Question Video: Determining the Coordinates of Vectors Mathematics

Given that 𝐀 β‹… 𝐁 = βˆ’2, 𝐀 β‹… 𝐂 = 3, 𝐀 β‹… 𝐃 = 3, 𝐁 = <βˆ’3, 2, βˆ’2>, 𝐂 = <3, βˆ’3, βˆ’1> and 𝐃 = <βˆ’1, 2, 2>, determine 𝐀.

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Video Transcript

Given that the dot product of vectors 𝐀 and 𝐁 equals negative two; the dot product of vectors 𝐀 and 𝐂 equals three; the dot product of vectors 𝐀 and 𝐃 equals three; vector 𝐁 is equal to negative three, two, negative two; vector 𝐂 is equal to three, negative three, negative one; and vector 𝐃 is equal to negative one, two, two, determine vector 𝐀.

In this question, we are given vectors 𝐁, 𝐂, and 𝐃 in the three-dimensional component form. For example, vector 𝐁 has an π‘₯-component equal to negative three, a 𝑦-component equal to two, and a 𝑧-component equal to negative two. We will let vector 𝐀 have components π‘₯, 𝑦 and 𝑧. And it is these three values we’re trying to calculate. We are also told in the question the value of the dot or scalar product of vectors 𝐀 and 𝐁, 𝐀 and 𝐂, and 𝐀 and 𝐃. We can use this information to create three equations in terms of the variables π‘₯, 𝑦, and 𝑧.

To calculate the dot product of any two vectors, we multiply the corresponding components and find the sum of these values. The dot product of vector 𝐀 and vector 𝐁 is therefore equal to negative three π‘₯ plus two 𝑦 minus two 𝑧. We know that this is equal to negative two. The dot product of vectors 𝐀 and 𝐂 is equal to three π‘₯ minus three 𝑦 minus 𝑧. We are told this is equal to three. Finally, the dot product of vectors 𝐀 and 𝐃 is equal to negative π‘₯ plus two 𝑦 plus two 𝑧. And once again, this is equal to three.

We now have three equations and three unknowns π‘₯, 𝑦, and 𝑧. In order to help us eliminate the variable 𝑧, we will multiply both sides of the second equation by two. This gives us the equation six π‘₯ minus six 𝑦 minus two 𝑧 is equal to six. Our next step is to number the three equations one, two, and three. We can eliminate 𝑧 from the first two equations by subtracting equation one from equation two. Six π‘₯ minus negative three π‘₯ is equal to nine π‘₯. Negative six 𝑦 minus two 𝑦 is equal to negative eight 𝑦. Negative two 𝑧 minus negative two 𝑧 is equal to zero. Finally, six minus negative two is equal to eight.

We are left with the equation nine π‘₯ minus eight 𝑦 is equal to eight. We can also eliminate the variable 𝑧 by adding equation one and equation three. This gives us the equation negative four π‘₯ plus four 𝑦 is equal to one. We now have two equations with two unknowns π‘₯ and 𝑦. Multiplying both sides of the second equation by two gives us negative eight π‘₯ plus eight 𝑦 is equal to two. We will call this equation five and the equation nine π‘₯ minus eight 𝑦 equals eight equation four.

Our next step is to add equation four and equation five to eliminate the variable 𝑦. Nine π‘₯ plus negative eight π‘₯ is equal to π‘₯. Adding negative eight 𝑦 and eight 𝑦 gives us zero. And adding eight and two gives us 10. Therefore, π‘₯ is equal to 10. We can now substitute this value of π‘₯ equals 10 into either equation four or five. This will enable us to calculate the variable 𝑦. Substituting π‘₯ equals 10 into equation five gives us negative 80 plus eight 𝑦 is equal to two. Adding 80 to both sides of this equation, we see that eight 𝑦 is equal to 82. We can then divide both sides of this equation by eight such that 𝑦 is equal to 82 over eight which simplifies to 41 over four.

We now have values of π‘₯ and 𝑦 that we can substitute into either equation one, two, or three to help us calculate the variable 𝑧. In this question, we will choose equation three. Substituting our values of π‘₯ and 𝑦 gives us negative 10 plus two multiplied by 41 over four plus two 𝑧 is equal to three. Adding 10 to both sides of this equation, we have 41 over two plus two 𝑧 is equal to 13. We can then multiply each term in this equation by two to eliminate the fraction. This gives us 41 plus four 𝑧 is equal to 26. Subtracting 41 from both sides gives us four 𝑧 is equal to negative 15. And finally, dividing both sides by four gives us 𝑧 is equal to negative 15 over four.

We now have values for π‘₯, 𝑦, and 𝑧 that satisfy the initial three equations. π‘₯ is equal to 10, 𝑦 is equal to 41 over four, and 𝑧 is equal to negative 15 over four. We can therefore conclude that vector 𝐀 has components 10, 41 over four, and negative 15 over four.

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