Question

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors each of magnitude $\sqrt{3}$ and the angle between each pair of them is $\frac{\pi}{3}$. If $\vec{b} \times \vec{c} + \vec{c} \times \vec{a} = x\vec{a} + y\vec{b} + z\vec{c}$ where $x, y, z$ are scalars, then

• A. volume of parallelpiped formed by $\vec{a}, \vec{b}, \vec{c}$ is $\frac{3\sqrt{3}}{\sqrt{2}}$
• B. the value of $x$ is $\frac{\sqrt{3}}{\sqrt{2}}$
• C. $x^2 + y^2 + z^2 \leq \sqrt{21}$

Answer 1

Answer: (A), (B), (C)

All three options are correct.

Answer 2

The problem provides three vectors $\vec{a}, \vec{b}, \vec{c}$ with equal magnitude $|\vec{a}| = |\vec{b}| = |\vec{c}| = \sqrt{3}$, and the angle between each pair is $\frac{\pi}{3}$. We have the relation $\vec{b} \times \vec{c} + \vec{c} \times \vec{a} = x\vec{a} + y\vec{b} + z\vec{c}$.

First, let's calculate the scalar products of these vectors: $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = (\sqrt{3})^2 = 3$. Similarly, $\vec{b} \cdot \vec{b} = 3$ and $\vec{c} \cdot \vec{c} = 3$. $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(\frac{\pi}{3}) = \sqrt{3} \cdot \sqrt{3} \cdot \frac{1}{2} = \frac{3}{2}$. Similarly, $\vec{b} \cdot \vec{c} = \frac{3}{2}$ and $\vec{c} \cdot \vec{a} = \frac{3}{2}$.

Option (1): Volume of the parallelpiped formed by $\vec{a}, \vec{b}, \vec{c}$ is $|[\vec{a} \vec{b} \vec{c}]|$. The square of the scalar triple product is given by the Gram determinant: $[\vec{a} \vec{b} \vec{c}]^2 = \begin{vmatrix} \vec{a} \cdot \vec{a} & \vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ \vec{b} \cdot \vec{a} & \vec{b} \cdot \vec{b} & \vec{b} \cdot \vec{c} \\ \vec{c} \cdot \vec{a} & \vec{c} \cdot \vec{b} & \vec{c} \cdot \vec{c} \end{vmatrix} = \begin{vmatrix} 3 & 3/2 & 3/2 \\ 3/2 & 3 & 3/2 \\ 3/2 & 3/2 & 3 \end{vmatrix}$ $= 3(3 \cdot 3 - \frac{3}{2} \cdot \frac{3}{2}) - \frac{3}{2}(\frac{3}{2} \cdot 3 - \frac{3}{2} \cdot \frac{3}{2}) + \frac{3}{2}(\frac{3}{2} \cdot \frac{3}{2} - 3 \cdot \frac{3}{2})$ $= 3(9 - \frac{9}{4}) - \frac{3}{2}(\frac{9}{2} - \frac{9}{4}) + \frac{3}{2}(\frac{9}{4} - \frac{9}{2})$ $= 3(\frac{27}{4}) - \frac{3}{2}(\frac{9}{4}) + \frac{3}{2}(-\frac{9}{4}) = \frac{81}{4} - \frac{27}{8} - \frac{27}{8} = \frac{81}{4} - \frac{54}{8} = \frac{81}{4} - \frac{27}{4} = \frac{54}{4} = \frac{27}{2}$. The volume is $|[\vec{a} \vec{b} \vec{c}]| = \sqrt{\frac{27}{2}} = \frac{\sqrt{27}}{\sqrt{2}} = \frac{3\sqrt{3}}{\sqrt{2}}$. Option (1) states the volume is $\frac{3\sqrt{3}}{\sqrt{2}}$, which is correct.

To find $x, y, z$, take the dot product of the given equation $\vec{b} \times \vec{c} + \vec{c} \times \vec{a} = x\vec{a} + y\vec{b} + z\vec{c}$ with $\vec{a}, \vec{b}, \vec{c}$. Dot product with $\vec{a}$: $\vec{a} \cdot (\vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = \vec{a} \cdot (x\vec{a} + y\vec{b} + z\vec{c})$ $\vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{c} \times \vec{a}) = x(\vec{a} \cdot \vec{a}) + y(\vec{a} \cdot \vec{b}) + z(\vec{a} \cdot \vec{c})$ $[\vec{a} \vec{b} \vec{c}] + 0 = 3x + \frac{3}{2}y + \frac{3}{2}z$ $\frac{3\sqrt{3}}{\sqrt{2}} = 3x + \frac{3}{2}y + \frac{3}{2}z$. Dividing by 3: $\frac{\sqrt{3}}{\sqrt{2}} = x + \frac{1}{2}y + \frac{1}{2}z$. (Equation 1)

Dot product with $\vec{b}$: $\vec{b} \cdot (\vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = \vec{b} \cdot (x\vec{a} + y\vec{b} + z\vec{c})$ $\vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{c} \times \vec{a}) = x(\vec{b} \cdot \vec{a}) + y(\vec{b} \cdot \vec{b}) + z(\vec{b} \cdot \vec{c})$ $0 + [\vec{b} \vec{c} \vec{a}] = \frac{3}{2}x + 3y + \frac{3}{2}z$ $[\vec{a} \vec{b} \vec{c}] = \frac{3}{2}x + 3y + \frac{3}{2}z$ $\frac{3\sqrt{3}}{\sqrt{2}} = \frac{3}{2}x + 3y + \frac{3}{2}z$. Dividing by 3: $\frac{\sqrt{3}}{\sqrt{2}} = \frac{1}{2}x + y + \frac{1}{2}z$. (Equation 2)

Dot product with $\vec{c}$: $\vec{c} \cdot (\vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = \vec{c} \cdot (x\vec{a} + y\vec{b} + z\vec{c})$ $\vec{c} \cdot (\vec{b} \times \vec{c}) + \vec{c} \cdot (\vec{c} \times \vec{a}) = x(\vec{c} \cdot \vec{a}) + y(\vec{c} \cdot \vec{b}) + z(\vec{c} \cdot \vec{c})$ $0 + 0 = \frac{3}{2}x + \frac{3}{2}y + 3z$ $0 = \frac{3}{2}x + \frac{3}{2}y + 3z$. Dividing by 3/2: $0 = x + y + 2z$. (Equation 3)

We have the system of equations:

1. $x + \frac{1}{2}y + \frac{1}{2}z = \frac{\sqrt{3}}{\sqrt{2}}$
2. $\frac{1}{2}x + y + \frac{1}{2}z = \frac{\sqrt{3}}{\sqrt{2}}$
3. $x + y + 2z = 0$

From (1) and (2), $x + \frac{y+z}{2} = \frac{x+z}{2} + y$, which implies $x + \frac{y}{2} = \frac{x}{2} + y$, so $\frac{x}{2} = \frac{y}{2}$, which gives $x=y$. Substitute $y=x$ into (3): $x + x + 2z = 0 \implies 2x + 2z = 0 \implies x + z = 0 \implies z = -x$. Substitute $y=x$ and $z=-x$ into (1): $x + \frac{1}{2}x + \frac{1}{2}(-x) = \frac{\sqrt{3}}{\sqrt{2}} \implies x = \frac{\sqrt{3}}{\sqrt{2}}$. So, $x = \frac{\sqrt{3}}{\sqrt{2}}$, $y = \frac{\sqrt{3}}{\sqrt{2}}$, $z = -\frac{\sqrt{3}}{\sqrt{2}}$.

Option (2): the value of $x$ is $\frac{\sqrt{3}}{\sqrt{2}}$. This is correct.

Option (3): $x^2 + y^2 + z^2 \leq \sqrt{21}$. $x^2 + y^2 + z^2 = (\frac{\sqrt{3}}{\sqrt{2}})^2 + (\frac{\sqrt{3}}{\sqrt{2}})^2 + (-\frac{\sqrt{3}}{\sqrt{2}})^2 = \frac{3}{2} + \frac{3}{2} + \frac{3}{2} = \frac{9}{2} = 4.5$. We need to check if $4.5 \leq \sqrt{21}$. Squaring both sides, we check if $(4.5)^2 \leq (\sqrt{21})^2$. $4.5^2 = 20.25$. $\sqrt{21}^2 = 21$. Since $20.25 \leq 21$, the inequality $4.5 \leq \sqrt{21}$ is true. Option (3) is correct.

All three options are correct.