How do you find the cross product and verify that the resulting vectors are perpendicular to the given vectors #<1,-3,2>times<5,1,-2>#?

Answer 1

The vector is #=〈4,12,16〉#

The cross product is calculated by the determinant

#vecu.vecv=#
#〈1,-3,2〉#x#〈5,1,-2〉#
#∣ ((i,j,k) , (1,-3,2) , (5,1,-2)) ∣#
#=i(6-2)-j(-2-10)+k(1+15)#
#vecw=〈4,12,16〉#

To verify, we do the dot products

#〈4,12,16〉.〈1,-3,2〉=4-36+32=0#
#〈4,12,16〉.〈5,1,-2〉=20+12-32=0#
Therefore #vecw# is perpendicular to #vecu# and #vecv#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2
To find the cross product of two vectors, you can use the following formula: \[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} \] where \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \) are unit vectors along the x, y, and z axes respectively, and \( A_x, A_y, A_z \) and \( B_x, B_y, B_z \) are the components of vectors \( \mathbf{A} \) and \( \mathbf{B} \) respectively. Using the given vectors \( \mathbf{A} = <1, -3, 2> \) and \( \mathbf{B} = <5, 1, -2> \), you can compute their cross product as follows: \[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -3 & 2 \\ 5 & 1 & -2 \end{vmatrix} \] Expanding the determinant: \[ \mathbf{A} \times \mathbf{B} = \mathbf{i} \begin{vmatrix} -3 & 2 \\ 1 & -2 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 2 \\ 5 & -2 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & -3 \\ 5 & 1 \end{vmatrix} \] \[ \mathbf{A} \times \mathbf{B} = \mathbf{i} \left( (-3)(-2) - (2)(1) \right) - \mathbf{j} \left( (1)(-2) - (2)(5) \right) + \mathbf{k} \left( (1)(1) - (-3)(5) \right) \] \[ \mathbf{A} \times \mathbf{B} = \mathbf{i}(6 - 2) - \mathbf{j}(-2 - 10) + \mathbf{k}(1 + 15) \] \[ \mathbf{A} \times \mathbf{B} = 4\mathbf{i} + 12\mathbf{j} + 16\mathbf{k} \] So, the cross product of \( \mathbf{A} \) and \( \mathbf{B} \) is \( \mathbf{A} \times \mathbf{B} = <4, 12, 16> \). To verify that the resulting vector is perpendicular to the given vectors \( \mathbf{A} \) and \( \mathbf{B} \), you can take the dot product of \( \mathbf{A} \times \mathbf{B} \) with each of \( \mathbf{A} \) and \( \mathbf{B} \). If the dot product is zero, then the vectors are perpendicular. \[ \mathbf{A} \cdot (\mathbf{A} \times \mathbf{B}) = <1, -3, 2> \cdot <4, 12, 16> = (1)(4) + (-3)(12) + (2)(16) = 4 - 36 + 32 = 0 \] \[ \mathbf{B} \cdot (\mathbf{A} \times \mathbf{B}) = <5, 1, -2> \cdot <4, 12, 16> = (5)(4) + (1)(12) + (-2)(16) = 20 + 12 - 32 = 0 \] Since both dot products are zero, \( \mathbf{A} \times \mathbf{B} \) is perpendicular to both \( \mathbf{A} \) and \( \mathbf{B} \).
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7