A-6 Vectors Standard A6a: Find and draw a resultant vector from other component vectors. Standard A6b: Find the direction angle of a resultant vector from.

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Presentation transcript:

A-6 Vectors Standard A6a: Find and draw a resultant vector from other component vectors. Standard A6b: Find the direction angle of a resultant vector from other component vectors.

A vector is in standard position if the initial point is at the origin. x y The component form of this vector is: What if we only knew the length of the vector and the angle? Before anyone panics, this is just SOHCAHTOA… Just watch… or

A vector is in standard position if the initial point is at the origin. x y Think of it as a hypotenuse of the right triangle above because it’s the length of the vector x component Remember what this really means: x component or x component

A vector is in standard position if the initial point is at the origin. x y See? Just SOHCAHTOA y component x component y component Think of it as a hypotenuse of the right triangle above because it’s the length of the vector x component Remember what this really means: or x component y component

The direction of a vector v is found this way: So this is just the x component divided by the magnitude How would we determine which one? If it’s the angle that you need to find, then you need to know this: x y x component y component Remember that the magnitude and components form a right triangle The direction v is the angle θ

If it’s the angle that you need to find, then you need to know this: y x component y component Remember that the magnitude and components form a right triangle The direction angle here is negative because the y component is in a lower quadrant The direction of a vector v is found this way: The direction v is the angle θ So this is just the x component divided by the magnitude So the sign is determined by the sign of the y component

A boat is sailing at 20 mph with a direction of 52° north of east. The current of the water at this time is 8 mph with a direction of 18° north of east 18° 52° How does the current affect the speed and direction of the boat? In other words find the vector that results from adding the first two vectors

A boat is sailing at 20 mph with a direction of 52° north of east. The current of the water at this time is 8 mph with a direction of 18° north of east How does the current affect the speed and direction of the boat? In other words find the vector that results from adding the first two vectors. The sum of these two vectors will look like this. In order to do add them we will have to break them into their x and y components one vector at a time.

A boat is sailing at 20 mph with a direction of 52° north of east. 52° 20 y x So the vector written in component form would look like this: In order to add them we need to break each down into their x and y component form. Notice the right triangle so let’s use SOHCAHTOA

The current of the water at this time is 8 mph with a direction of 18° north of east y x So the vector written in component form would look like this: In order to add them we need to break each down into their x and y component form. Notice the right triangle so let’s use SOHCAHTOA 18° 8

A boat is sailing at 20 mph with a direction of 52° north of east. The current of the water at this time is 8 mph with a direction of 18° north of east How does the current affect the speed and direction of the boat? In other words find the vector that results from adding the first two vectors. Now we just add their corresponding components

A boat is sailing at 20 mph with a direction of 52° north of east. The current of the water at this time is 8 mph with a direction of 18° north of east How does the current affect the speed and direction of the boat? In other words find the vector that results from adding the first two vectors. We’ll use the calculator to find the actual values Note that we stored the values which will be explained shortly

We’ll use the calculator to find the actual values we don’t want to round anything until we have our final answer. Enter your calculations then press this button followed by any letter. The calculator stores the exact values here in A and B so we can use them later.

A boat is sailing at 20 mph with a direction of 52° north of east. The current of the water at this time is 8 mph with a direction of 18° north of east How does the current affect the speed and direction of the boat? In other words find the vector that results from adding the first two vectors. The speed is the length of this new vector which we can find using the Pythagorean Theorem and the stored values in the calculator. But what about it’s new direction? mph

A boat is sailing at 20 mph with a direction of 52° north of east. The current of the water at this time is 8 mph with a direction of 18° north of east How does the current affect the speed and direction of the boat? In other words find the vector that results from adding the first two vectors. Remembering the slide about inverse cosine, we just take the x component (adjacent) and divide it by the speed we just found (hypotenuse) And we know it’s positive because the y component is positive ° We can also see this because it is in the first quadrant ° ‘Ans’ means enter the previously calculated answer which in this case is the hypotenuse.

A boat is sailing at 20 mph with a direction of 52° north of east. This time the current of the water is 8 mph with a direction of 18° north of west 18° 52° How does the current affect the speed and direction of the boat? This time we will have to do a little more work with the direction angles? °

This time the current of the water is 8 mph with a direction of 18° north of west 18° 52° How does the current affect the speed and direction of the boat? This time we will have to do a little more work with the direction angles? °

A boat is sailing at 20 mph with a direction of 52° north of east. This time the current of the water is 8 mph with a direction of 18° north of west ° How does the current affect the speed and direction of the boat? This time we will have to do a little more work with the direction angles? mph