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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are three vectors such that $\mid\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\mid=0$ and $\mid\overrightarrow{a}\mid=.2, \mid\overrightarrow{b}\mid=3,\mid\overrightarrow{c}\mid=5$,then value of $\overrightarrow{a}.\overrightarrow{b}+\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{c}.\overrightarrow{a}$ is

\[(A)\;0\quad(B)\;1\quad(C)\;-19\quad(D)\;38\]
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1 Answer

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Toolbox:
  • $ (\overrightarrow a+\overrightarrow b)^2=|\overrightarrow a|^2+|\overrightarrow b|^2+2 \overrightarrow a.\overrightarrow b$
Step 1:
Given
$ |\overrightarrow a+\overrightarrow b+\overrightarrow c|=0$,$|\overrightarrow a|=2$, $|\overrightarrow b|=3$ and $|\overrightarrow c|=5$
Since $ \overrightarrow a+\overrightarrow b+\overrightarrow c=0$
$=>\overrightarrow a+\overrightarrow b=-\overrightarrow c$
Squaring on both sides we get,
$(\overrightarrow a+\overrightarrow b).(\overrightarrow a+\overrightarrow b)=(-\overrightarrow c).(\overrightarrow c)$
$=> |\overrightarrow a+\overrightarrow b|^2=|\overrightarrow c|^2$
$=>|\overrightarrow a|^2+|\overrightarrow b |^2+2.\overrightarrow a. \overrightarrow b|=|c|^2$
Substituting the given values we get
$2^2+3^2+2. \overrightarrow a. \overrightarrow b=5^2$
$=>2\overrightarrow a.\overrightarrow b=25-4-9$
$=>2\overrightarrow a.\overrightarrow b=12$
Step 2:
Similarly, $\overrightarrow b.\overrightarrow c=-\overrightarrow a$
Squaring on both sides we get,
$(\overrightarrow b+\overrightarrow c).(\overrightarrow b+\overrightarrow c)=(-\overrightarrow a).(-\overrightarrow a)$
$=>|\overrightarrow b|^2+|\overrightarrow c |^2+2.\overrightarrow b. \overrightarrow c=|\overrightarrow a|^2$
$=>9+25+2 \overrightarrow b.\overrightarrow c=4$
$2 \overrightarrow b.\overrightarrow c=4-9-25$
$\qquad\quad=-30$
Similarly, $\overrightarrow a+\overrightarrow c=-\overrightarrow b$
Squaring on both sides we get,
$(\overrightarrow a+\overrightarrow c).(\overrightarrow a+\overrightarrow c)=(-\overrightarrow b).(-\overrightarrow b)$
$=>|\overrightarrow a|^2+|\overrightarrow c |^2+2.\overrightarrow a. \overrightarrow c=|\overrightarrow b|^2$
$=>4+25+2 \overrightarrow a.\overrightarrow c=9$
$2 \overrightarrow a.\overrightarrow c=9-4-25$
$\qquad\quad=-20$
Therefore $2(\overrightarrow a.\overrightarrow b+\overrightarrow b.\overrightarrow c+\overrightarrow a.\overrightarrow c)=12-30-20$
$=38$
$\overrightarrow a.\overrightarrow b+\overrightarrow b.\overrightarrow c+\overrightarrow a.\overrightarrow c=\large\frac{-38}{2}$
$=-19$
Hence the correct option is $C$
answered Jun 3, 2013 by meena.p
 

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