### Video Transcript

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

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

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 π and vector π is therefore equal to negative three π₯ plus two π¦ minus two π§. We know that this is equal to negative two. The dot product of vectors π and π is equal to three π₯ minus three π¦ minus π§. We are told this is equal to three. Finally, the dot product of vectors π and π is equal to negative π₯ plus two π¦ plus two π§. And once again, this is equal to three.

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

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

Our next step is to add equation four and equation five to eliminate the variable π¦. Nine π₯ plus negative eight π₯ is equal to π₯. Adding negative eight π¦ and eight π¦ 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 π¦. Substituting π₯ equals 10 into equation five gives us negative 80 plus eight π¦ is equal to two. Adding 80 to both sides of this equation, we see that eight π¦ is equal to 82. We can then divide both sides of this equation by eight such that π¦ is equal to 82 over eight which simplifies to 41 over four.

We now have values of π₯ and π¦ that we can substitute into either equation one, two, or three to help us calculate the variable π§. In this question, we will choose equation three. Substituting our values of π₯ and π¦ gives us negative 10 plus two multiplied by 41 over four plus two π§ is equal to three. Adding 10 to both sides of this equation, we have 41 over two plus two π§ 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 π§ is equal to 26. Subtracting 41 from both sides gives us four π§ is equal to negative 15. And finally, dividing both sides by four gives us π§ is equal to negative 15 over four.

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