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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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The vector having initial and terminal points as $(2,5,0)$ and $(-3,7,4)$ respectively is

\[\begin{array}{1 1}(A)\;\hat i+12\hat j+4\hat k & (B)\;5\hat i+2\hat j+4\hat k\\(C)\;-5\hat i+2\hat j+4\hat k & (D)\;\hat i-\hat j+\hat k\end{array}\]
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1 Answer

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  • The vector joining of the terminal point B to the initial point A given by $\overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow{OA}$
Let $\overrightarrow {OA}=2 \hat i+5 \hat j+0 \hat k$ and $\overrightarrow {OB}=-3\hat i+7 \hat j+4 \hat k$
Therefore $\overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow {OA}$
Now substitutie for $\overrightarrow {OA}$ and $\overrightarrow {OB}$
$=(-3\hat i+7 \hat j+4 \hat k)-(2 \hat i+5 \hat j+0 \hat k)$
On simplifying we get,
$=- 5\hat i + 2\hat j + 4\hat k$
Hence the correct option is $C$
answered May 28, 2013 by meena.p
 

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