Question Video: Using the Properties of Vectors to Solve a Problem | Nagwa Question Video: Using the Properties of Vectors to Solve a Problem | Nagwa

# Question Video: Using the Properties of Vectors to Solve a Problem Mathematics • Third Year of Secondary School

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Find the values of π, π, and π that make the two vectors π = β¨π β 2, πΒ² β 2, 4β© and π = β¨3, 2, πΒ²β© equal.

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### Video Transcript

Find the values of π, π, and π that make the two vectors the vector π: π minus two, π squared minus two, four and the vector π: three, two, π squared equal.

In this question, weβre given two vectors, vector π and vector π. And these two vectors are given component-wise. We need to determine the values of the unknowns π, π, and π which make these equal. To answer this question, letβs start by recalling what we mean when we say that two vectors are equal. Remember that vectors are an object with both a magnitude and direction. So we say that two vectors are the same vector, in other words, theyβre equal, if they have the same magnitude and the same direction. So one way of answering this question would be to find the values of π, π, and π which make these vectors have the same magnitude and the same direction.

However, this would be quite a long process. In fact, thereβs a much easier method. We see that weβre given vector π and vector π in terms of their components. And we know an equivalent definition for checking whether two vectors are equal in terms of their components. We can also say that two vectors are equal if they have the same dimension and all of their corresponding components are equal. And this is a much-easier definition to work with if weβre given two vectors component-wise, like we are in this question. So weβll use this definition to find the values of π, π, and π. Letβs start with checking the dimensions of vector π and vector π.

Remember, the dimensions of a vector given component-wise is just the number of components in our vector. We can see that both vector π and vector π have three components. This means both of these are three-dimensional vectors, so they do have the same dimension. Next, for vector π to be equal to vector π, all of their corresponding components must be equal. In other words, the first component of vector π must be equal to the first component of vector π. The second component of vector π must be equal to the second component of vector π. And the third component of vector π must be equal to the third component of vector π. Therefore, if vector π is equal to vector π, by setting the corresponding components to be equal to each other, we get three equations which must be true.

By setting the first components to be equal, we get π minus two should be equal to three. By setting the second components to be equal, we get π squared minus two should be equal to two. And by setting the third components to be equal, we get four is equal to π squared. We can find the values of π, π, and π by solving these equations, letβs start with our first equation. We can solve this by just adding two to both sides of the equation. We get that π has to be equal to five. And itβs worth pointing out we could substitute π is equal to five into our vector π. And we would see that our first component will be five minus two, which is equal to three, the same as the first component in vector π.

So weβve solved our first equation. Letβs now solve our second equation. Weβll start by adding two to both sides. This gives us π squared is equal to four. We need to find all of the values of π which solve this equation. One way of solving this is to take the square root of both sides. But remember, weβll get a positive and a negative square root. We get π is positive or negative the square root of four. And of course, we know that the square root of four is equal to two, so π is equal to positive or negative two. Once again, we could substitute π is equal to two or negative two into our vector π, and we would see that our second component evaluates to give us two.

Finally, we want to solve our last equation, four is equal to π squared. And we do this in exactly the same way we did above. We need to take the square root of both sides of our equation. And remember, we get a positive and a negative square root. So we get π is positive or negative root four, and we know that root four is equal to two. So π is positive or negative two. And we can check this answer by substituting π is equal to two or π is equal to negative two into our vector π and checking that our third component evaluates to give us four. And this gives us our final answer.

Therefore, for vector π: π minus two, π squared minus two, four to be equal to the vector π: three, two, π squared, we showed by checking the corresponding components needed to be equal that π needs to be equal to five, π can be equal to two or negative two, and π can be equal to two or negative two.

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