Question Statement We are tasked with verifying the following vector identity for three vectors a \mathbf{a} a b \mathbf{b} b c \mathbf{c} c
a ⋅ ( b × c ) = b ⋅ ( c × a ) = c ⋅ ( a × b ) \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) a ⋅ ( b × c ) = b ⋅ ( c × a ) = c ⋅ ( a × b ) Given the vectors:
a = 3 i ^ − j ^ + 5 k ^ , b = 4 i ^ + 3 j ^ − 2 k ^ , c = 2 i ^ + 5 j ^ + k ^ \mathbf{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}, \quad \mathbf{b} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}, \quad \mathbf{c} = 2 \hat{i} + 5 \hat{j} + \hat{k} a = 3 i ^ − j ^ + 5 k ^ , b = 4 i ^ + 3 j ^ − 2 k ^ , c = 2 i ^ + 5 j ^ + k ^ We need to verify this identity by calculating the scalar triple product for each permutation of the vectors.
Background and Explanation The scalar triple product is a fundamental operation in vector algebra. It involves a dot product and a cross product, and can be represented as:
a ⋅ ( b × c ) \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) a ⋅ ( b × c ) This product gives the volume of the parallelepiped formed by the vectors a \mathbf{a} a b \mathbf{b} b c \mathbf{c} c
a ⋅ ( b × c ) = b ⋅ ( c × a ) = c ⋅ ( a × b ) \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) a ⋅ ( b × c ) = b ⋅ ( c × a ) = c ⋅ ( a × b ) We will verify this property by explicitly calculating each of the three scalar triple products.
Solution We will compute each side of the equation separately by calculating the determinants for the cross products.
1. Compute a ⋅ ( b × c ) \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) a ⋅ ( b × c )
Given the vectors:
a = 3 i ^ − j ^ + 5 k ^ , b = 4 i ^ + 3 j ^ − 2 k ^ , c = 2 i ^ + 5 j ^ + k ^ \mathbf{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}, \quad \mathbf{b} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}, \quad \mathbf{c} = 2 \hat{i} + 5 \hat{j} + \hat{k} a = 3 i ^ − j ^ + 5 k ^ , b = 4 i ^ + 3 j ^ − 2 k ^ , c = 2 i ^ + 5 j ^ + k ^ We calculate the determinant:
a ⋅ ( b × c ) = ∣ 3 − 1 54 3 − 22 5 1 ∣ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \left| \begin{array}{ccc}
3 & -1 & 5
4 & 3 & -2
2 & 5 & 1
\end{array} \right| a ⋅ ( b × c ) = 3 − 1 54 3 − 22 5 1 Now expand this determinant along the first row:
= 3 ( ∣ 3 − 25 1 ∣ ) − ( − 1 ) ( ∣ 4 − 22 1 ∣ ) + 5 ( ∣ 4 32 5 ∣ ) = 3 \left( \begin{vmatrix} 3 & -2 5 & 1 \end{vmatrix} \right) - (-1) \left( \begin{vmatrix} 4 & -2 2 & 1 \end{vmatrix} \right) + 5 \left( \begin{vmatrix} 4 & 3 2 & 5 \end{vmatrix} \right) = 3 ( 3 − 25 1 ) − ( − 1 ) ( 4 − 22 1 ) + 5 ( 4 32 5 ) Calculating each 2x2 determinant:
∣ 3 − 25 1 ∣ = ( 3 ) ( 1 ) − ( − 2 ) ( 5 ) = 3 + 10 = 13 \begin{vmatrix} 3 & -2 5 & 1 \end{vmatrix} = (3)(1) - (-2)(5) = 3 + 10 = 13 3 − 25 1 = ( 3 ) ( 1 ) − ( − 2 ) ( 5 ) = 3 + 10 = 13 ∣ 4 − 22 1 ∣ = ( 4 ) ( 1 ) − ( − 2 ) ( 2 ) = 4 + 4 = 8 \begin{vmatrix} 4 & -2 2 & 1 \end{vmatrix} = (4)(1) - (-2)(2) = 4 + 4 = 8 4 − 22 1 = ( 4 ) ( 1 ) − ( − 2 ) ( 2 ) = 4 + 4 = 8 ∣ 4 32 5 ∣ = ( 4 ) ( 5 ) − ( 3 ) ( 2 ) = 20 − 6 = 14 \begin{vmatrix} 4 & 3 2 & 5 \end{vmatrix} = (4)(5) - (3)(2) = 20 - 6 = 14 4 32 5 = ( 4 ) ( 5 ) − ( 3 ) ( 2 ) = 20 − 6 = 14 Thus, we have:
= 3 ( 13 ) + 1 ( 8 ) + 5 ( 14 ) = 39 + 8 + 70 = 117 cubic units (1) = 3(13) + 1(8) + 5(14) = 39 + 8 + 70 = 117 \text{ cubic units} \tag{1} = 3 ( 13 ) + 1 ( 8 ) + 5 ( 14 ) = 39 + 8 + 70 = 117 cubic units ( 1 ) 2. Compute b ⋅ ( c × a ) \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) b ⋅ ( c × a )
Now we calculate the scalar triple product for the second permutation of the vectors:
b ⋅ ( c × a ) = ∣ 4 3 − 22 5 13 − 1 5 ∣ \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \left| \begin{array}{ccc}
4 & 3 & -2
2 & 5 & 1
3 & -1 & 5
\end{array} \right| b ⋅ ( c × a ) = 4 3 − 22 5 13 − 1 5 Expand this determinant along the first row:
= 4 ( ∣ 5 1 − 1 5 ∣ ) − 3 ( ∣ 2 13 5 ∣ ) + ( − 2 ) ( ∣ 2 53 − 1 ∣ ) = 4 \left( \begin{vmatrix} 5 & 1 -1 & 5 \end{vmatrix} \right) - 3 \left( \begin{vmatrix} 2 & 1 3 & 5 \end{vmatrix} \right) + (-2) \left( \begin{vmatrix} 2 & 5 3 & -1 \end{vmatrix} \right) = 4 ( 5 1 − 1 5 ) − 3 ( 2 13 5 ) + ( − 2 ) ( 2 53 − 1 ) Calculating each 2x2 determinant:
∣ 5 1 − 1 5 ∣ = ( 5 ) ( 5 ) − ( 1 ) ( − 1 ) = 25 + 1 = 26 \begin{vmatrix} 5 & 1 -1 & 5 \end{vmatrix} = (5)(5) - (1)(-1) = 25 + 1 = 26 5 1 − 1 5 = ( 5 ) ( 5 ) − ( 1 ) ( − 1 ) = 25 + 1 = 26 ∣ 2 13 5 ∣ = ( 2 ) ( 5 ) − ( 1 ) ( 3 ) = 10 − 3 = 7 \begin{vmatrix} 2 & 1 3 & 5 \end{vmatrix} = (2)(5) - (1)(3) = 10 - 3 = 7 2 13 5 = ( 2 ) ( 5 ) − ( 1 ) ( 3 ) = 10 − 3 = 7 ∣ 2 53 − 1 ∣ = ( 2 ) ( − 1 ) − ( 5 ) ( 3 ) = − 2 − 15 = − 17 \begin{vmatrix} 2 & 5 3 & -1 \end{vmatrix} = (2)(-1) - (5)(3) = -2 - 15 = -17 2 53 − 1 = ( 2 ) ( − 1 ) − ( 5 ) ( 3 ) = − 2 − 15 = − 17 Thus, we have:
= 4 ( 26 ) − 3 ( 7 ) + ( − 2 ) ( − 17 ) = 104 − 21 + 34 = 117 cubic units (2) = 4(26) - 3(7) + (-2)(-17) = 104 - 21 + 34 = 117 \text{ cubic units} \tag{2} = 4 ( 26 ) − 3 ( 7 ) + ( − 2 ) ( − 17 ) = 104 − 21 + 34 = 117 cubic units ( 2 ) 3. Compute c ⋅ ( a × b ) \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) c ⋅ ( a × b )
Finally, we calculate the scalar triple product for the third permutation of the vectors:
c ⋅ ( a × b ) = ∣ 2 5 13 − 1 54 3 − 2 ∣ \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) = \left| \begin{array}{ccc}
2 & 5 & 1
3 & -1 & 5
4 & 3 & -2
\end{array} \right| c ⋅ ( a × b ) = 2 5 13 − 1 54 3 − 2 Expand this determinant along the first row:
= 2 ( ∣ − 1 53 − 2 ∣ ) − 5 ( ∣ 3 54 − 2 ∣ ) + 1 ( ∣ 3 − 14 3 ∣ ) = 2 \left( \begin{vmatrix} -1 & 5 3 & -2 \end{vmatrix} \right) - 5 \left( \begin{vmatrix} 3 & 5 4 & -2 \end{vmatrix} \right) + 1 \left( \begin{vmatrix} 3 & -1 4 & 3 \end{vmatrix} \right) = 2 ( − 1 53 − 2 ) − 5 ( 3 54 − 2 ) + 1 ( 3 − 14 3 ) Calculating each 2x2 determinant:
∣ − 1 53 − 2 ∣ = ( − 1 ) ( − 2 ) − ( 5 ) ( 3 ) = 2 − 15 = − 13 \begin{vmatrix} -1 & 5 3 & -2 \end{vmatrix} = (-1)(-2) - (5)(3) = 2 - 15 = -13 − 1 53 − 2 = ( − 1 ) ( − 2 ) − ( 5 ) ( 3 ) = 2 − 15 = − 13 ∣ 3 54 − 2 ∣ = ( 3 ) ( − 2 ) − ( 5 ) ( 4 ) = − 6 − 20 = − 26 \begin{vmatrix} 3 & 5 4 & -2 \end{vmatrix} = (3)(-2) - (5)(4) = -6 - 20 = -26 3 54 − 2 = ( 3 ) ( − 2 ) − ( 5 ) ( 4 ) = − 6 − 20 = − 26 ∣ 3 − 14 3 ∣ = ( 3 ) ( 3 ) − ( − 1 ) ( 4 ) = 9 + 4 = 13 \begin{vmatrix} 3 & -1 4 & 3 \end{vmatrix} = (3)(3) - (-1)(4) = 9 + 4 = 13 3 − 14 3 = ( 3 ) ( 3 ) − ( − 1 ) ( 4 ) = 9 + 4 = 13 Thus, we have:
= 2 ( − 13 ) − 5 ( − 26 ) + 1 ( 13 ) = − 26 + 130 + 13 = 117 cubic units (3) = 2(-13) - 5(-26) + 1(13) = -26 + 130 + 13 = 117 \text{ cubic units} \tag{3} = 2 ( − 13 ) − 5 ( − 26 ) + 1 ( 13 ) = − 26 + 130 + 13 = 117 cubic units ( 3 ) a ⋅ ( b × c ) = Determinant of the matrix formed by the vectors \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \text{Determinant of the matrix formed by the vectors} a ⋅ ( b × c ) = Determinant of the matrix formed by the vectors
The property of cyclic invariance of the scalar triple product:
a ⋅ ( b × c ) = b ⋅ ( c × a ) = c ⋅ ( a × b ) \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}) = \mathbf{c} \cdot (\mathbf{a} \times \mathbf{b}) a ⋅ ( b × c ) = b ⋅ ( c × a ) = c ⋅ ( a × b ) Summary of Steps
Write down the given vectors a \mathbf{a} a b \mathbf{b} b c \mathbf{c} c
Set up the determinant for each scalar triple product.
Expand each determinant along the first row.
Compute the 2x2 minors for each determinant.
Sum the terms to find the value of each scalar triple product.
Verify that the three scalar triple products are equal.
