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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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Find the value of $\lambda$ such that the vectors $\overrightarrow{a}=2\hat i+\lambda\hat j+\hat k$ and $\overrightarrow{b}=\hat i+2\hat j+3\hat k$ are orthogonal

\[(A)\;0\quad(B)\;1\quad(C)\;\frac{3}{2}\quad(D)\;\frac{-5}{2}\]
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1 Answer

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  • If $\overrightarrow a \perp \overrightarrow b,$ then $\overrightarrow a.\overrightarrow b=0$
Let $\overrightarrow a=2\hat i+\lambda\hat j+\hat k\:and\:\overrightarrow b=\hat i+2\hat j+3\hat k$
It is given $\overrightarrow a$ and $\overrightarrow b=0$ are orthogonal (ie) $\overrightarrow a \perp \overrightarrow b,$
$=>\overrightarrow a.\overrightarrow b=0$
Now substituting the values,
$=>(2\hat i+\lambda\hat j+\hat k).(\hat i+2\hat j+3\hat k)=0$
On simplifying we get,
$=>2+2\lambda+3=0$
$2\lambda+5=0$
$\lambda=\large\frac{-5}{2}$
Hence the correct option is $D$
answered May 28, 2013 by meena.p
 

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