Let overrightarrow{OA} = hat{i} - 3hat{j} + h | vedclass

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(D) The position vector of point $P$ which divides $AB$ in the ratio $2:1$ is given by the section formula:$\vec{r} = \frac{2(\overrightarrow{OB}) + 1

Vedclass · Accepted Answer

$3$

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(D) The position vector of point $P$ which divides $AB$ in the ratio $2:1$ is given by the section formula:$\vec{r} = \frac{2(\overrightarrow{OB}) + 1(\overrightarrow{OA})}{2+1} = \frac{2(\hat{i} + 3\hat{j} - 2\hat{k}) + 1(\hat{i} - 3\hat{j} + \hat{k})}{3}$$= \frac{2\hat{i} + 6\hat{j} - 4\hat{k} + \hat{i} - 3\hat{j} + \hat{k}}{3} = \frac{3\hat{i} + 3\hat{j} - 3\hat{k}}{3} = \hat{i} + \hat{j} - \hat{k}$Now,the vector $\overrightarrow{PC} = \overrightarrow{OC} - \vec{r} = (4\hat{i} + 3\hat{j} + 5\hat{k}) - (\hat{i} + \hat{j} - \hat{k}) = 3\hat{i} + 2\hat{j} + 6\hat{k}$The magnitude of $\overrightarrow{PC}$ is $|\overrightarrow{PC}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$The direction cosines $(l, m, n)$ of $\overrightarrow{PC}$ are given by $\frac{\vec{v}}{|\vec{v}|}$:$l = \frac{3}{7}, m = \frac{2}{7}, n = \frac{6}{7}$Therefore,$l + 3m + 2n = \frac{3}{7} + 3(\frac{2}{7}) + 2(\frac{6}{7}) = \frac{3 + 6 + 12}{7} = \frac{21}{7} = 3$

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