Question

The three vectors $\hat{i}+\hat{j},\hat{j}+\hat{k},\hat{k}+\hat{i}$ taken two at a time form three planes The three unit vectors drawn perpendicular to these planes form a parallelopiped of volume

• A. $\frac{1}{3}$
• B. 4
• C. $\frac{3\sqrt{3}}{4}$
• D. $\frac{4}{3\sqrt{3}}$

Answer 1

Answer: (D)

$\frac{4}{3\sqrt{3}}$

The normal vector of the first plane is given by

$(i+j)\times (j+k)$
$=i-j+k$

Hence unit vector will be

$=\frac{i-j+k}{\sqrt{3}}=\overrightarrow{n_1}$

The normal vector of the second plane is given by

$(i+j)\times (k+i)$

$=i-j-k$

Hence unit vector will be

$=\frac{i-j-k}{\sqrt{3}}=\overrightarrow{n_2}$

The normal vector of the third plane is given by

$(j+k)\times (k+i)$

$=i+j-k$

Hence unit vector will be

$=\frac{i+j-k}{\sqrt{3}}=\overrightarrow{n_3}$

Hence volume of the parallel-pipe will be

$\frac{1}{3}\left[\overrightarrow{n_1}\overrightarrow{n_2}\overrightarrow{n_3}\right]$

$=\frac{4}{3\sqrt{3}}$