Question

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

• A. $-\frac{1}{\sqrt{3}},-\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}$
• B. $\frac{-1}{3},\frac{-2}{3},\frac{2}{3}$
• C. $\frac{1}{\sqrt{3}},\frac{2}{\sqrt{3}},\frac{2}{\sqrt{3}}$
• D. None of these

Answer 1

Answer: (B)

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
$|AB|=\sqrt{x^2+y^2+z^2}=\sqrt{(-2{)}^2+(-4{)}^2+4^2}$
$=\sqrt{4+16+16}=\sqrt{36}=6$
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 $-\frac{1}{3},-\frac{2}{3},\frac{2}{3}$.