Find the direction cosines of the vector join | vedclass

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Question

(A) The given points are $A(1, 2, -3)$ and $B(-1, -2, 1)$.The vector $\overrightarrow{AB}$ is given by $(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2

Vedclass · Accepted Answer

$(-1/3, -2/3, 2/3)$

Vedclass · Answer

(A) The given points are $A(1, 2, -3)$ and $B(-1, -2, 1)$.The vector $\overrightarrow{AB}$ is given by $(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.$\overrightarrow{AB} = (-1 - 1)\hat{i} + (-2 - 2)\hat{j} + (1 - (-3))\hat{k} = -2\hat{i} - 4\hat{j} + 4\hat{k}$.The magnitude of $\overrightarrow{AB}$ is $|\overrightarrow{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.The direction cosines are given by $\left(\frac{a}{|\overrightarrow{AB}|}, \frac{b}{|\overrightarrow{AB}|}, \frac{c}{|\overrightarrow{AB}|}\right)$,where $\overrightarrow{AB} = a\hat{i} + b\hat{j} + c\hat{k}$.Thus,the direction cosines are $\left(-\frac{2}{6}, -\frac{4}{6}, \frac{4}{6}\right) = \left(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right)$.

Find the direction cosines of the vector joining the points $A(1, 2, -3)$ and $B(-1, -2, 1)$,directed from $A$ to $B$.

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