Answer on Question #85462, Physics / Other
If vector A = A x i + A y j + A z k \mathbf{A} = A_{x}\mathbf{i} + A_{y}\mathbf{j} + A_{z}\mathbf{k} A = A x i + A y j + A z k , B = B x i + B y j + B z k \mathbf{B} = B_{x}\mathbf{i} + B_{y}\mathbf{j} + B_{z}\mathbf{k} B = B x i + B y j + B z k and C = C x i + C y j + C z k \mathbf{C} = C_{x}\mathbf{i} + C_{y}\mathbf{j} + C_{z}\mathbf{k} C = C x i + C y j + C z k .
Show that the scalar triple product of the vector is [ A x A y A z ] [ B x B y B z ] [ C x C y C z ] [A_x A_y A_z][B_x B_y B_z ][C_x C_y C_z] [ A x A y A z ] [ B x B y B z ] [ C x C y C z ] .
Solution:
The vector product by definition
[ B × C ] = ∣ i j k B x B y B z C x C y C z ∣ = i ( B y C z − B z C y ) + j ( B z C x − B x C z ) + k ( B x C y − B y C x ) [ \mathbf {B} \times \mathbf {C} ] = \left| \begin{array}{c c c} \mathbf {i} & \mathbf {j} & \mathbf {k} \\ B _ {x} & B _ {y} & B _ {z} \\ C _ {x} & C _ {y} & C _ {z} \end{array} \right| = \mathbf {i} \big (B _ {y} C _ {z} - B _ {z} C _ {y} \big) + \mathbf {j} (B _ {z} C _ {x} - B _ {x} C _ {z}) + \mathbf {k} (B _ {x} C _ {y} - B _ {y} C _ {x}) [ B × C ] = ∣ ∣ i B x C x j B y C y k B z C z ∣ ∣ = i ( B y C z − B z C y ) + j ( B z C x − B x C z ) + k ( B x C y − B y C x )
The scalar product by definition
( A ⋅ B ) = A x B x + A y B y + A z B z (\mathbf {A} \cdot \mathbf {B}) = A _ {x} B _ {x} + A _ {y} B _ {y} + A _ {z} B _ {z} ( A ⋅ B ) = A x B x + A y B y + A z B z
So, for the triple product we obtain
( A ⋅ [ B × C ] ) = A x ( B y C z − B z C y ) + A y ( B z C x − B x C z ) + A z ( B x C y − B y C x ) = ∣ A x A y A z B x B y B z C x C y C z ∣ \begin{array}{l}
(\mathbf {A} \cdot [ \mathbf {B} \times \mathbf {C} ]) = A _ {x} \big (B _ {y} C _ {z} - B _ {z} C _ {y} \big) + A _ {y} (B _ {z} C _ {x} - B _ {x} C _ {z}) + A _ {z} \big (B _ {x} C _ {y} - B _ {y} C _ {x} \big) \\
= \left| \begin{array}{c c c} A _ {x} & A _ {y} & A _ {z} \\ B _ {x} & B _ {y} & B _ {z} \\ C _ {x} & C _ {y} & C _ {z} \end{array} \right| \\
\end{array} ( A ⋅ [ B × C ]) = A x ( B y C z − B z C y ) + A y ( B z C x − B x C z ) + A z ( B x C y − B y C x ) = ∣ ∣ A x B x C x A y B y C y A z B z C z ∣ ∣
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