Question

Let a=i+j+k and c=j-k . If b is a vector satisfying a×b=c and a.b=3, then find b.

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

Answer 1

Answer: (A)

$\frac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k})$

Let$b=xi+yj+zk$

Now

$a.b=x+y+z=3$

Now $a\times b=(i+j+k)(xi+yj+zk)=(z-y)i+(x-z)j+(y-x)k=j-k$

Now $z.y=0$ hence $k=y$

Now $x-z=1$ and $y-x=-1$

Now

$x+y+z=31+z+z+z=3z=\frac{2}{3}$

Hence $y=\frac{2}{3}x=\frac{5}{3}$

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