Question Video: Finding the Cross Product of Vectors Mathematics

If 𝐀 = ⟨−5, 0, 1⟩ and 𝐁 = ⟨3, 1, −3⟩, find 𝐀 × (𝐀 − 2𝐁).

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Video Transcript

If vector 𝐀 is equal to negative five, zero, one and vector 𝐁 is equal to three, one, negative three, find 𝐀 cross 𝐀 minus two 𝐁.

In this question, we need to find the cross product of two vectors. We recall that when finding the cross product, the answer is a vector that is perpendicular to the original two vectors. Given two three-dimensional vectors 𝐂 and 𝐃, their cross product is equal to the determinant of the three-by-three matrix shown, where the top row is made up of the unit vectors 𝐢, 𝐣, and 𝐤. The second row is made up of the components of vector 𝐂, denoted 𝐂 sub 𝑥, 𝐂 sub 𝑦, and 𝐂 sub 𝑧. The bottom row of the matrix is made up of the components of vector 𝐃.

In this question, we need to find the cross product of vector 𝐀 and the vector 𝐀 minus two 𝐁. We can begin by working out the vector two 𝐁. We do this by multiplying vector 𝐁 by the scalar or constant two. Multiplying each of the components by the scalar two gives us the vector six, two, negative six.

We need to subtract this vector from vector 𝐀. When subtracting two vectors, we simply subtract the corresponding components, in this case negative five minus six, zero minus two, and one minus negative six. This gives us the vector negative 11, negative two, seven.

We can now find the cross product of vector 𝐀 and the vector 𝐀 minus two 𝐁. This will be equal to the determinant of the three-by-three matrix 𝐢, 𝐣, 𝐤, negative five, zero, one, negative 11, negative two, seven. We calculate the determinant of a three-by-three matrix in three parts. Firstly, we multiply the unit vector 𝐢 by the determinant of the two-by-two matrix zero, one, negative two, seven. This gives us the unit vector 𝐢 multiplied by zero minus negative two, which in turn is equal to two 𝐢.

Next, we multiply the negative of the unit vector 𝐣 by the determinant of the two-by-two matrix negative five, one, negative 11, seven. This is equal to negative 𝐣 multiplied by negative 35 minus negative 11, which is equal to 24𝐣.

Finally, we multiply the unit vector 𝐤 by the determinant of the two-by-two matrix negative five, zero, negative 11, negative two. This is equal to 10𝐤.

The determinant of our three-by-three matrix is two 𝐢 plus 24𝐣 plus 10𝐤. We can therefore conclude that this is the cross product of vector 𝐀 and the vector 𝐀 minus two 𝐁. Our final answer is two 𝐢 plus 24𝐣 plus 10𝐤. This could also be written in component form: two, 24, 10.

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