A parallelopiped is formed using three non-collinear vectors $\vec{a}$, $\vec{b}$ and $\vec{c}$ with fixed magnitudes. Angles between any of the vector with normal of the plane determined by the other two is $\alpha$ and volume of the parallelopiped is $T$ and its surface area is $Y$. If
$$\frac{Y}{T} = 4\left(\frac{1}{|\vec{a}|} + \frac{1}{|\vec{b}|} + \frac{1}{|\vec{c}|}\right)$$
then:
$\cos^2\alpha + \cos\alpha = \dfrac{3}{4}$
$\sin^2\alpha + \sin^4\alpha = \dfrac{21}{16}$
$\cos^2\alpha + \cos\alpha = \dfrac{3 + 2\sqrt{3}}{4}$
$\sin^2\alpha + \sin^4\alpha = \dfrac{5}{16}$
Step-by-Step Solution
Key Concept: The key idea involves expressing the volume of the parallelopiped using the scalar triple product ($T = |\vec{a} \cdot (\vec{b} \times \vec{c})|$) and relating it to the area of the base faces using the given angle $\alpha$. Each face area (e.g., $|\vec{b} \times \vec{c}|$) can then be written in terms of the total volume $T$, the magnitude of the third vector ($|\vec{a}|$), and $|\cos\alpha|$.
Step 1:
Let $a = |\vec{a}|$, $b = |\vec{b}|$, $c = |\vec{c}|$.
The volume of the parallelopiped is $T = |\vec{a} \cdot (\vec{b} \times \vec{c})|$.
Given that $\alpha$ is the angle between $\vec{a}$ and the normal to the plane determined by $\vec{b}$ and $\vec{c}$, the volume can be expressed as:
$$T = |\vec{a}| |\vec{b} \times \vec{c}| \cos\alpha = a |\vec{b} \times \vec{c}| \cos\alpha$$
By cyclic permutation, we also have:
$$T = b |\vec{c} \times \vec{a}| \cos\alpha$$
$$T = c |\vec{a} \times \vec{b}| \cos\alpha$$
Step 2:
The surface area $Y$ of the parallelopiped is the sum of the areas of its six faces. These faces come in three pairs of identical parallelograms.
$$Y = 2(|\vec{a} \times \vec{b}| + |\vec{b} \times \vec{c}| + |\vec{c} \times \vec{a}|)$$
Step 3:
From the volume expressions in Step 1, we can write the magnitudes of the cross products:
$$|\vec{b} \times \vec{c}| = \frac{T}{a\cos\alpha}$$
$$|\vec{c} \times \vec{a}| = \frac{T}{b\cos\alpha}$$
$$|\vec{a} \times \vec{b}| = \frac{T}{c\cos\alpha}$$
Substitute these into the expression for $Y$:
$$Y = 2\left(\frac{T}{c\cos\alpha} + \frac{T}{a\cos\alpha} + \frac{T}{b\cos\alpha}\right)$$
$$Y = \frac{2T}{\cos\alpha}\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$$
Dividing by $T$, we obtain:
$$\frac{Y}{T} = \frac{2}{\cos\alpha}\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$$
Step 4:
The problem provides the condition:
$$\frac{Y}{T} = 4\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$$
Equating this with the derived expression for $\frac{Y}{T}$:
$$\frac{2}{\cos\alpha}\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) = 4\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$$
Since $\vec{a}$, $\vec{b}$, $\vec{c}$ are non-collinear, their magnitudes $a, b, c$ are non-zero, and thus $\left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \neq 0$. We can divide both sides by this term:
$$\frac{2}{\cos\alpha} = 4$$
$$\cos\alpha = \frac{2}{4} = \frac{1}{2}$$
Step 5:
Using the value $\cos\alpha = \frac{1}{2}$, we can evaluate related trigonometric expressions.
For the expression $\cos^2\alpha + \cos\alpha$:
$$\cos^2\alpha + \cos\alpha = \left(\frac{1}{2}\right)^2 + \frac{1}{2} = \frac{1}{4} + \frac{1}{2} = \frac{1+2}{4} = \frac{3}{4}$$
Step 6:
Using $\cos\alpha = \frac{1}{2}$, we find $\sin\alpha$:
$$\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$$
Now, we evaluate the expression $\sin^2\alpha + \sin^4\alpha$:
$$\sin^2\alpha + \sin^4\alpha = \left(\frac{\sqrt{3}}{2}\right)^2 + \left(\left(\frac{\sqrt{3}}{2}\right)^2\right)^2$$
$$= \frac{3}{4} + \left(\frac{3}{4}\right)^2$$
$$= \frac{3}{4} + \frac{9}{16}$$
$$= \frac{12}{16} + \frac{9}{16} = \frac{21}{16}$$
Correct Answer: 1, 4