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Let aˉ=i^+2j^−k^\bar{a}= \hat{i}+2\hat{j}-\hat{k} and bˉ=i^+j^−k^\bar{b}= \hat{i}+\hat{j}-\hat{k} be two vectors. If cˉ\bar{c} is a vector such that bˉ×cˉ=bˉ×aˉ\bar{b}\times \bar{c}= \bar{b}\times \bar{a} and cˉ⋅aˉ=0\bar{c}\cdot \bar{a}=0, then cˉ⋅bˉ\bar{c}\cdot \bar{b} is

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Full solution

Q. Let aˉ=i^+2j^−k^\bar{a}= \hat{i}+2\hat{j}-\hat{k} and bˉ=i^+j^−k^\bar{b}= \hat{i}+\hat{j}-\hat{k} be two vectors. If cˉ\bar{c} is a vector such that bˉ×cˉ=bˉ×aˉ\bar{b}\times \bar{c}= \bar{b}\times \bar{a} and cˉ⋅aˉ=0\bar{c}\cdot \bar{a}=0, then cˉ⋅bˉ\bar{c}\cdot \bar{b} is
  1. Identify Cross Product Equation: Identify the cross product equation. bˉ×cˉ=bˉ×aˉ\bar{b}\times \bar{c}= \bar{b}\times \bar{a}
  2. Calculate Cross Product: Calculate the cross product bˉ×aˉ\bar{b}\times \bar{a}.bˉ×aˉ=(i^+j^−k^)×(i^+2j^−k^)\bar{b}\times \bar{a} = (\hat{i}+ \hat{j}- \hat{k})\times(\hat{i}+2 \hat{j}- \hat{k})bˉ×aˉ=i^×i^+i^×2j^+i^×(−k^)+j^×i^+j^×2j^+j^×(−k^)−k^×i^−k^×2j^−k^×(−k^)\bar{b}\times \bar{a} = \hat{i}\times \hat{i} + \hat{i}\times2 \hat{j} + \hat{i}\times(- \hat{k}) + \hat{j}\times \hat{i} + \hat{j}\times2 \hat{j} + \hat{j}\times(- \hat{k}) - \hat{k}\times \hat{i} - \hat{k}\times2 \hat{j} - \hat{k}\times(- \hat{k})bˉ×aˉ=0+2k^−j^−j^+0+i^+i^−2k^+0\bar{b}\times \bar{a} = 0 + 2 \hat{k} - \hat{j} - \hat{j} + 0 + \hat{i} + \hat{i} - 2 \hat{k} + 0bˉ×aˉ=2i^−2j^\bar{b}\times \bar{a} = 2 \hat{i} - 2 \hat{j}
  3. Find Cross Product Equality: Since bˉ×cˉ=bˉ×aˉ\bar{b}\times \bar{c} = \bar{b}\times \bar{a}, we have bˉ×cˉ=2i^−2j^.\bar{b}\times \bar{c} = 2\hat{i} - 2\hat{j}.
  4. Identify Dot Product Equation: Identify the dot product equation. cˉ⋅aˉ=0\bar{c} \cdot \bar{a} = 0
  5. Use Dot Product Equation: Use the dot product equation to find components of cˉ\bar{c}. Let cˉ=xi^+yj^+zk^\bar{c} = x \hat{i} + y \hat{j} + z \hat{k}. cˉ⋅aˉ=(xi^+yj^+zk^)⋅(i^+2j^−k^)\bar{c}\cdot \bar{a} = (x \hat{i} + y \hat{j} + z \hat{k})\cdot(\hat{i}+2 \hat{j}- \hat{k}) cˉ⋅aˉ=x⋅1+y⋅2+z⋅(−1)\bar{c}\cdot \bar{a} = x\cdot 1 + y\cdot 2 + z\cdot (-1) cˉ⋅aˉ=x+2y−z\bar{c}\cdot \bar{a} = x + 2y - z Since cˉ⋅aˉ=0\bar{c}\cdot \bar{a}=0, we have x+2y−z=0x + 2y - z = 0.
  6. Calculate Dot Product of cc and bb: Now, calculate cˉ⋅bˉ\bar{c} \cdot \bar{b} using the components of cˉ\bar{c}.cˉ⋅bˉ=(xi^+yj^+zk^)⋅(i^+j^−k^)\bar{c} \cdot \bar{b} = (x \hat{i} + y \hat{j} + z \hat{k})\cdot(\hat{i}+ \hat{j}- \hat{k})cˉ⋅bˉ=x⋅1+y⋅1+z⋅(−1)\bar{c} \cdot \bar{b} = x\cdot1 + y\cdot1 + z\cdot(-1)cˉ⋅bˉ=x+y−z\bar{c} \cdot \bar{b} = x + y - z
  7. Calculate Dot Product of cc and bb: Now, calculate cˉ⋅bˉ\bar{c}\cdot \bar{b} using the components of cˉ\bar{c}.cˉ⋅bˉ=(xi^+yj^+zk^)⋅(i^+j^−k^)\bar{c}\cdot \bar{b} = (x \hat{i} + y \hat{j} + z \hat{k})\cdot(\hat{i}+ \hat{j}- \hat{k})cˉ⋅bˉ=x⋅1+y⋅1+z⋅(−1)\bar{c}\cdot \bar{b} = x\cdot 1 + y\cdot 1 + z\cdot (-1)cˉ⋅bˉ=x+y−z\bar{c}\cdot \bar{b} = x + y - zWe know bˉ×cˉ=2i^−2j^\bar{b}\times \bar{c} = 2 \hat{i} - 2 \hat{j}, which means the ii and jj components of cˉ\bar{c} must be such that when crossed with bb11, they give bb22. However, we made a mistake in the cross product calculation earlier. Let's correct it.

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