# 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 $\overrightarrow{a}.\overrightarrow{c}=3$.

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

Toolbox:
• $\overrightarrow a\times\overrightarrow b=\begin{vmatrix} \hat i & \hat j & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
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 $\hat i,\hat j \;and\; \hat k$
$z-y= 0$
$-(z-x)=1\;and\;y-x=-1$
$x-z=1$
$=>x-1=z$ -----(1)
$x-y=1$
$x-1=y$-----(2)
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
$x+x-1+x-1=3$
$3x=5=>x=\large\frac{5}{3}$
$y=x-1=>y=\large\frac{5}{3}$$-1=\large\frac{2}{3} z=y=>z=\large\frac{2}{3} Hence \overrightarrow c=\large\frac{5}{3}$$ \hat i+\large\frac{2}{3}$$\hat j+\large\frac{2}{3}$$ \hat k$
edited Feb 6, 2014