Question Video: Finding the Cross Product of Vectors Mathematics

Given that ๐€ = โŸจ9, โˆ’5, โˆ’1โŸฉ, ๐ = โŸจ7, โˆ’๐‘˜, โˆ’5โŸฉ, ๐‚ = โŸจ10, โˆ’55, ๐‘š โˆ’ 3โŸฉ, and ๐€๐ โˆฅ ๐‚, find ๐‘˜ โˆ’ ๐‘š.


Video Transcript

Given that vector ๐€ equals nine, negative five, negative one; vector ๐ equals seven, negative ๐‘˜, negative five; vector ๐‚ equals 10, negative 55, ๐‘š minus three; and vector ๐€๐ is parallel to vector ๐‚, find ๐‘˜ minus ๐‘š.

In these three vectors weโ€™re given ๐€, ๐, and ๐‚, two of these vectors, ๐ and ๐‚, have unknown components. The ๐‘ฆ-component of vector ๐ is negative ๐‘˜ and the ๐‘ง-component of vector ๐‚ is ๐‘š minus three. We donโ€™t know the values of either ๐‘˜ or ๐‘š, but we do know that this vector ๐€๐ is parallel to our vector ๐‚. This vector ๐€๐ is the fourth vector that we donโ€™t yet know, but we can solve for.

Graphically, if vector ๐€ looked like this and starting from the same point vector ๐ looked like this, then this vector ๐€๐ would go from the tip of vector ๐€ to the tip of vector ๐. The way to compute the components of vector ๐€๐ is to subtract the components of vector ๐€ from those of vector ๐. Since weโ€™re working with vectors, we do the subtraction component by component. For the ๐‘ฅ-component, we have seven minus nine or negative two; for the ๐‘ฆ-component, negative ๐‘˜ minus negative five or negative ๐‘˜ plus five; and for the ๐‘ง-component, negative five minus negative one or negative five plus one. The components of this vector ๐€๐ then are negative two, negative ๐‘˜ plus five, and negative four.

Recalling that vector ๐€๐ is parallel to vector ๐‚, we can remember that, in general, if we have two three-dimensional vectors โ€” weโ€™ll call them ๐ฎ and ๐ฏ โ€” and those vectors are parallel, then we can say this about their components. The ratio of their ๐‘ฅ-components equals the ratio of their ๐‘ฆ-components equals the ratio of their ๐‘ง-components. The fact that ๐€๐ is parallel to vector ๐‚ means we can say the same thing about the components of these vectors. That is, that the ratio of the ๐‘ฅ-component of ๐€๐ to the ๐‘ฅ-component of ๐‚ equals the ratio of the ๐‘ฆ-component of ๐€๐ to the ๐‘ฆ-component of ๐‚ and also the ๐‘ง-component of ๐€๐ to the ๐‘ง-component of ๐‚.

In this whole expression here, we essentially have two unknowns and two independent equations. The unknowns are ๐‘˜ and ๐‘š. And our first independent equation is that the ๐‘ฅ-value ratio equals the ๐‘ฆ-value ratio and the second is that the ๐‘ฅ-value ratio equals the ๐‘ง-value ratio. By cross multiplying, we can solve both of these equations for ๐‘˜ and ๐‘š. In our first equation, by multiplying both sides by negative 55 and 10, we get 110 is equal to negative 10๐‘˜ plus 50. This implies that ๐‘˜ is equal to negative six. In our second equation, multiplying both sides by ๐‘š minus three and 10 tells us that negative two ๐‘š plus six equals negative 40. That means that negative two ๐‘š equals negative 46 or that ๐‘š equals positive 23.

Knowing the values of ๐‘˜ and ๐‘š, we can move on towards our solution. Since ๐‘˜ is negative six and ๐‘š is positive 23, ๐‘˜ minus ๐‘š equals negative 29. This is the value of the constant ๐‘˜ minus the constant ๐‘š.

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