Question:
The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vector $\widehat{a},\widehat{b},\widehat{c}$ such that $\widehat{a}.\widehat{b}=\widehat{b}.\widehat{c}=\widehat{c}.\widehat{a}=\frac{1}{2}.$Then, the volume of the parallelopiped is
The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vector $\widehat{a},\widehat{b},\widehat{c}$ such that $\widehat{a}.\widehat{b}=\widehat{b}.\widehat{c}=\widehat{c}.\widehat{a}=\frac{1}{2}.$Then, the volume of the parallelopiped is
Updated On: Jun 14, 2022
- $\frac{1}{\sqrt{2}}$
- $\frac{1}{2\sqrt{2}} $
- $\frac{\sqrt{3}}{2}$
- $\frac{1}{\sqrt{3}}$
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The Correct Option is A
Solution and Explanation
The volume of the parallelopiped with coterminus edges
as $\widehat{a},\widehat{b},\widehat{c}$ is given by $[\widehat{a},\widehat{b},\widehat{c}]=\widehat{a}.(\widehat{b}\times\widehat{c})$
$Now,\ [\widehat{a}.\widehat{b}\widehat{c}]^2=\begin{array}| \widehat{a}.\widehat{a}&\widehat{a}.\widehat{b}&\widehat{a}.\widehat{c}\\\widehat{b}.\widehat{a}&\widehat{b}.\widehat{b}&\widehat{b}.\widehat{c}\\\widehat{c}.\widehat{a}&\widehat{c}.\widehat{b}&\widehat{c}.\widehat{c}\\\end{array}=\begin{array}|1&1/2&1/2\\1/2&1&1/2\\1/2&1/2&1\\\end{array}$
$\Rightarrow \ \ \ \ \ \ [\widehat{a}.\widehat{b}\widehat{c}]^2= 1\Bigg(1-\frac{1}{4}\Bigg)-\frac{1}{2}\Bigg(\frac{1}{2}\frac{1}{2}\Bigg)+\frac{1}{2}\Bigg(\frac{1}{4}\frac{1}{2}\Bigg)=\frac{1}{2}$
Thus, the required volume of the parallelopiped
$=\frac{1}{\sqrt{2}} cu unit$
as $\widehat{a},\widehat{b},\widehat{c}$ is given by $[\widehat{a},\widehat{b},\widehat{c}]=\widehat{a}.(\widehat{b}\times\widehat{c})$
$Now,\ [\widehat{a}.\widehat{b}\widehat{c}]^2=\begin{array}| \widehat{a}.\widehat{a}&\widehat{a}.\widehat{b}&\widehat{a}.\widehat{c}\\\widehat{b}.\widehat{a}&\widehat{b}.\widehat{b}&\widehat{b}.\widehat{c}\\\widehat{c}.\widehat{a}&\widehat{c}.\widehat{b}&\widehat{c}.\widehat{c}\\\end{array}=\begin{array}|1&1/2&1/2\\1/2&1&1/2\\1/2&1/2&1\\\end{array}$
$\Rightarrow \ \ \ \ \ \ [\widehat{a}.\widehat{b}\widehat{c}]^2= 1\Bigg(1-\frac{1}{4}\Bigg)-\frac{1}{2}\Bigg(\frac{1}{2}\frac{1}{2}\Bigg)+\frac{1}{2}\Bigg(\frac{1}{4}\frac{1}{2}\Bigg)=\frac{1}{2}$
Thus, the required volume of the parallelopiped
$=\frac{1}{\sqrt{2}} cu unit$
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