2 hat{i}-3 hat{j}+hat{k} and hat{i}+2 hat{j}- | vedclass

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(C) Let the position vectors of points $A$ and $B$ be $\vec{a} = 2 \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \hat{i}+2 \hat{j}-3 \hat{k}$ respectively

Vedclass · Accepted Answer

$\frac{1}{3 \sqrt{42}}(8 \hat{i}-5 \hat{j}+17 \hat{k})$

Vedclass · Answer

(C) Let the position vectors of points $A$ and $B$ be $\vec{a} = 2 \hat{i}-3 \hat{j}+\hat{k}$ and $\vec{b} = \hat{i}+2 \hat{j}-3 \hat{k}$ respectively.Point $C$ divides $AB$ in the ratio $3:2$. Using the section formula, the position vector of $C$ is:$\vec{c} = \frac{3\vec{b} + 2\vec{a}}{3+2} = \frac{3(\hat{i}+2 \hat{j}-3 \hat{k}) + 2(2 \hat{i}-3 \hat{j}+\hat{k})}{5}$$\vec{c} = \frac{(3+4)\hat{i} + (6-6)\hat{j} + (-9+2)\hat{k}}{5} = \frac{7}{5} \hat{i} + 0 \hat{j} - \frac{7}{5} \hat{k}$Given the position vector of point $D$ is $\vec{d} = 3 \hat{i}-\hat{j}+2 \hat{k}$.The vector $\overrightarrow{CD} = \vec{d} - \vec{c} = (3 \hat{i}-\hat{j}+2 \hat{k}) - (\frac{7}{5} \hat{i} + 0 \hat{j} - \frac{7}{5} \hat{k})$$\overrightarrow{CD} = (3 - \frac{7}{5}) \hat{i} + (-1 - 0) \hat{j} + (2 + \frac{7}{5}) \hat{k} = \frac{8}{5} \hat{i} - \hat{j} + \frac{17}{5} \hat{k} = \frac{1}{5} (8 \hat{i} - 5 \hat{j} + 17 \hat{k})$The magnitude of $\overrightarrow{CD}$ is $|\overrightarrow{CD}| = \frac{1}{5} \sqrt{8^2 + (-5)^2 + 17^2} = \frac{1}{5} \sqrt{64 + 25 + 289} = \frac{1}{5} \sqrt{378} = \frac{1}{5} \sqrt{9 	imes 42} = \frac{3 \sqrt{42}}{5}$.The unit vector in the direction of $\overrightarrow{CD}$ is $\frac{\overrightarrow{CD}}{|\overrightarrow{CD}|} = \frac{\frac{1}{5} (8 \hat{i} - 5 \hat{j} + 17 \hat{k})}{\frac{3 \sqrt{42}}{5}} = \frac{1}{3 \sqrt{42}} (8 \hat{i} - 5 \hat{j} + 17 \hat{k})$.

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