The question given is
The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors $\hat{a}, \hat{b}, \hat{c}$ such that $a.b = b.c = c.a = 1/2$ (writing $\hat{a},\hat{b},\hat{c}$ as a,b,c). Then find the volume of the parallelopiped
My approach:
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The first approach I thought of was geometrical. Since dot product of unit vectors is 1/2 I considered an angle 60 between them. So I imagined a plane with $a$ and $b$ making 60 with each other and $c$ above this plane making 60 with both $a$ and $b$ and so the area of base is $|a\times b|$ which is $\sin{60} = \frac{\sqrt{3}}{2}$ now $a \times b$ gives a vector perpendicular to their plane and making angle $30$ with c so the I did dot product of this with c and the final area i got was $\frac{3}{4}$
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The second approach I did was like this. The area of the parallelopiped is given as $[a b c] = a.(b \times c)$ which on expanding gives $(a.c)\vec{b} – (a.b)\vec{c}$ now since cross product amongst them is given as $1/2$ this gives me the area as $\frac{1}{2}(\vec{b}-\vec{c})$ the modulus of which is $1/2$
However none of the methods I have mentioned give the correct answer and the correct answer given is:
$\frac{1}{\sqrt{2}}$
Please help me in getting to know what I am doing wrong in my approach and how I should go about doing it the "right" way ?
Best Answer
HINT: The right way (one of them, atleast) is to use the square of the scalar triple product to get a $3×3$ matrix with only magnitude and dot product terms.
Your second approach is wrong because you confused between vector triple product and scalar triple product.
You first approach is wrong because it is not necessary for $c$ to subtend $\frac {\pi}{3}$ radians with the plane of $a$ and $b$.