WEBVTT
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In this video, we will learn how to find the magnitude of a position vector in space.
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We will begin by recalling what we mean by a 3D vector.
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A three-dimensional vector has an 𝐢-, 𝐣-, and 𝐤-component.
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If we consider the point 𝑃 with coordinates two, three, five, we can write the vector 𝐎𝐏 in numerous ways.
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Firstly, we can consider the 𝐢-, 𝐣-, and 𝐤-components.
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The vector 𝐎𝐏 is equal to two 𝐢 plus three 𝐣 plus five 𝐤.
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Vectors are also sometimes written similar to coordinates.
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With triangular brackets, we have two, three, five.
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A third way of writing the same vector is as a column inside parentheses, with the 𝐢-component followed by the 𝐣-component and then the 𝐤-component.
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The magnitude of any vector is the distance between two points in three-dimensional space.
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If we consider a vector 𝐀 written in the general form 𝑥𝐢 plus 𝑦𝐣 plus 𝑧𝐤, then the magnitude of vector 𝐀 can be calculated using an application of the Pythagorean theorem.
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The magnitude of vector 𝐀 denoted by two vertical lines is equal to the square root of 𝑥 squared plus 𝑦 squared plus 𝑧 squared.
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We find the sum of the squares of the 𝐢-, 𝐣-, and 𝐤-components and then square root the answer.
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As well as seeing our unit vectors 𝐢, 𝐣, and 𝐤 written with a hat, you may also have seen them underlined.
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These notations for handwritten vectors will vary from location to location.
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Generally when typed, either in a text book or on the internet, vectors will appear in bold.
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We will now look at some questions where we need to calculate the magnitude of a vector.
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If vector 𝐀 is equal to two, negative five, two, find the magnitude of vector 𝐀.
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The 𝐢-, 𝐣-, and 𝐤-components of vector 𝐀 are two, negative five, and two, respectively.
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Therefore, vector 𝐀 could be rewritten as two 𝐢 minus five 𝐣 plus two 𝐤.
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We recall that the magnitude of any vector could be calculated by square rooting 𝑥 squared plus 𝑦 squared plus 𝑧 squared, where 𝑥, 𝑦, and 𝑧 are the 𝐢-, 𝐣-, and 𝐤-components, respectively.
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The magnitude of vector 𝐀 is, therefore, equal to the square root of two squared plus negative five squared plus two squared.
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Two squared is equal to four.
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Squaring a negative number gives a positive answer.
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Therefore, negative five squared is 25.
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The magnitude of 𝐀 is equal to the square root of four plus 25 plus four.
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This is equal to the square root of 33.
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Whilst we could work out this answer on the calculator, as a general rule, we will leave our answers as radicals or surds.
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The magnitude of vector 𝐀 is the square root of 33.
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In our next question, the vector will be written in a different format.
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If vector 𝐀 is equal to two 𝐢 plus three 𝐣 minus 𝐤, find the magnitude of vector 𝐀.
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For any vector written in the form 𝑥𝐢 plus 𝑦𝐣 plus 𝑧𝐤, the magnitude of the vector is equal to the square root of 𝑥 squared plus 𝑦 squared plus 𝑧 squared.
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The 𝐢-component of our vector is equal to two, the 𝐣-component is equal to three, and the 𝐤-component is equal to negative one.
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This means that the magnitude of vector 𝐀 is equal to the square root of two squared plus three squared plus negative one squared.
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Two squared is equal to four.
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Three squared is equal to nine.
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Squaring a negative number gives us a positive answer.
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Therefore, negative one squared is one.
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As four plus nine plus one equals 14, the magnitude of vector 𝐀 is the square root of 14.
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In our next question, we will be given the magnitude and need to calculate one of the components of the vector.
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If vector 𝐀 is equal to 𝑎𝐢 plus 𝐣 minus 𝐤 and the magnitude of vector 𝐀 is equal to the square root of six, find all the possible values of 𝑎.
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Before starting this question, it is worth noting that the wording says find all possible values of 𝑎.
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This suggests there will be more than one correct answer.
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We are given two pieces of information.
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We are told vector 𝐀 is equal to 𝑎𝐢 plus 𝐣 minus 𝐤 and the magnitude of vector 𝐀 is the square root of six.
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We know that for any vector written in the form 𝑥𝐢 plus 𝑦𝐣 plus 𝑧𝐤, then its magnitude is equal to the square root of 𝑥 squared plus 𝑦 squared plus 𝑧 squared.
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In this question, the square root of six is equal to the square root of 𝑎 squared plus one squared plus negative one squared.
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This is because the 𝐢-, 𝐣-, and 𝐤-components are 𝑎, one, and negative one, respectively.
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We can begin to solve this equation by squaring both sides.
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As squaring is the inverse or opposite of square rooting, the square root of six squared is equal to six.
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In the same way, the right-hand side becomes 𝑎 squared plus one squared plus negative one squared.
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Both one squared and negative one squared are equal to one.
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Therefore, this simplifies to six is equal to 𝑎 squared minus two.
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We can then subtract two from both sides of this equation so that 𝑎 squared is equal to four.
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Our final step is to square root both sides.
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The square root of 𝑎 squared is 𝑎.
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The square root of four is equal to two.
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But we must take the positive or negative of this.
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Therefore, 𝑎 is equal to positive or negative two.
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The possible values of 𝑎 such that the magnitude of vector 𝐀 is the square root of six are two and negative two.
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This is because when we square both of these, we get an answer of four.
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Our next few questions will also involve the addition and subtraction of vectors.
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Given that vector 𝐀 plus vector 𝐁 is equal to negative two, four, three and vector 𝐀 is equal to three, five, three, determine the magnitude of vector 𝐁.
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We recall that when adding two vectors, we simply add the 𝐢-, 𝐣-, and 𝐤-components separately.
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The vector 𝐀 plus 𝐁 is equal to the vector 𝐀 plus the vector 𝐁.
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If we let vector 𝐁 have 𝐢-, 𝐣-, and 𝐤-components 𝑥, 𝑦, and 𝑧, respectively, then negative two, four, three is equal to three, five, three plus 𝑥, 𝑦, 𝑧.
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We can then subtract vector 𝐀 from both sides of this equation.
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The left-hand side becomes negative two, four, three minus three, five, three.
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Negative two minus three is equal to negative five.
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Therefore, our 𝐢-component of vector 𝐁 is negative five.
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Four minus five is equal to negative one, so the 𝐣-component is negative one.
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Finally, three minus three is equal to zero.
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Vector 𝐁 is, therefore, equal to negative five, negative one, zero.
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We can calculate the magnitude of this vector by squaring each of the components, finding their sum, and then square rooting.
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The magnitude of vector 𝐁 is equal to the square root of negative five squared plus negative one squared plus zero squared.
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Negative five squared is 25, negative one squared is one, and zero squared is equal to zero.
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The magnitude of vector 𝐁 is, therefore, equal to the square root of 26.
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In our next question, we’ll find the magnitude of a vector joining the endpoints of two other vectors.
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Given that vector 𝐀𝐁 is equal to negative five 𝐢 plus two 𝐣 minus four 𝐤 and vector 𝐁𝐂 is equal to four 𝐢 plus four 𝐣 plus six 𝐤, determine the magnitude of vector 𝐀𝐂.
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In this type of question, it is worth drawing a diagram first.
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This will hopefully ensure that our direction and signs are correct.
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We are given three points 𝐴, 𝐵, and 𝐶.
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We can join these to form a triangle.
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In the question, we are given the value of vector 𝐀𝐁.
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We are also given the value of vector 𝐁𝐂.
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Our aim is to calculate the magnitude of vector 𝐀𝐂.
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Therefore, our first step is to work out vector 𝐀𝐂.
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We can see that one way to get from point 𝐴 to point 𝐶 is via point 𝐵.
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Therefore, vector 𝐀𝐂 is equal to vector 𝐀𝐁 plus vector 𝐁𝐂.
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Vector 𝐀𝐂 is, therefore, equal to negative five 𝐢 plus two 𝐣 minus four 𝐤 plus four 𝐢 plus four 𝐣 plus six 𝐤.
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We can add two vectors by adding the individual components separately.
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Negative five 𝐢 plus four 𝐢 is equal to negative 𝐢.
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Two 𝐣 plus four 𝐣 is equal to six 𝐣.
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Finally, negative four 𝐤 plus six 𝐤 is equal to two 𝐤.
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Vector 𝐀𝐂 is equal to negative 𝐢 plus six 𝐣 plus two 𝐤.
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The magnitude of any vector can be found by squaring the 𝐢-, 𝐣-, and 𝐤-components, finding their sum, and then square rooting the answer.
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This means that the magnitude of vector 𝐀𝐂 is the square root of negative one squared plus six squared plus two squared.
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Negative one squared is equal to one, six squared is 36, and two squared is equal to four.
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One, 36, and four sum to 41.
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Therefore, the magnitude of vector 𝐀𝐂 is the square root of 41.
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In our final question, we will find the magnitude of the difference of two vectors.
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If vector 𝐀 is equal to four 𝐢 plus four 𝐣 minus five 𝐤 and vector 𝐁 is equal to three 𝐢 minus 𝐤, determine the magnitude of vector 𝐀 minus vector 𝐁.
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Our first step in this question is to calculate vector 𝐀 minus vector 𝐁.
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This is important as a common mistake here would be to think that the magnitude of vector 𝐀 minus vector 𝐁 is equal to the magnitude of vector 𝐀 minus the magnitude of vector 𝐁.
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This, however, is not true.
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To work out vector 𝐀 minus vector 𝐁, we simply subtract the individual components separately.
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Four 𝐢 minus three 𝐢 is equal to one 𝐢, or just 𝐢.
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There is no 𝐣-component in vector 𝐁.
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Therefore, we are left with four 𝐣.
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Negative five 𝐤 minus negative 𝐤 is equal to negative four 𝐤.
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This is because negative five minus negative one is the same as negative five plus one, which is equal to negative four.
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We now need to calculate the magnitude of this vector.
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We know that for any vector written in the form 𝑥𝐢 plus 𝑦𝐣 plus 𝑧𝐤, its magnitude is equal to the square root of 𝑥 squared plus 𝑦 squared plus 𝑧 squared.
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This means that the magnitude of vector 𝐀 minus vector 𝐁 is the square root of one squared plus four squared plus negative four squared.
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Both four squared and negative four squared are equal to 16, so we are left with the square root of one plus 16 plus 16.
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This is equal to the square root of 33.
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The magnitude of vector 𝐀 minus vector 𝐁 is the square root of 33.
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We will now summarize the key points from this video.
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For any vector written in the form 𝑥𝐢 plus 𝑦𝐣 plus 𝑧𝐤, the magnitude of the vector is the square root of 𝑥 squared plus 𝑦 squared plus 𝑧 squared.
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We square the individual components of 𝐢, 𝐣, and 𝐤, find their sum, and then square root the answer.
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The answer for the magnitude will always be positive.
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When adding or subtracting two vectors, we add or subtract the individual components separately.
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We also saw that the magnitude of the sum or difference of two vectors is not the same as the magnitude of their individual parts.
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The magnitude of vector 𝐀 plus vector 𝐁 is not equal to the magnitude of vector 𝐀 plus the magnitude of vector 𝐁.