Question Video: Finding the Norm of the Cross Product of Vectors | Nagwa Question Video: Finding the Norm of the Cross Product of Vectors | Nagwa

# Question Video: Finding the Norm of the Cross Product of Vectors Mathematics • Third Year of Secondary School

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If 𝐀 = −10𝐣 + 5𝐤 and 𝐁 = −4𝐢 + 9𝐣 + 𝐤, find |5𝐁 × 𝐀|.

04:35

### Video Transcript

If vector 𝐀 is equal to negative 10𝐣 plus five 𝐤 and vector 𝐁 is equal to negative four 𝐢 plus nine 𝐣 plus 𝐤, find the magnitude of five 𝐁 cross 𝐀.

We recall that when calculating the cross product of two vectors in three dimensions, our answer is also a vector that is perpendicular to the original two vectors. The cross product of two vectors 𝐂 and 𝐃 is equal to the determinant of the three-by-three matrix shown where the top row contains 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, and the bottom row is made up of the components of vector 𝐃.

In this question, we need to find the cross product of five multiplied by vector 𝐁 and vector 𝐀. We can begin by rewriting vectors 𝐀 and 𝐁 in terms of their components. Vector 𝐀 is equal to zero, negative 10, five. The first component is equal to zero as there is no 𝐢-component in vector 𝐀. Vector 𝐁 is equal to negative four, nine, one. We can now use this to calculate five 𝐁. When multiplying any vector by a scalar, we simply multiply each of the individual components by that scalar. Five multiplied by negative four is negative 20. Five multiplied by nine is 45. And five multiplied by one is equal to five. The vector five 𝐁 is therefore equal to negative 20, 45, five.

We can now calculate the cross product of five 𝐁 and 𝐀. This is equal to the determinant of the three-by-three matrix shown. We can calculate this determinant in three steps. Firstly, we multiply the unit vector 𝐢 by the determinant of the two-by-two matrix 45, five, negative 10, five. This is equal to 𝐢 multiplied by 225 minus negative 50, which in turn simplifies to 275𝐢.

Next, we multiply the negative of the unit vector 𝐣 by the determinant of the two-by-two matrix negative 20, five, zero, five. This is equal to negative 𝐣 multiplied by negative 100 minus zero, which equals 100𝐣. Finally, we multiply the unit vector 𝐤 by the determinant of the two-by-two matrix negative 20, 45, zero, negative 10. This is equal to 200𝐤. The cross product of five 𝐁 and 𝐀 is 275𝐢 plus 100𝐣 plus 200𝐤. This, in turn, can be written in terms of its components: 275, 100, 200.

This is not the final answer, however, as we need to find the magnitude of this vector. Recalling that the magnitude of any vector is equal to the square root of the sum of the squares of its individual components, this is equal to the square root of 275 squared plus 100 squared plus 200 squared. Typing this into our calculator and keeping our answer in radical or surd form, we have 25 root 201. If vector 𝐀 is equal to negative 10𝐣 plus five 𝐤 and vector 𝐁 is equal to negative four 𝐢 plus nine 𝐣 plus 𝐤, then the magnitude of the cross product of five 𝐁 and 𝐀 is 25 multiplied by the square root of 201.

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