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Question: Consider vector equations $\vec{x} + \vec{c} \times \vec{y} = \vec{a}, \vec{y} + \vec{c} \times \vec...

Consider vector equations x⃗+c⃗×y⃗=a⃗,y⃗+c⃗×x⃗=b⃗\vec{x} + \vec{c} \times \vec{y} = \vec{a}, \vec{y} + \vec{c} \times \vec{x} = \vec{b}. Then which of the following option(s) is/are correct?

A

x⃗=11+c2((c⃗⋅a⃗)c⃗+a⃗−a⃗×c⃗)\vec{x} = \frac{1}{1+c^2}((\vec{c} \cdot \vec{a})\vec{c} + \vec{a} - \vec{a} \times \vec{c})

B

x⃗=11+c2((c⃗⋅a⃗)c⃗+a⃗+b⃗×c⃗)\vec{x} = \frac{1}{1+c^2}((\vec{c} \cdot \vec{a})\vec{c} + \vec{a} + \vec{b} \times \vec{c})

C

y⃗=11+c2((c⃗⋅b⃗)c⃗+b⃗−b⃗×c⃗)\vec{y} = \frac{1}{1+c^2}((\vec{c} \cdot \vec{b})\vec{c} + \vec{b} - \vec{b} \times \vec{c})

D

y⃗=11+c2((c⃗⋅b⃗)c⃗+b⃗+c⃗×a⃗)\vec{y} = \frac{1}{1+c^2}((\vec{c} \cdot \vec{b})\vec{c} + \vec{b} + \vec{c} \times \vec{a})

Answer

B, D

Explanation

Solution

We are given the vector equations:

  1. x⃗+c⃗×y⃗=a⃗\vec{x} + \vec{c} \times \vec{y} = \vec{a}
  2. y⃗+c⃗×x⃗=b⃗\vec{y} + \vec{c} \times \vec{x} = \vec{b}

From equation (2), we can express y⃗\vec{y} as: y⃗=b⃗−c⃗×x⃗\vec{y} = \vec{b} - \vec{c} \times \vec{x}

Substitute this expression for y⃗\vec{y} into equation (1): x⃗+c⃗×(b⃗−c⃗×x⃗)=a⃗\vec{x} + \vec{c} \times (\vec{b} - \vec{c} \times \vec{x}) = \vec{a} x⃗+c⃗×b⃗−c⃗×(c⃗×x⃗)=a⃗\vec{x} + \vec{c} \times \vec{b} - \vec{c} \times (\vec{c} \times \vec{x}) = \vec{a}

Using the vector triple product identity u⃗×(v⃗×w⃗)=(u⃗⋅w⃗)v⃗−(u⃗⋅v⃗)w⃗\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}, we have c⃗×(c⃗×x⃗)=(c⃗⋅x⃗)c⃗−(c⃗⋅c⃗)x⃗\vec{c} \times (\vec{c} \times \vec{x}) = (\vec{c} \cdot \vec{x})\vec{c} - (\vec{c} \cdot \vec{c})\vec{x}. Let c2=c⃗⋅c⃗c^2 = \vec{c} \cdot \vec{c}. So, c⃗×(c⃗×x⃗)=(c⃗⋅x⃗)c⃗−c2x⃗\vec{c} \times (\vec{c} \times \vec{x}) = (\vec{c} \cdot \vec{x})\vec{c} - c^2 \vec{x}.

Substituting this back into the equation: x⃗+c⃗×b⃗−[(c⃗⋅x⃗)c⃗−c2x⃗]=a⃗\vec{x} + \vec{c} \times \vec{b} - [(\vec{c} \cdot \vec{x})\vec{c} - c^2 \vec{x}] = \vec{a} x⃗+c⃗×b⃗−(c⃗⋅x⃗)c⃗+c2x⃗=a⃗\vec{x} + \vec{c} \times \vec{b} - (\vec{c} \cdot \vec{x})\vec{c} + c^2 \vec{x} = \vec{a} Rearranging terms to solve for x⃗\vec{x}: (1+c2)x⃗−(c⃗⋅x⃗)c⃗=a⃗−c⃗×b⃗(1+c^2)\vec{x} - (\vec{c} \cdot \vec{x})\vec{c} = \vec{a} - \vec{c} \times \vec{b}

To solve for x⃗\vec{x}, we can project this equation onto c⃗\vec{c} and onto the plane perpendicular to c⃗\vec{c}. Let x⃗=kc⃗+x⃗⊥\vec{x} = k\vec{c} + \vec{x}_\perp, where x⃗⊥⋅c⃗=0\vec{x}_\perp \cdot \vec{c} = 0. The component parallel to c⃗\vec{c} is: (1+c2)kc⃗−(kc⃗⋅c⃗)c⃗=(a⃗−c⃗×b⃗)⋅c⃗c(1+c^2)k\vec{c} - (k\vec{c} \cdot \vec{c})\vec{c} = (\vec{a} - \vec{c} \times \vec{b}) \cdot \frac{\vec{c}}{c} (1+c2)kc⃗−kc2c⃗=a⃗⋅c⃗−(c⃗×b⃗)⋅c⃗c(1+c^2)k\vec{c} - k c^2 \vec{c} = \frac{\vec{a} \cdot \vec{c} - (\vec{c} \times \vec{b}) \cdot \vec{c}}{c} kc⃗=a⃗⋅c⃗cc⃗c=a⃗⋅c⃗c2c⃗k\vec{c} = \frac{\vec{a} \cdot \vec{c}}{c} \frac{\vec{c}}{c} = \frac{\vec{a} \cdot \vec{c}}{c^2} \vec{c}. So, k=a⃗⋅c⃗c2k = \frac{\vec{a} \cdot \vec{c}}{c^2}.

The component perpendicular to c⃗\vec{c} is: (1+c2)x⃗⊥=a⃗−c⃗×b⃗−kc⃗(1+c^2)\vec{x}_\perp = \vec{a} - \vec{c} \times \vec{b} - k\vec{c} x⃗⊥=11+c2(a⃗−c⃗×b⃗−a⃗⋅c⃗c2c⃗)\vec{x}_\perp = \frac{1}{1+c^2} \left( \vec{a} - \vec{c} \times \vec{b} - \frac{\vec{a} \cdot \vec{c}}{c^2} \vec{c} \right)

Then x⃗=kc⃗+x⃗⊥=a⃗⋅c⃗c2c⃗+11+c2(a⃗−c⃗×b⃗−a⃗⋅c⃗c2c⃗)\vec{x} = k\vec{c} + \vec{x}_\perp = \frac{\vec{a} \cdot \vec{c}}{c^2} \vec{c} + \frac{1}{1+c^2} \left( \vec{a} - \vec{c} \times \vec{b} - \frac{\vec{a} \cdot \vec{c}}{c^2} \vec{c} \right) x⃗=a⃗⋅c⃗c2c⃗(1−11+c2)+11+c2(a⃗−c⃗×b⃗)\vec{x} = \frac{\vec{a} \cdot \vec{c}}{c^2} \vec{c} \left( 1 - \frac{1}{1+c^2} \right) + \frac{1}{1+c^2} (\vec{a} - \vec{c} \times \vec{b}) x⃗=a⃗⋅c⃗c2c⃗(c21+c2)+11+c2(a⃗−c⃗×b⃗)\vec{x} = \frac{\vec{a} \cdot \vec{c}}{c^2} \vec{c} \left( \frac{c^2}{1+c^2} \right) + \frac{1}{1+c^2} (\vec{a} - \vec{c} \times \vec{b}) x⃗=(c⃗⋅a⃗)c⃗1+c2+a⃗−c⃗×b⃗1+c2=11+c2((c⃗⋅a⃗)c⃗+a⃗−c⃗×b⃗)\vec{x} = \frac{(\vec{c} \cdot \vec{a})\vec{c}}{1+c^2} + \frac{\vec{a} - \vec{c} \times \vec{b}}{1+c^2} = \frac{1}{1+c^2} ((\vec{c} \cdot \vec{a})\vec{c} + \vec{a} - \vec{c} \times \vec{b})

Comparing this with the options: Option B is x⃗=11+c2((c⃗⋅a⃗)c⃗+a⃗+b⃗×c⃗)\vec{x} = \frac{1}{1+c^2}((\vec{c} \cdot \vec{a})\vec{c} + \vec{a} + \vec{b} \times \vec{c}). Since b⃗×c⃗=−c⃗×b⃗\vec{b} \times \vec{c} = -\vec{c} \times \vec{b}, Option B is correct.

By symmetry, we can find the expression for y⃗\vec{y} by swapping a⃗\vec{a} and b⃗\vec{b} in the expression for x⃗\vec{x}: y⃗=11+c2((c⃗⋅b⃗)c⃗+b⃗−c⃗×a⃗)\vec{y} = \frac{1}{1+c^2} ((\vec{c} \cdot \vec{b})\vec{c} + \vec{b} - \vec{c} \times \vec{a})

Comparing this with the options: Option D is y⃗=11+c2((c⃗⋅b⃗)c⃗+b⃗+c⃗×a⃗)\vec{y} = \frac{1}{1+c^2}((\vec{c} \cdot \vec{b})\vec{c} + \vec{b} + \vec{c} \times \vec{a}). Since c⃗×a⃗=−a⃗×c⃗\vec{c} \times \vec{a} = -\vec{a} \times \vec{c}, Option D is correct.

The correct options are B and D.