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Q.
The direction cosines of the vector joining the points $A(1,2,-3)$ and $B(-1,-2,1)$, directed from $A$ to $B$ are
Vector Algebra
Solution:
The given points are $A(1,2,-3)$ and $B(-1,-2,1)$.
i.e.,$ x_1=1, y_1=2, z_1=-3 $
and $ x_2=-1, y_2=-2, z_2=1 $
Vector $ AB =\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}$
$=(-1-1) \hat{i}+(-2-2) \hat{j}+[1-(-3)] \hat{k} $
$ =-2 \hat{i}-4 \hat{j}+4 \hat{k} $
Comparing with $X=x \hat{i}+y \hat{j}+z \hat{k}$, we get
$x=-2, y=-4, z=4$
Now, magnitude
$|A B| =\sqrt{x^2+y^2+z^2}=\sqrt{(-2)^2+(-4)^2+4^2} $
$ =\sqrt{4+16+16}=\sqrt{36}=6$
Direction cosines of a vector $X=x \hat{i}+y \hat{j}+z \hat{k}$ are $\frac{x}{|X|}, \frac{y}{|X|}, \frac{z}{|X|}$
$\therefore$ Direction cosines of $AB$ are $\frac{-2}{6}, \frac{-4}{6}, \frac{4}{6}$ or $-\frac{1}{3},-\frac{2}{3}, \frac{2}{3}$.