Answer on Question #52361 – Math – Vector Calculus
Question
For the following vectors
a = ( 2 , 3 , 5 ) , b = ( 4 , 6 , 8 ) , c = ( − 1 , − 5 , 9 ) a = (2, 3, 5), b = (4, 6, 8), c = (-1, -5, 9) a = ( 2 , 3 , 5 ) , b = ( 4 , 6 , 8 ) , c = ( − 1 , − 5 , 9 )
Calculate
a ⋅ ( b × c ) a \cdot (b \times c) a ⋅ ( b × c )
Solution
Method 1
The cross (or vector) product is
b × c = ∣ i j k 4 6 8 − 1 − 5 9 ∣ = ∣ 6 8 − 5 9 ∣ i − ∣ 4 8 − 1 9 ∣ j + ∣ 4 6 − 1 − 5 ∣ k = ( 6 ⋅ 9 − ( − 5 ) ⋅ 8 ) i − ( − 4 ⋅ 9 − ( − 1 ) ⋅ 8 ) j + ( 4 ⋅ ( − 5 ) − ( − 1 ) ⋅ 6 ) k = 94 i − 44 j − 14 k = ( 94 , − 44 , − 14 ) b \times c = \left| \begin{array}{ccc} i & j & k \\ 4 & 6 & 8 \\ -1 & -5 & 9 \end{array} \right| = \left| \begin{array}{cc} 6 & 8 \\ -5 & 9 \end{array} \right| i - \left| \begin{array}{cc} 4 & 8 \\ -1 & 9 \end{array} \right| j + \left| \begin{array}{cc} 4 & 6 \\ -1 & -5 \end{array} \right| k = (6 \cdot 9 - (-5) \cdot 8) i - (-4 \cdot 9 - (-1) \cdot 8) j + (4 \cdot (-5) - (-1) \cdot 6) k = 94i - 44j - 14k = (94, -44, -14) b × c = ∣ ∣ i 4 − 1 j 6 − 5 k 8 9 ∣ ∣ = ∣ ∣ 6 − 5 8 9 ∣ ∣ i − ∣ ∣ 4 − 1 8 9 ∣ ∣ j + ∣ ∣ 4 − 1 6 − 5 ∣ ∣ k = ( 6 ⋅ 9 − ( − 5 ) ⋅ 8 ) i − ( − 4 ⋅ 9 − ( − 1 ) ⋅ 8 ) j + ( 4 ⋅ ( − 5 ) − ( − 1 ) ⋅ 6 ) k = 94 i − 44 j − 14 k = ( 94 , − 44 , − 14 )
The dot (or scalar) product of vectors a a a b × c b \times c b × c
a ⋅ ( b × c ) = ( 2 , 3 , 5 ) ⋅ ( 94 , − 44 , − 14 ) = 2 ⋅ 94 − 3 ⋅ 44 − 5 ⋅ 14 = − 14 a \cdot (b \times c) = (2, 3, 5) \cdot (94, -44, -14) = 2 \cdot 94 - 3 \cdot 44 - 5 \cdot 14 = -14 a ⋅ ( b × c ) = ( 2 , 3 , 5 ) ⋅ ( 94 , − 44 , − 14 ) = 2 ⋅ 94 − 3 ⋅ 44 − 5 ⋅ 14 = − 14
Method 2
The scalar triple product is
a ⋅ ( b × c ) = ∣ a x a y a z b x b y b z c x c y c z ∣ = ∣ 2 3 5 4 6 8 − 1 − 5 9 ∣ = 2 ∣ 6 8 − 5 9 ∣ − 3 ∣ 4 8 − 1 9 ∣ + 5 ∣ 4 6 − 1 − 5 ∣ = 2 ( 6 ⋅ 9 − ( − 5 ) ⋅ 8 ) − 3 ( 4 ⋅ 9 − ( − 1 ) ⋅ 8 ) + 5 ⋅ ( 4 ⋅ ( − 5 ) − ( − 1 ) ⋅ 6 ) = 2 ⋅ ( 54 + 40 ) − 3 ( 36 + 8 ) + 5 ( − 20 + 6 ) = 2 ⋅ 94 − 3 ⋅ 44 − 5 ⋅ 14 = − 14. a \cdot (b \times c) = \left| \begin{array}{ccc} a_x & a_y & a_z \\ b_x & b_y & b_z \\ c_x & c_y & c_z \end{array} \right| = \left| \begin{array}{ccc} 2 & 3 & 5 \\ 4 & 6 & 8 \\ -1 & -5 & 9 \end{array} \right| = 2 \left| \begin{array}{cc} 6 & 8 \\ -5 & 9 \end{array} \right| - 3 \left| \begin{array}{cc} 4 & 8 \\ -1 & 9 \end{array} \right| + 5 \left| \begin{array}{cc} 4 & 6 \\ -1 & -5 \end{array} \right| = 2 (6 \cdot 9 - (-5) \cdot 8) - 3 (4 \cdot 9 - (-1) \cdot 8) + 5 \cdot (4 \cdot (-5) - (-1) \cdot 6) = 2 \cdot (54 + 40) - 3 (36 + 8) + 5 (-20 + 6) = 2 \cdot 94 - 3 \cdot 44 - 5 \cdot 14 = -14. a ⋅ ( b × c ) = ∣ ∣ a x b x c x a y b y c y a z b z c z ∣ ∣ = ∣ ∣ 2 4 − 1 3 6 − 5 5 8 9 ∣ ∣ = 2 ∣ ∣ 6 − 5 8 9 ∣ ∣ − 3 ∣ ∣ 4 − 1 8 9 ∣ ∣ + 5 ∣ ∣ 4 − 1 6 − 5 ∣ ∣ = 2 ( 6 ⋅ 9 − ( − 5 ) ⋅ 8 ) − 3 ( 4 ⋅ 9 − ( − 1 ) ⋅ 8 ) + 5 ⋅ ( 4 ⋅ ( − 5 ) − ( − 1 ) ⋅ 6 ) = 2 ⋅ ( 54 + 40 ) − 3 ( 36 + 8 ) + 5 ( − 20 + 6 ) = 2 ⋅ 94 − 3 ⋅ 44 − 5 ⋅ 14 = − 14.
Answer: -14.
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#339914 on Dec 2023