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Exercise 6.2 · Q1

Q.If a⃗=i^−2j^+3k^, b⃗=2i^+j^−2k^, c⃗=3i^+2j^+k^\vec a=\hat i-2\hat j+3\hat k,\ \vec b=2\hat i+\hat j-2\hat k,\ \vec c=3\hat i+2\hat j+\hat k, find a⃗⋅(b⃗×c⃗)\vec a\cdot(\vec b\times\vec c).

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✓ Free question

The scalar triple product a⃗⋅(b⃗×c⃗)\vec a\cdot(\vec b\times\vec c) is exactly the determinant of the 3×33\times3 matrix whose rows are a⃗,b⃗,c⃗\vec a,\vec b,\vec c.

Step 1. Compute b⃗×c⃗\vec b\times\vec c.

b⃗×c⃗=∣i^j^k^21−2321∣=i^(1(1)−(−2)(2))−j^(2(1)−(−2)(3))+k^(2(2)−1(3))=5i^−8j^+k^.\vec b\times\vec c=\begin{vmatrix}\hat i&\hat j&\hat k\\ 2&1&-2\\ 3&2&1\end{vmatrix}=\hat i\big(1(1)-(-2)(2)\big)-\hat j\big(2(1)-(-2)(3)\big)+\hat k\big(2(2)-1(3)\big)=5\hat i-8\hat j+\hat k.

Step 2. Dot with a⃗\vec a.

a⃗⋅(b⃗×c⃗)=(1)(5)+(−2)(−8)+(3)(1)=5+16+3=24.\vec a\cdot(\vec b\times\vec c)=(1)(5)+(-2)(-8)+(3)(1)=5+16+3=24.

✓Final answer

a⃗⋅(b⃗×c⃗)=24\vec a\cdot(\vec b\times\vec c)=\boxed{24}.

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