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In this video, we will learn how to represent a vector in space using a three-dimensional coordinate system.
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We will begin by looking at unit vectors in the direction of the π₯-, π¦-, and π§-axis.
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We will then move on to find the components of a vector which connects two points in 3D space.
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We will do this both algebraically and graphically.
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Find the unit vector in the direction of the π¦-axis.
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We recall that a unit vector has a magnitude equal to one.
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Letβs consider the three-dimensional coordinate system or grid with center or origin π.
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We are told that the vector is moving in the direction of the π¦-axis.
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This means that its π₯- and π§-components must be equal to zero.
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In order for the vector to have a magnitude of one, the π¦-component must also be one.
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The unit vector in the direction of the π¦-axis is equal to zero, one, zero.
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We can check this as a magnitude equal to one by finding the square root of the sum of the squares of the individual components.
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This is the square root of zero squared plus one squared plus zero squared.
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This is equal to the square root of one.
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And as the magnitude must be positive, this is equal to one.
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We can use this information to find the unit vector in the direction of the π₯- and π§-axis.
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The unit vector in the direction of the π₯-axis is one, zero, zero.
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It has an π₯-component of one and a π¦- and π§-component of zero.
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We have just seen that the unit vector in the direction of the π¦-axis is zero, one, zero.
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Finally, the unit vector in the direction of the π§-axis is zero, zero, one.
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This time, we have a π§-component equal to one and an π₯- and π¦-component equal to zero.
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We will now look at some questions where we need to find the vector between two given points.
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Which of the following is equal to the vector ππ?
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Is it (A) π plus π, (B) π minus π, (C) π minus π, or (D) π times π?
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Letβs begin by considering the two points π΄ and π΅ on a two-dimensional coordinate grid.
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The vector π will take us from the origin π to the point π΄.
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Likewise, vector π will take us from the origin to the point π΅.
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We need to work out how we can get from point π΄ to point π΅.
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One way of doing this is going via the origin π.
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This means that vector ππ is equal to vector ππ plus vector ππ.
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As the vector ππ equals π, then the vector ππ will be negative π, as it is going in the opposite direction but has the same magnitude.
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The vector ππ is equal to π.
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Negative π plus π can be rewritten as π minus π.
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This means that the correct answer is option (C).
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The vector ππ is equal to vector π minus vector π.
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We will now use this rule to find a vector between two given points.
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Given vector π is equal to six, one, four and vector π is equal to three, one, two, find vector ππ.
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We recall the general rule when finding vectors between two points states that vector ππ is equal to vector π minus vector π.
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In this question, we want to subtract the vector six, one, four from the vector three, one, two.
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When subtracting vectors, we subtract the corresponding components.
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In this case, we need to subtract six from three.
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This gives us negative three.
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One minus one is equal to zero.
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Finally, two minus four is equal to negative two.
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The π₯-component is negative three, the π¦-component is zero, and the π§-component is negative two.
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The vector ππ is equal to negative three, zero, negative two.
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In our next question, we will need to find the position vector of a point given the vector joining it to another point.
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Given ππ is equal to negative one, negative three, zero and vector π is equal to negative four, negative five, negative five, express vector π in terms of the fundamental unit vectors.
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We recall that when finding the vector between two points, vector ππ is equal to vector π minus vector π.
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If we let vector π have components π₯, π¦, π§, then negative one, negative three, zero is equal to π₯, π¦, π§ minus negative four, negative five, negative five.
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Adding vector π to both sides of this equation gives us negative one, negative three, zero plus negative four, negative five, negative five is equal to π₯, π¦, π§.
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When adding and subtracting vectors, we can look at each component separately.
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This means that π₯ is equal to negative one plus negative four.
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This is the same as negative one minus four, which is equal to negative five.
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π¦ is equal to negative three plus negative five.
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This equals negative eight.
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Finally, π§ is equal to negative five.
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Vector π is, therefore, equal to negative five, negative eight, negative five.
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We were asked to write vector π in terms of the fundamental unit vectors.
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This means we need to write it in the form π₯π’ plus π¦π£ plus π§π€.
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Vector π is, therefore, equal to negative five π’ minus eight π£ minus five π€.
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In our next question, we will find the components of a 3D position vector which is represented graphically.
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Using the graph, write the vector π in terms of its components.
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As we have a three-dimensional coordinate grid, vector π will have three components, π₯, π¦, and π§.
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Moving along the π₯-axis, we can see that the π₯-component is two.
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The π¦-component is equal to three.
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Finally, the π§-component is equal to four.
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This means that vector π is equal to two, three, four.
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The vector π in terms of its components is the displacement of point π΄ from the origin in the π₯-, π¦-, and π§-directions.
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In our final question, we will find the components of a 3D vector which is again represented graphically.
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Find the vector ππ using the graph.
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One way of answering this question would be to recall that we can find the vector ππ by subtracting vector π from vector π.
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This means that in our question, we need to subtract vector π from vector π.
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Vector π is the displacement of point π΄ from the origin.
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This has an π₯-component of one, a π¦-component of one, and a π§-component of zero.
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This means that vector π is equal to one, one, zero.
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Vector π has an π₯-component of four, a π¦-component of four, and a π§-component of three.
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This means that vector π is equal to four, four, three.
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To calculate vector ππ, we need to subtract one, one, zero from four, four, three.
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When subtracting vectors, we subtract each component separately.
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Four minus one is equal to three.
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When subtracting the π¦-components, we also get three.
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The same is true of the π§-components as three minus zero is equal to three.
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Vector ππ is, therefore, equal to three, three, three.
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An alternative method here would be to recognize that we have a cube of side length three.
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Vertices π΄ and πΊ are opposite corners of the cube.
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This means that we need to move three units in the π₯-, π¦-, and π§-direction to get from point π΄ to point πΊ.
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This confirms that vector ππ is equal to three, three, three.
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We will now summarize the key points from this video.
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We saw in our first question that a unit vector has a magnitude of one.
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A vector in three-dimensional space can be written in terms of its three components, in triangular brackets π₯, π¦, π§ or π₯π’ plus π¦π£ plus π§π€.
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The vector connecting two points π΄ and π΅ in 3D space is written vector ππ.
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This is equal to vector π minus vector π.
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We can, therefore, calculate the components of vector ππ given the coordinates of points π΄ and π΅.
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We also saw that we can find the coordinates of an unknown point using the coordinates of a known point and the components of a known vector.
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Finally, we saw in the last two questions that we can find the components of a 3D vector which is represented graphically.