^2 + 30s - 275 = 0 \implies s^2 + 6s - 55 = 0$.Factoring gives $(s+11)(s-5) = 0$, so $s=5$ (assuming positive orientation or magnitude).Using the triangle law $\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}$, we have $\vec{AB} = \vec{AC} - \vec{BC} = \vec{AC} + \vec{CB}$.$p \hat{i} + q \hat{j} + r \hat{k} = (s \hat{i} + 3 \hat{j} + 4 \hat{k}) + (3 \hat{i} + \hat{j} - 2 \hat{k}) = (s+3) \hat{i} + 4 \hat{j} + 2 \hat{k}$.Comparing coefficients with $s=5$: $p = 5+3 = 8, q = 4, r = 2$. Thus, $p=8, q=4, r=2, s=5$.

If the vectors $\vec{AB} = p \hat{i} + q \hat{j} + r \hat{k}$, $\vec{AC} = s \hat{i} + 3 \hat{j} + 4 \hat{k}$, and $\vec{CB} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ form a $\triangle ABC$, then the values of $p, q, r$ and $s$ such that the area of that $\triangle ABC$ is $5 \sqrt{6}$ are:

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