Question

Given that $\overline{a}=(1,1,1)$ and $\overline{c}=(0,1,-1)$. If $\overline{a}.\overline{b}=3$ and $\overline{a}\times \overline{b}=\overline{c}$, then $\overline{b}=$

• A. $\frac{1}{3}(5\overline{i}+2\overline{j}+2\overline{k})$
• B. $\frac{1}{3}(5\overline{i}-2\overline{j}-2\overline{k})$
• C. $\frac{1}{3}(5\overline{i}+2\overline{j}-2\overline{k})$
• D. $\frac{1}{4}5\overline{i}+2\overline{j}+2\overline{k}$

Answer 1

The correct option is A $\frac{1}{3}(5\overline{i}+2\overline{j}+2\overline{k})$

Let $\overrightarrow{b}=(x,y,z)=xi+yj+zk$ also $\overrightarrow{a}=i+j+k$ and $\overrightarrow{c}=j-k$

From $\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}$ on equating the coefficients of $i,j,k$ we get

$z-y=0,x-z=1$ and $y-x=-1$

These three equation are equivalent to two equation only

i.e. $z-y=0\Rightarrow y=z$ ...(1)

and $4x-z=1$ ...(2)

Also $\overrightarrow{a}.\overrightarrow{b}=3$ gives $4x=y+z=3$ ...(3)

Solving (1), (2) and (3), we get

$x=\frac{5}{3},y=z=\frac{2}{3}$

Hence, $\overrightarrow{b}=\frac{5}{3}i+\frac{2}{3}j+\frac{2}{3}k=\frac{1}{3}(5i+2j+2k)$