Video: Evaluating Expressions Involving the Addition and Subtraction of Given Vectors

True or false: If 𝐀 = ⟨1, 1, 2⟩ and 𝐁 = ⟨2, 1, 2⟩, then |𝐀 + 𝐁| > |𝐀 − 𝐁|.

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Video Transcript

True or false: If vector 𝐀 is equal to one, one, two and vector 𝐁 is equal to two, one, two, then the magnitude of vector 𝐀 plus vector 𝐁 is greater than the magnitude of vector 𝐀 minus vector 𝐁.

We begin here by recalling that when we add or subtract two vectors, we simply add or subtract their corresponding components. Vector 𝐀 plus vector 𝐁 is equal to one, one, two plus two, one, two. One plus two is equal to three, one plus one is equal to two, and two plus two is equal to four. Vector 𝐀 plus vector 𝐁 is equal to three, two, four.

We calculate the magnitude of any vector by square rooting the sum of the squares of its individual components. The magnitude of vector 𝐀 plus vector 𝐁 is therefore equal to the square root of three squared plus two squared plus four squared. As three squared is equal to nine, two squared is equal to four, and four squared equals 16, this is equal to the square root of 29.

We will now repeat this process with the right-hand side of our inequality. We begin by subtracting vector 𝐁 from vector 𝐀. One minus two is equal to negative one. One minus one is equal to zero. And two minus two is also equal to zero. Vector 𝐀 minus vector 𝐁 is equal to negative one, zero, zero. The magnitude of this vector is equal to the square root of negative one squared plus zero squared plus zero squared. This is equal to the square root of one. And as we are looking for the magnitude, this is equal to positive one.

We have now calculated the magnitude of vector 𝐀 plus vector 𝐁 together with the magnitude of vector 𝐀 minus vector 𝐁. The statement in the question said that the magnitude of vector 𝐀 plus vector 𝐁 was greater than the magnitude of vector 𝐀 minus vector 𝐁. From our calculations, we see that the square root of 29 is greater than one. This means that given the vectors 𝐀 and 𝐁 in the question, the statement is true.

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