If P hat{i}-2 hat{j}+3 hat{k}, 2 hat{i}+3 hat | vedclass

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(D) Let the position vectors be $\vec{a} = P \hat{i} - 2 \hat{j} + 3 \hat{k}$, $\vec{b} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$, and $\vec{c} = 4 \hat{i}

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$\frac{1}{5 \sqrt{3}}(\hat{i}+5 \hat{j}-7 \hat{k})$

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(D) Let the position vectors be $\vec{a} = P \hat{i} - 2 \hat{j} + 3 \hat{k}$, $\vec{b} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}$, and $\vec{c} = 4 \hat{i} + 13 \hat{j} - 18 \hat{k}$.Since the points $A$, $B$, and $C$ are collinear, the vectors $\vec{AB}$ and $\vec{BC}$ must be parallel.$\vec{AB} = \vec{b} - \vec{a} = (2-P) \hat{i} + 5 \hat{j} - 7 \hat{k}$$\vec{BC} = \vec{c} - \vec{b} = (4-2) \hat{i} + (13-3) \hat{j} + (-18+4) \hat{k} = 2 \hat{i} + 10 \hat{j} - 14 \hat{k}$Since $\vec{AB} = k \vec{BC}$, we have $(2-P) \hat{i} + 5 \hat{j} - 7 \hat{k} = k(2 \hat{i} + 10 \hat{j} - 14 \hat{k})$.Comparing the coefficients of $\hat{j}$, we get $5 = 10k \Rightarrow k = 0.5$.Comparing the coefficients of $\hat{i}$, we get $2-P = 2k = 2(0.5) = 1 \Rightarrow P = 1$.Thus, $\vec{AB} = (2-1) \hat{i} + 5 \hat{j} - 7 \hat{k} = \hat{i} + 5 \hat{j} - 7 \hat{k}$.The magnitude $|\vec{AB}| = \sqrt{1^2 + 5^2 + (-7)^2} = \sqrt{1 + 25 + 49} = \sqrt{75} = 5 \sqrt{3}$.The unit vector in the direction of $\vec{AB}$ is $\hat{u} = \frac{\vec{AB}}{|\vec{AB}|} = \frac{1}{5 \sqrt{3}} (\hat{i} + 5 \hat{j} - 7 \hat{k})$.Since $|P| = |1| = 1$, the required vector is $1 	imes \hat{u} = \frac{1}{5 \sqrt{3}} (\hat{i} + 5 \hat{j} - 7 \hat{k})$.

If $P \hat{i}-2 \hat{j}+3 \hat{k}$, $2 \hat{i}+3 \hat{j}-4 \hat{k}$, and $4 \hat{i}+13 \hat{j}-18 \hat{k}$ are the position vectors of three collinear points $A$, $B$, and $C$ respectively, then the vector in the direction of $AB$ of length $|P|$ units is

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