The direction cosines of the vector joining p | vedclass

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Question

(A) Given points are $A(1, 2, -3)$ and $B(-1, -2, 1)$.First,find the vector $\vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.

Vedclass · Accepted Answer

$-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}$

Vedclass · Answer

(A) Given points are $A(1, 2, -3)$ and $B(-1, -2, 1)$.First,find the vector $\vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.$\vec{AB} = (-1 - 1)\hat{i} + (-2 - 2)\hat{j} + (1 - (-3))\hat{k} = -2\hat{i} - 4\hat{j} + 4\hat{k}$.Next,calculate the magnitude of $\vec{AB}$:$|\vec{AB}| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$.The direction cosines are given by $\frac{x}{|\vec{AB}|}, \frac{y}{|\vec{AB}|}, \frac{z}{|\vec{AB}|}$.$l = \frac{-2}{6} = -\frac{1}{3}$,$m = \frac{-4}{6} = -\frac{2}{3}$,$n = \frac{4}{6} = \frac{2}{3}$.Thus,the direction cosines are $(-\frac{1}{3}, -\frac{2}{3}, \frac{2}{3})$.

The direction cosines of the vector joining point $A$ to point $B$,where the coordinates are $A(1, 2, -3)$ and $B(-1, -2, 1)$,are . . . . . . .

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