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JEE Main 2024 (6 Apr, Shift 1) · Vector (Cross) Product

Vector (Cross) Product question from JEE Main 2024 (6 Apr, Shift 1)

Q7JEE Main 2024 · 6 APR · SHIFT 1Medium

Vector (Cross) Product

For three vectors A⃗=(−xi^−6j^−2k^), B⃗=(−i^+4j^+3k^)\vec{A} = (-x\hat{i} - 6\hat{j} - 2\hat{k}), \ \vec{B} = (-\hat{i} + 4\hat{j} + 3\hat{k}) and C⃗=(−8i^−j^+3k^)\vec{C} = (-8\hat{i} - \hat{j} + 3\hat{k}), if A⃗⋅(B⃗×C⃗)=0\vec{A} \cdot (\vec{B} \times \vec{C}) = 0, then value of xx is ________.

Numerical answer type. Work it out, then open the answer.

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Answer: 4

  1. B⃗×C⃗=i^[(4)(3)−(3)(−1)]−j^[(−1)(3)−(3)(−8)]+k^[(−1)(−1)−(4)(−8)]\vec{B} \times \vec{C} = \hat{i}[(4)(3) - (3)(-1)] - \hat{j}[(-1)(3) - (3)(-8)] + \hat{k}[(-1)(-1) - (4)(-8)]
  2. =15i^−21j^+33k^= 15\hat{i} - 21\hat{j} + 33\hat{k}
  3. A⃗⋅(B⃗×C⃗)=−15x+126−66=60−15x=0\vec{A} \cdot (\vec{B} \times \vec{C}) = -15x + 126 - 66 = 60 - 15x = 0
  4. So x=4x = 4.
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