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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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The value of $\lambda$ for which the vectors $3\hat i+6\hat j+\hat k $ and $ 2\hat i+4\hat j+\lambda\hat k$ are parallel is

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

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  • $\overrightarrow a.\overrightarrow b=|\;a\;|\;|\;b\;| \cos \theta$
Let $\overrightarrow a=3\hat i+6\hat j+\hat k\:and\:\overrightarrow b=2\hat i+4\hat j+\lambda\hat k$
It is given that $\overrightarrow a$ is parallel to $\overrightarrow b$
$=>\theta=0\qquad \cos 0=1$
On substituting for $\theta$
$\overrightarrow a.\overrightarrow b=|\;a\;|\;|\;b\;| \cos \theta$
$\overrightarrow a.\overrightarrow b=|\; \overrightarrow a\;|\;|\; \overrightarrow b\;|\; \times 1$
On substituting for $\overrightarrow a$ and $\overrightarrow b$
(ie) $(3\hat i+6\hat j+\hat k).(2\hat i+4\hat j+\lambda k)$
$=\sqrt {(3)^2+(6)^2+(1)^2}.\sqrt {(2)^2+(4)^2+(\lambda)^2}$
$=(6+24+\lambda)=\sqrt {9+36+1}.\sqrt {4+16+\lambda^2}$
$(30+\lambda)=\sqrt {46}. $ $\sqrt {20+\lambda^2}$
Squaring on bothsides we get,
$(30+ \lambda)^2=46(20+\lambda^2)$
$=>900+60 \lambda +\lambda^2=920+46 \lambda^2$
On simplifying we get,
$45\lambda^2-60 \lambda+20=0$
Divide throughout by $5$
$9\lambda^2-12 \lambda+4=0$
$=>(3 \lambda-2)^2=0$
$=>3 \lambda-2=0$
Therefore $\lambda =\large\frac{2}{3}$
Hence the correct option is $A$
answered May 28, 2013 by meena.p
 

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