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

# Question Video: Finding the Cross Product of Vectors Mathematics

If π = β©3, 4, β4βͺ, π = β©2, 5, β4βͺ, and π = β©β4, β4, 2βͺ, find (π β π ) Γ (π β π).

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### Video Transcript

If vector π equals three, four, negative four; vector π equals two, five, negative four; and vector π equals negative four, negative four, two, find π minus π cross π minus π.

Okay, so here we have these three vectors π, π, and π written out by their components. So, for example, considering vector π, this is its π’-component, this is its π£-component, and this is its π€-component. Our goal is to calculate this cross product, where the two vectors involved are differences between two of our π, π, and π vectors.

What we can do here is give these differences their own names. Weβll call π minus π vector π, and π minus π weβll call vector π. What weβll eventually do then is cross π and π. But before that, weβll need to figure out what each of these vectors are. Starting with vector π, we know that thatβs equal to vector π, which is three, four, negative four, minus vector π, which is two, five, negative four. Working component by component, three minus two is one, four minus five is negative one, and negative four minus negative four is zero. So thatβs vector π.

And vector π is equal to vector π here minus vector π here. Negative four minus three is negative seven, negative four minus four is negative eight, and then two minus negative four is positive six. So we now have the two vectors π and π. And weβll write them off to the side that will combine in our cross product. To do that, letβs recall this rule for the three-dimensional cross product of two vectors. Weβve called them π and π here. This cross product equals the determinant of a three-by-three matrix.

In the first row, we have our unit vectors, in the second the corresponding components of vector π, and in the third those of vector π. So when we set up the cross product of π and π, weβll once again have π’, π£, and π€ unit vectors at the top of each column. And in the next row, weβll write in the components of vector π. Those are one, negative one, and zero. Lastly, the components of vector π: negative seven, negative eight, positive six.

Computing this cross product is now a matter of calculating the determinant of this matrix. Beginning with the π’-component, if we cross out the row and column that contain this element, then this component is equal to the determinant of the two-by-two matrix remaining. This is equal to negative one times six or negative six minus zero times negative eight, or zero. So the π’-component of our resulting vector is negative six.

And then we move on to calculate the π£-component. This has a value of negative one times one times six, or six, minus zero times negative seven, or zero. This comes out to negative six π£. And lastly, for our π€-component, this is equal to one times negative eight, or negative eight, minus negative one times negative seven, which comes out to negative seven. Our π€-component then simplifies to negative 15.

And now weβve computed all three components of a resulting cross product. The vector we end up with is negative six π’ minus six π£ minus 15π€. Going back to our original three vectors π, π, and π, this equals π minus π cross π minus π.