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MHT CET 2025 (26 Apr, Shift 1) · Vector (Cross) Product

Vector (Cross) Product question from MHT CET 2025 (26 Apr, Shift 1)

Q4MHT CET 2025 · 26 APR · SHIFT 1Easy

Vector (Cross) Product

If ∣a⃗∣=26, ∣b⃗∣=7|\vec{a}| = \sqrt{26}, \ |\vec{b}| = 7

∣a⃗×b⃗∣=35|\vec{a} \times \vec{b}| = 35, find a⃗⋅b⃗\vec{a} \cdot \vec{b}

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Answer: (d)

  1. ∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=a2b2|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = a^2b^2
  2. (a⃗⋅b⃗)2=26×49−352=1274−1225=49(\vec{a} \cdot \vec{b})^2 = 26 \times 49 - 35^2 = 1274 - 1225 = 49
  3. a⃗⋅b⃗=7\vec{a} \cdot \vec{b} = 7, answer (d).
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