Let a⃗=î+2 ĵ+k̂, b⃗=3 î-3 ĵ+3 k̂, c⃗=2 î-ĵ+2 k̂ and d⃗ be a vector such that b⃗ × d⃗=c⃗ × d⃗ and a⃗ · d⃗=4. Then |(a⃗ × d⃗)|² is equal to___________.

Solution

Maths sir · 4 steps

Subtract: (b⃗-c⃗)×d⃗=0⃗, so d⃗ is parallel to b⃗-c⃗.

1. Direction of d. From b⃗×d⃗=c⃗×d⃗ we get (b⃗-c⃗)×d⃗=0⃗. So d⃗ is parallel to b⃗-c⃗=î-2ĵ+k̂. Write d⃗=λ(î-2ĵ+k̂).
2. Fix lambda. a⃗·d⃗=λ(1-4+1)=-2λ=4, so λ=-2. Then d⃗=-2î+4ĵ-2k̂ and |d⃗|²=4+16+4=24.

   |a⃗×d⃗|²=|a⃗|²|d⃗|²-(a⃗·d⃗)²

3. Put in numbers. |a⃗|²=1+4+1=6, so |a⃗×d⃗|²=6× 24-4²=144-16.
4. So. |a⃗×d⃗|²= 128.

The common slip: 144. Forgetting to subtract (a⃗·d⃗)² in Lagrange's identity gives 6× 24=144, not 128.

The idea to keep. When two cross products with the same vector are equal, subtract them: a zero cross product means parallel vectors. And Lagrange's identity gives |a⃗×d⃗|² without computing the cross product.

Answer from the NTA final answer key. Solution by PrepWiser.