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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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If $\overrightarrow{a}=\hat i+\hat j+\hat k$ and $\overrightarrow{b}=\hat j-\hat k$,find a vector $\overrightarrow{c}$ such that $\overrightarrow{a}\times\overrightarrow{c}=\overrightarrow{b}$ and $\overrightarrow{a}.\overrightarrow{c}=3$.

$\begin{array}{1 1} (A)\;\large\frac{3}{5} \hat i+\large\frac{3}{2} \hat j+\large\frac{3}{2} \hat k \\(B)\;\large\frac{5}{8} \hat i+\large\frac{2}{8} \hat j+\large\frac{2}{7} \hat k \\ (C)\;\large\frac{5}{3} \hat i+\large\frac{2}{3} \hat j+\large\frac{2}{3} \hat k \\ (D)\;\large\frac{1}{3} \hat i+\large\frac{1}{3} \hat j+\large\frac{1}{3} \hat k \end{array} $

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1 Answer

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  • $ \overrightarrow a\times\overrightarrow b=\begin{vmatrix} \hat i & \hat j & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Step 1:
Given $\overrightarrow a=\hat i+\hat j+\hat k$ and $\overrightarrow b=\hat j-\hat k$
$\overrightarrow a \times \overrightarrow c=\overrightarrow b$ and $\overrightarrow a.\overrightarrow c=3$
Let $ \overrightarrow c =x\hat i+y\hat j+z\hat k$
$ \overrightarrow a\times\overrightarrow c=\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix}=\hat j -\hat k$
$ =>(z-y)\hat i - (z-x)\hat j + (y-x)\hat k=\hat j-\hat k $
Now equating the coefficient of $ \hat i,\hat j \;and\; \hat k$
$z-y= 0 $
$=>x-1=z$ -----(1)
Also given that $\overrightarrow a.\overrightarrow c=3$
$i.e., \:\:(\hat i+\hat j+\hat k).(x \hat i+y \hat j +z \hat k)=3$
$ \Rightarrow x+y+z=3 $-----(3)
Step 2:
Now let us solve equations (1) , (2) and (3) to get
Hence $\overrightarrow c=\large\frac{5}{3}$$ \hat i+\large\frac{2}{3}$$ \hat j+\large\frac{2}{3}$$ \hat k$
answered Jun 3, 2013 by meena.p
edited Feb 6, 2014 by rvidyagovindarajan_1

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