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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find the direction cosines of the vector joining the points $A(1, 2, –3)$ and $B(–1, –2, 1),$ directed from $A$ to $B$.

$\begin{array}{1 1}(A) (\large\frac{-1}{3},\frac{-2}{3},\frac{2}{3}) \\(B) (-1,-2,2) \\ (C) (\large\frac{1}{3},\frac{2}{3},\frac{-2}{3}) \\(D) (1,2,-2) \end{array} $

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1 Answer

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Toolbox:
  • Th cosines of the angle made by the vector with the coordinate axes is called direction cosines.
  • D.C of vector, $x\hat i+y\hat j+z\hat k\:is\:(\large\frac{x}{\sqrt{x^2+y^2+z^2}},\frac{y}{\sqrt{x^2+y^2+z^2}},\frac{z}{\sqrt{x^2+y^2+z^2}})$
  • $\overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow {OA}$ where $\overrightarrow {OB}$ is position vector of the point A and $\overrightarrow {OA}$ is position vector of A
Step 1:
A(1,2,-3) , B(-1,-2,1)
Position vector of A is $\overrightarrow {OA}=\hat i+2\hat j-3\hat k $
Position vector of B is $\overrightarrow {OB}=-\hat i-2\hat j+\hat k $
We know that $\overrightarrow{AB} =\overrightarrow {OB}-\overrightarrow {OA}$
$=(-\hat i-2\hat j+\hat k)- (\hat i+2\hat j-3\hat k)= -2\hat i - 4\hat j + 4\hat k$
Step 2:
Now let us find the magnitude of $\overrightarrow{AB}=|\overrightarrow{AB}|=\sqrt{(-2)^2+(-4)^2+(4)^2}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;=\sqrt{4+16+16}=\sqrt{36}$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\;\;=6$
Step 3:
We know D.C of vector, $x\hat i+y\hat j+z\hat k\:is\:(\large\frac{x}{\sqrt{x^2+y^2+z^2}},\frac{y}{\sqrt{x^2+y^2+z^2}},\frac{z}{\sqrt{x^2+y^2+z^2}})$
Now the direction cosines made by this vector with the coordinate axes $OX,OY$ and $OZ$ are
$\bigg( \large\frac{-2}{6},\large\frac{-4}{6},\large\frac{4}{6} \bigg) = \bigg( \large\frac{-1}{3}, \large\frac{-2}{3},\large \frac{2}{3} \bigg) $
answered May 17, 2013 by sreemathi.v
 

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