Questions: Calculate the cross product. [ (5 mathbfi-3 mathbfj+4 mathbfk) times(mathbfi+mathbfj-5 mathbfk)=a mathbfi+b mathbfj+c mathbfk ] (Give your answer using component form or standard basis vectors. Express numbers in exact form. Use symbolic notation and fractions where needed.) [ langle a, b, crangle= ]

Solution Steps

To calculate the cross product of two vectors, we use the determinant of a 3x3 matrix. The first row of the matrix consists of the unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\). The second row contains the components of the first vector, and the third row contains the components of the second vector. The cross product is then determined by calculating the determinant of this matrix.

We are given two vectors: \[ \mathbf{A} = 5\mathbf{i} - 3\mathbf{j} + 4\mathbf{k} \] \[ \mathbf{B} = \mathbf{i} + \mathbf{j} - 5\mathbf{k} \]

The cross product \(\mathbf{A} \times \mathbf{B}\) is calculated using the determinant of the following matrix: \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & -3 & 4 \\ 1 & 1 & -5 \end{vmatrix} \]

The cross product is given by: \[ \mathbf{A} \times \mathbf{B} = \left( (-3)(-5) - (4)(1) \right)\mathbf{i} - \left( (5)(-5) - (4)(1) \right)\mathbf{j} + \left( (5)(1) - (-3)(1) \right)\mathbf{k} \]

Calculating each component:

• The \(\mathbf{i}\) component: \((-3)(-5) - (4)(1) = 15 - 4 = 11\)
• The \(\mathbf{j}\) component: \(-( (5)(-5) - (4)(1) ) = -(-25 - 4) = 29\)
• The \(\mathbf{k}\) component: \((5)(1) - (-3)(1) = 5 + 3 = 8\)

Final Answer