Show that Three Vectors are Coplanar *

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Derick Stokes
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1 OpenStax-CNX module: m Show that Three Vectors are Coplanar * John Taylor This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License 4.0 Abstract Demonstrates that three vectors are coplanar by forming the scalar triple product to see if the volume of the related parallelopiped is zero. The determinant method is used. 1 Three Vectors Coplanar (The small steps below can be used for a self test. To do so, Scroll in small increments.) (Updated 5/19/14) Determine whether the vectors specied by the following four points are coplanar: A (6, 0, 2), B (2, 0, 4), C (6, 6, 1), and D (2, 6, 3). To visualize the problem, let's draw a diagram. On paper, see if you can plot the point A. Use an x-axis out of the plane of the diagram. Then check your graph by scrolling down. The three components are shown in blue, yellow and green. * Version 1.4: May 20, :04 pm

2 OpenStax-CNX module: m Figure 1 Add point B. Then scroll down to check. The three components are shown in blue, yellow and green. Figure 2

3 OpenStax-CNX module: m Add point C in a similar way. The three components are shown in blue, yellow and green. Figure 3 Add point D. The three components are shown in blue, yellow and green.

4 OpenStax-CNX module: m Figure 4 How do we proceed? Since any three points, such as A, B, and C, determine a plane, we can determine two vectors, AB and AC, in that plane. By taking their cross product, we can obtain a vector perpendicular to their plane. Then we can determine the dot product of that vector and the vector AD. How will that help? That dot product is the volume of the parallelepiped determined by the three vectors. What should we look for? If that dot product is zero, the volume is zero. And then? We can conclude that the three vectors lie in the same plane. For the vector B. Add the vector AB, should the arrow point be at A or B? AB to your diagram. Then scroll down to check.

5 OpenStax-CNX module: m Figure 5 Add the vectors AC and AD to your diagram. Then scroll down to check.

6 OpenStax-CNX module: m Figure 6 Can we use the magnitudes of these vectors and the angle between them to get the cross product? No. Why? We don't know the angle between them. Then how can we get the cross product? We can determine the components of each vector and use the determinant method. How do we get the x-component of AB? We simply subtract the x-coordinates of the two points. What is the order of the subtraction? For the x-component of AB, we do the subtraction AB x = B x A x. Substitute values. We get AB x = B x A x = 2 6 = 4. Find the other components of this vector. We get AB y = B y A y = 0 0 = 0 and AB z = B z A z = 4 2 = 2 AC. Determine the 3 components of AC x = C x A x = 6 6 = 0 AC y = C y A y = 6 0 = 6 AC z = C z A z = 1 2 = 1 Set up the determinant for the cross product of AB x, AB y, AB z, AC x, AC y, and AC z. i j k AB AC= AB x AB y AB z AC x AC y AC z AB and AC in terms of the literal components

OpenStax-CNX module: m47413 7 Substitute the values. i j k AB AC= 4 0 2 0 6 1 Set up the multiplication.

7 OpenStax-CNX module: m Substitute the values. i j k AB AC= Set up the multiplication. AB AC= i [0 ( 1) 6 2] - j [( 4) ( 1) 0 2] + k [( 4) 6 0 0] Note the sign before the j -component. Simplify. AB AC= i ( 12) + j ( 4) + k ( 24) To get an idea of the direction of this vector, scale it down by a factor of 4 to 3 i j 6 k it will t better on the diagram. Add this vector to the last diagram, with its base at point A. so that Figure 7 Notice that the colored lines showing the components of this vector start at point A. They correspond to the vector we found: 3 i j 6 k = 0.25 AB AC Hence the blue line is 3 units in the negative x-direction, the yellow line is one unit in the negative y-direction, and the green line is 6 units in the negative z-direction. Are we done? No. What else do we need to do?

8 OpenStax-CNX module: m We need to determine dot product of this vector with Determine the 3 components of AD. AD x = D x A x = 2 6 = 4 AD y = D y A y = 6 0 = 6 AD z = D z A z = 3 2 = 1 Now set up the ( dot product. ) We need AD AB AC Substitute ( the vectors. ) AB AD AC = ( 4 i +6 j +1 k AD. ) ( 12 i 4 j 24 ) k Set up( the products ) of the components. AB AD AC = ( 4) ( 12) + 6 ( 4) + 1 ( 24) Simplify. ( ) AB We get AD AC = = 0 What is our conclusion? The volume is zero and the three vectors are coplanar. If you found this helpful and would recommend that I create more pages like this one, please let me know using the link at the top of the page.

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