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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Vector Algebra

If $\overrightarrow a=\hat i+2\hat j+\hat k,\:\overrightarrow b=\hat i-\hat j+\hat k$ and $\overrightarrow c=\hat i+\hat j-\hat k$ and a vector $\overrightarrow d$ which is in the plane of $\overrightarrow a\:and\:\overrightarrow b$ and whose projection on $\overrightarrow c$ is $\large\frac{1}{\sqrt 3}, $ then $\overrightarrow d=?$

1 Answer

Let $\overrightarrow d=x\hat i+y\hat j+z\hat k$ be vector in the plane of $\overrightarrow a\:and\:\overrightarrow b$
$\Rightarrow\:[\overrightarrow a\:\overrightarrow b\:\overrightarrow d]=0$
$\Rightarrow\:(\overrightarrow a\times\overrightarrow b).\overrightarrow d=0$
$\Rightarrow\:3x-3z=0,\:\:\Rightarrow\:x-z=0.........(i)$
Given projection of $\overrightarrow d$ in $\overrightarrow c =\large\frac{1}{\sqrt 3}$
$\Rightarrow\:\large\frac{\overrightarrow d.\overrightarrow c}{|\overrightarrow c|}=\frac{1}{\sqrt3}$
$\Rightarrow\:\large\frac{x+y-z}{\sqrt 3}=\frac{1}{\sqrt 3}$
$\Rightarrow\:x+y-z=1............(ii)$
Solving (i) and (ii) we get
$x=z$ and $y=1$
which satisfies the vector $-2\hat i+\hat j-2\hat k$

 

answered Nov 10, 2013 by rvidyagovindarajan_1
edited Nov 10, 2013 by rvidyagovindarajan_1
 

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