Use the triple scalar product to find the volume of the parallelepiped having adjacent edges u, v, and w. u = 1, 3, 1 v = 0, 6,
6 w = −4, 0, −4
1 answer:
Triple scalar product is applicable and was derived originally for determining the volume of a parallelpiped. On a 3-dimensional plane, there are three coordinate axes, usually denoted as i, j and k. Such that side u has component vectors of ui, uj and uk.
According to the triple scalar product, you can find the magnitude of the three-component vector by applying the dot product of one variable, together with a cross product of the other two variables. It is written as u·(v×w). In other solution, the magnitude is simply the determinant of the matrix of the component vectors. The matrix is formed as shown in the picture. The upper table acts as the guide for the second table.
Determinant = ∑(product of numbers along diagonal arrow down) - ∑(product of numbers along diagonal arrow up)
Determinant = [(1*6*-4)+(3*6*-4)+(1*0*0)] - [(-4*6*1)+(0*6*1)+(-4*0*3)]
Determinant = -72
Take the absolute value, then this will be the value for the volume. Thus, the parallelpiped has a volume of 72 cubic units.
According to the triple scalar product, you can find the magnitude of the three-component vector by applying the dot product of one variable, together with a cross product of the other two variables. It is written as u·(v×w). In other solution, the magnitude is simply the determinant of the matrix of the component vectors. The matrix is formed as shown in the picture. The upper table acts as the guide for the second table.
Determinant = ∑(product of numbers along diagonal arrow down) - ∑(product of numbers along diagonal arrow up)
Determinant = [(1*6*-4)+(3*6*-4)+(1*0*0)] - [(-4*6*1)+(0*6*1)+(-4*0*3)]
Determinant = -72
Take the absolute value, then this will be the value for the volume. Thus, the parallelpiped has a volume of 72 cubic units.
You might be interested in