0580 · 7.3Magnitude of a vector

Vector magnitude and Pythagoras

0:00Imagine moving three metres to the right and four metres upwards. Your final position is not seven metres from your starting point. Seven metres describes the route along those two separate movements. The direct distance is shorter. Finding the magnitude of a vector means calculating the length of that direct arrow. A vector has both magnitude and direction. Magnitude means its length, without including which way it points. In this video, we will connect that length to the horizontal and vertical components.

0:27You will understand why the formula works, how negative components affect it, and how to present your calculations. The exam prints vector names in bold. We will call this vector p, written as the column vector with x component three and y component four. The top entry describes horizontal movement, while the bottom entry describes vertical movement. Positive components mean right or up, while negative components mean left or down.

0:52The vertical bars around the vector name are called modulus signs. Here, they tell us to find the magnitude of vector p. The result is a single number, not another column vector. If the components represent lengths in metres, the magnitude is also measured in metres. Horizontal and vertical movements meet at a right angle. Together with the vector arrow, they form a right-angled triangle. The arrow is the hypotenuse because it lies opposite the right angle.

1:18Pythagoras tells us that the hypotenuse squared equals the sum of the squares of the other two sides. For a column vector with components x and y, the magnitude squared equals x² + y². Taking the square root gives the magnitude itself. This is why we square both components, add the results, and then take the square root. The formula is simply Pythagoras applied to the vector arrow.

Calculating magnitudes with negative components

1:42Our first exam-style question asks us to calculate the magnitude of vector p, with x component three and y component four. Substitute the components to obtain the square root of three squared plus four squared. Squaring the components gives the square root of nine plus sixteen. Adding inside the square root gives √25. √25 is five, so the magnitude is five. This is the direct distance represented by the arrow.

2:10We do not add three and four because that measures the two sides rather than the diagonal. This calculation should be manageable without a calculator on Paper two. Next, calculate the magnitude of vector q, the column vector with x component negative five and y component twelve. Substitute to obtain √−5 in brackets, squared, plus 12². Squaring negative five gives twenty-five, so the expression becomes the square root of twenty-five plus one hundred and forty-four.

2:37The sum is one hundred and sixty-nine. Taking the square root gives a magnitude of thirteen. The negative horizontal component means the arrow points left rather than right. It does not make the arrow have a negative length. Squaring negative five means multiplying negative five by itself, which produces a positive result. On a calculator, put a negative component inside brackets before squaring it. Otherwise, the calculator may interpret the minus sign as applying after the square.

3:06Also, never forget the final square root. One hundred and sixty-nine is the magnitude squared, whereas thirteen is the magnitude.

Exact answers and rounding

3:14A third exam-style question gives vector OP as the column vector with x component two and y component negative three. Calculate its magnitude to three significant figures. Substitute to obtain the square root of two squared plus negative three in brackets, squared. Squaring the components gives the square root of four plus nine. Adding gives the exact answer, √13. A calculator shows three point six zero five and further digits.

3:40Rounding to three significant figures gives three point six one. The exam writes vector OP with an arrow above the letters, inside modulus signs. Those modulus signs mean the length of the arrow from O to P. On non-calculator Paper two, leave an answer such as √13 exact when a decimal approximation is not requested. On calculator Paper four, follow any accuracy instruction in the question. Otherwise, give answers that are not exact to three significant figures.

4:09Keep full calculator accuracy until the final answer. Magnitude is never negative, and it is zero only when both components are zero. Changing a component from positive to negative changes direction but leaves its square unchanged. To calculate any two-dimensional vector magnitude, square the components, add them, and take the square root. Remember the triangle behind those steps, and the formula becomes a statement about length rather than something to memorise.