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 three planes form a parallelopiped of volume (in cubic units)
Answer & explanation
Correct answer: option 4
Let $\vec a =\hat i +\hat j, \vec b =\hat j+\hat k$ and $\vec c =\hat k +\hat i$. Further, let
$\vec α=\frac{\vec a×\vec b}{|\vec a×\vec b|},\vec β=\frac{\vec b×\vec c}{|\vec b×\vec c|}$ and $\vec γ=\frac{\vec c×\vec a}{|\vec c×\vec a|}$
∴ Required volume
$=[α\,\,β\,\,γ]$
$=\frac{1}{|\vec a×\vec b||\vec b×\vec c||\vec c×\vec a|}\begin{bmatrix}\vec a×\vec b&\vec b×\vec c&\vec c×\vec a\end{bmatrix}$
$=\frac{[\vec a\,\,\vec b\,\,\vec c]^2}{|\vec a×\vec b||\vec b×\vec c||\vec c×\vec a|}$ $[∵\begin{bmatrix}\vec a×\vec b&\vec b×\vec c&\vec c×\vec a\end{bmatrix}=[\vec a\,\,\vec b\,\,\vec c]^2]$
Now,
$[\vec a\,\,\vec b\,\,\vec c]=\begin{vmatrix}1&1&0\\0&1&1\\1&0&1\end{vmatrix}=1-1 (0-1)=2$
And,
$\vec a×\vec b=(\hat i+\hat j)×(\hat j+\hat k)=\hat k-\hat j+\hat i$
$\vec b×\vec c=(\hat j+\hat k)×(\hat k+\hat i)=\hat i-\hat k+\hat j$
and, $\vec c×\vec a=(\hat k+\hat i)×(\hat i+\hat j)=\hat j-\hat i+\hat k$
$∴|\vec a×\vec b|=|\vec b×\vec c|=|\vec c×\vec a|=\sqrt{3}$
Hence, required volume = $\frac{2^2}{3\sqrt{3}}=\frac{4}{3\sqrt{3}}$