Answer on Question #46941, Math, Vector Calculus
Let
r 1 = 3 i − 2 j + k , r 2 = 2 i − 4 j − 3 k , r 3 = − i + 2 j + 2 k r _ {1} = 3 i - 2 j + k, r _ {2} = 2 i - 4 j - 3 k, r _ {3} = - i + 2 j + 2 k r 1 = 3 i − 2 j + k , r 2 = 2 i − 4 j − 3 k , r 3 = − i + 2 j + 2 k
Find r 2 ⋅ ( r 1 × r 3 ) r_2 \cdot (r_1 \times r_3) r 2 ⋅ ( r 1 × r 3 )
Solution
r 2 ⋅ ( r 1 × r 3 ) = ( r 2 i r 2 j r 2 k r 1 i r 1 j r 1 k r 3 i r 3 j r 3 k ) r _ {2} \cdot (r _ {1} \times r _ {3}) = \left( \begin{array}{ccc} r _ {2 i} & r _ {2 j} & r _ {2 k} \\ r _ {1 i} & r _ {1 j} & r _ {1 k} \\ r _ {3 i} & r _ {3 j} & r _ {3 k} \end{array} \right) r 2 ⋅ ( r 1 × r 3 ) = ⎝ ⎛ r 2 i r 1 i r 3 i r 2 j r 1 j r 3 j r 2 k r 1 k r 3 k ⎠ ⎞ r 2 ⋅ ( r 1 × r 3 ) = ∣ 2 − 4 − 3 3 − 2 1 − 1 2 2 ∣ = = 2 ⋅ ( − 2 ) ⋅ 2 + ( − 4 ) ⋅ 1 ⋅ ( − 1 ) + ( − 3 ) ⋅ 3 ⋅ 2 − ( − 3 ) ⋅ ( − 2 ) ⋅ ( − 1 ) − ( − 4 ) ⋅ 3 ⋅ 2 − 2 ⋅ 1 ⋅ 2 = = − 8 + 4 − 18 + 6 + 24 − 4 = 4 r _ {2} \cdot (r _ {1} \times r _ {3}) = \left| \begin{array}{ccc} 2 & -4 & -3 \\ 3 & -2 & 1 \\ -1 & 2 & 2 \end{array} \right| = \\
= 2 \cdot (-2) \cdot 2 + (-4) \cdot 1 \cdot (-1) + (-3) \cdot 3 \cdot 2 - (-3) \cdot (-2) \cdot (-1) - (-4) \cdot 3 \cdot 2 - 2 \cdot 1 \cdot 2 = \\
= -8 + 4 - 18 + 6 + 24 - 4 = 4 r 2 ⋅ ( r 1 × r 3 ) = ∣ ∣ 2 3 − 1 − 4 − 2 2 − 3 1 2 ∣ ∣ = = 2 ⋅ ( − 2 ) ⋅ 2 + ( − 4 ) ⋅ 1 ⋅ ( − 1 ) + ( − 3 ) ⋅ 3 ⋅ 2 − ( − 3 ) ⋅ ( − 2 ) ⋅ ( − 1 ) − ( − 4 ) ⋅ 3 ⋅ 2 − 2 ⋅ 1 ⋅ 2 = = − 8 + 4 − 18 + 6 + 24 − 4 = 4
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