Question

Let \[ \vec{r}=\sin x(\vec{a}\times\vec{b})+\cos y(\vec{b}\times\vec{c})+2(\vec{c}\times\vec{a}), \] where \(\vec{a},\vec{b}\) and \(\vec{c}\) are three non-coplanar vectors. It is given that \(\vec{r}\) is perpendicular to \((\vec{a}+\vec{b}+\vec{c})\). Then the possible value(s) of \((x^{2}+y^{2})\) is/are:

• A. $\frac{5\pi^{2}}{4}$
• B. $\frac{35\pi^{2}}{4}$
• C. $\frac{37\pi^{2}}{4}$
• D. $\frac{\pi^{2}}{4}$

Hint:

Cyclic order is everything in scalar triple products! Remember that $[\vec{a}\ \vec{b}\ \vec{c}] = [\vec{b}\ \vec{c}\ \vec{a}] = [\vec{c}\ \vec{a}\ \vec{b}]$. Keeping them in this correct order allows you to factor out the triple product as a single common term easily.

Answer 1

The Correct Option is A

Concept: If a vector $\vec{r}$ is perpendicular to another vector, their scalar dot product is exactly zero. When taking the dot product of a vector combination of cross products with the base vectors $(\vec{a}+\vec{b}+\vec{c})$, any term containing a repeated vector vanishes ($\vec{a} \cdot (\vec{a} \times \vec{b}) = 0$), leaving only the standard scalar triple products $[\vec{a}\ \vec{b}\ \vec{c}]$. Step 1: Set up the perpendicular dot product equation.
We are given that $\vec{r}$ is perpendicular to $(\vec{a} + \vec{b} + \vec{c})$, which means: \[ \vec{r} \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \quad \cdots (1) \] Substitute the full expanded expression for $\vec{r}$ into equation (1): \[ \Big[ \sin x(\vec{a}\times\vec{b}) + \cos y(\vec{b}\times\vec{c}) + 2(\vec{c}\times\vec{a}) \Big] \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]

Step 2: Distribute the dot product and simplify using scalar triple product properties.
Distribute the dot product across the three vectors. Recall that any scalar triple product with two identical vectors is zero. The only non-zero terms that survive are:
• For the first term: $\sin x(\vec{a}\times\vec{b})\cdot\vec{c} = \sin x [\vec{a}\ \vec{b}\ \vec{c}]$
• For the second term: $\cos y(\vec{b}\times\vec{c})\cdot\vec{a} = \cos y [\vec{b}\ \vec{c}\ \vec{a}] = \cos y [\vec{a}\ \vec{b}\ \vec{c}]$
• For the third term: $2(\vec{c}\times\vec{a})\cdot\vec{b} = 2 [\vec{c}\ \vec{a}\ \vec{b}] = 2 [\vec{a}\ \vec{b}\ \vec{c}]$ Substitute these simplified terms back into the equation: \[ \sin x [\vec{a}\ \vec{b}\ \vec{c}] + \cos y [\vec{a}\ \vec{b}\ \vec{c}] + 2 [\vec{a}\ \vec{b}\ \vec{c}] = 0 \]

Step 3: Isolate the trigonometric constraint equation.
Since $\vec{a}, \vec{b},$ and $\vec{c}$ are non-coplanar vectors, their scalar triple product is non-zero ($[\vec{a}\ \vec{b}\ \vec{c}] \neq 0$). This allows us to divide the entire equation by $[\vec{a}\ \vec{b}\ \vec{c}]$, yielding: \[ \sin x + \cos y + 2 = 0 \quad \Rightarrow \quad \sin x + \cos y = -2 \quad \cdots (2) \]

Step 4: Solve for the angle parameters and calculate $(x^2 + y^2)$.
For the sum of a sine function and a cosine function to equal $-2$, both individual functions must simultaneously reach their absolute minimum value of $-1$: \[ \sin x = -1 \quad \Rightarrow \quad x = 2n\pi - \frac{\pi}{2} \] \[ \cos y = -1 \quad \Rightarrow \quad y = (2k + 1)\pi \] Let us evaluate the baseline principal values to find the combinations matching our options:
• If we select the primary principal solutions $x = -\frac{\pi}{2}$ and $y = \pi$: \[ x^2 + y^2 = \left(-\frac{\pi}{2}\right)^2 + (\pi)^2 = \frac{\pi^2}{4} + \pi^2 = \frac{5\pi^2}{4} \quad \text{(Matches Option A)} \]
• If we select the next sequential solution branch $x = \frac{3\pi}{2}$ and $y = \pi$: \[ x^2 + y^2 = \left(\frac{3\pi}{2}\right)^2 + (\pi)^2 = \frac{\pi^2}{4} + 9\pi^2 = \frac{13\pi^2}{4} \]
• If we select the branch $x = -\frac{\pi}{2}$ and $y = 3\pi$: \[ x^2 + y^2 = \left(-\frac{\pi}{2}\right)^2 + (3\pi)^2 = \frac{\pi^2}{4} + 9\pi^2 = \frac{37\pi^2}{4} \quad \text{(Matches Option C)} \] Therefore, the valid possible values are given by choices (A) and (C).