Question Video: Finding the Magnitude of a Vector Expressed as a Sum and Scalar Multiple of Two Known Vectors | Nagwa Question Video: Finding the Magnitude of a Vector Expressed as a Sum and Scalar Multiple of Two Known Vectors | Nagwa

# Question Video: Finding the Magnitude of a Vector Expressed as a Sum and Scalar Multiple of Two Known Vectors Mathematics • First Year of Secondary School

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If 𝐀 = 〈5, −3〉 and 𝐁 = 〈2, 1〉, then |𝐀 + 3𝐁| = ＿ length units.

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### Video Transcript

If vector 𝐀 is equal to five, negative three and vector 𝐁 is equal to two, one, then the magnitude of 𝐀 plus three 𝐁 is equal to what length units.

In this question, we are given the two-dimensional vectors 𝐀 and 𝐁 in terms of their 𝑥- and 𝑦-components. We are asked to calculate the magnitude of vector 𝐀 plus three multiplied by vector 𝐁. We will do this in three steps, firstly by using scalar multiplication to calculate three 𝐁. When multiplying any vector by a scalar, we simply multiply each of the individual components by that scalar. This means that three multiplied by the vector two, one gives us the vector six, three. Our next step is to add this to vector 𝐀. And we will do this using the process of vector addition. We do this by adding the corresponding components separately, giving us the vector 11, zero.

Our final step is to find the magnitude of this vector. As the 𝑦-component of our vector is zero, there is a shortcut here. However, we will begin by looking at how we calculate the magnitude of any two-dimensional vector. If vector 𝐮 is equal to 𝑥, 𝑦, then the magnitude of vector 𝐮 is equal to the square root of 𝑥 squared plus 𝑦 squared. We find the sum of the squares of the individual components and then square root our answer. This means that the magnitude of the vector 11, zero is equal to the square root of 11 squared plus zero squared. This is equal to the square root of 121, which is equal to 11. If vector 𝐀 is equal to five, negative three, vector 𝐁 is equal to two, one, then the magnitude of 𝐀 plus three 𝐁 is equal to 11 length units.

As previously mentioned, there is a shortcut to calculate the magnitude when one of the components equals zero. In this question, the 𝑦-component was equal to zero. This means that the vector 11, zero moves 11 units in the positive 𝑥-direction. As the magnitude of a vector is its length, this confirms that the magnitude of the vector 11, zero is 11 length units. When a two-dimensional vector has one of its components equal to zero, then the magnitude of that vector will be equal to the nonzero component.

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