- WELCOME TO A VIDEO ON VECTOR APPLICATIONS.
LET'S GO AHEAD AND TAKE A LOOK AT OUR FIRST PROBLEM.
A SHIP LEAVES PORT ON A BEARING OF 28 DEGREES
AND TRAVELS 7.5 MILES.
THE SHIP THEN TURNS DUE EAST AND TRAVELS 4.1 MILES.
HOW FAR IS THE SHIP FROM THE PORT,
AND WHAT IS ITS BEARING?
REMEMBER, BEARINGS ARE MEASURED FROM DUE NORTH.
SO IF WE HAVE A BEARING OF 28 DEGREES,
THE SHIP WOULD BE TRAVELING IN THIS PATH,
WHERE THIS ANGLE HERE WOULD BE 28 DEGREES,
AND THIS WOULD HAVE A DISTANCE OF 7.5 MILES.
WHEN THE SHIP TURNS DUE EAST,
IT'LL BE TRAVELING IN THIS DIRECTION FOR 4.1 MILES.
SO IF WE CAN DETERMINE THE BEARING
AND THE MAGNITUDE OF THIS RESULTANT VECTOR,
WE WILL BE ABLE TO ANSWER THIS QUESTION.
LET'S GO AHEAD AND TAKE A LOOK AT OUR SKETCH
IN A LITTLE MORE DETAIL.
IF THIS IS 28 DEGREES,
AND WE KNOW THIS IS A RIGHT ANGLE,
THIS ANGLE HERE MUST BE 90 DEGREES - 28 DEGREES,
OR 62 DEGREES.
NEXT, THE X AXIS AND THIS VECTOR HERE ARE PARALLEL,
SO HERE WE HAVE INTERIOR ANGLES, AND THESE ARE SUPPLEMENTARY.
SO IF THIS IS 62 DEGREES,
THIS WOULD HAVE TO BE 180 DEGREES - 62 DEGREES,
OR 118 DEGREES.
OKAY, THE REASON THAT'S SO IMPORTANT
IS NOW WE CAN DETERMINE THE LENGTH OF THIS VECTOR HERE,
OR ITS MAGNITUDE,
AND THAT'LL BE THE DISTANCE FROM THE PORT.
SO LET'S GO AHEAD AND LABEL THIS VECTOR R
AND WE'LL TRY TO FIND THE MAGNITUDE OF IT
USING THE LAW OF COSINES.
THE SQUARE OF THE MAGNITUDE OF VECTOR R
IS GOING TO BE EQUAL TO 7.5 SQUARED
+ 4.1 SQUARED - 2 x 7.5 x 4.1 x THE COSINE OF 118 DEGREES.
OKAY, AND FOR THE SAKE OF TIME
I'VE ALREADY CALCULATED THESE VALUES
AND YOU MAY WANT TO CHECK THESE FOR YOURSELF.
BUT FOR THE SAKE OF TIME WE'RE GOING TO GO AHEAD
AND MOVE ALONG.
NOTICE HERE, WE'RE SUBTRACTING A NEGATIVE,
SO WE'LL END UP ADDING 28.87.
SO WE HAVE THE SQUARE ROOT OF THE MAGNITUDE OF R
IS EQUAL TO 101.93.
NOW WE CAN SQUARE ROOT BOTH SIDES OF THE EQUATION.
SO THE MAGNITUDE OF VECTOR R IS APPROXIMATELY EQUAL TO 10.1.
SO WHAT THAT MEANS IS THE SHIP IS 10.1 MILES FROM PORT.
NOW, TO DETERMINE THE BEARING OF THIS SHIP
WE ARE GOING TO HAVE TO FIND THIS ANGLE HERE.
SO IF WE CAN FIND THIS ACUTE ANGLE IN THIS TRIANGLE,
WE CAN ADD IT TO 28 DEGREES AND THEN FIND THE BEARING.
AND WE CAN DO THAT USING THE LAW OF SINES
NOW THAT WE KNOW THIS ANGLE
AND THE LENGTH OF THE OPPOSITE SIDE OF THAT TRIANGLE.
WE KNOW THIS IS 10.1 MILES HERE.
SO WE'LL HAVE THE SINE OF 118 DEGREES DIVIDED BY 10.1
MUST EQUAL THE SINE OF THETA DIVIDED BY 4.1.
NOW WE CROSS-MULTIPLY AND GO AHEAD AND DIVIDE BY 10.1.
WE HAVE SINE THETA EQUALS APPROXIMATELY .5384.
SO NOW WE'LL GO TO OUR CALCULATOR
AND FIND THE INVERSE SINE OF .5384
TO DETERMINE WHAT THAT ANGLE WOULD BE.
2nd, SINE, .5384,
AND SO OUR ANGLE IS APPROXIMATELY 21 DEGREES
IN THIS TRIANGLE.
SO, AGAIN, THIS ANGLE HERE IS 21 DEGREES.
SO THE BEARING WOULD BE THE SUM OF THESE TWO ANGLES,
OR 21 DEGREES + 28 DEGREES,
THAT WOULD BE A BEARING OF 49 DEGREES.
LET'S GO AHEAD AND TAKE A LOOK AT ONE MORE PROBLEM.
TWO TOW TRUCKS ARE PULLING ON A TRUCK STUCK IN THE MUD.
TOW TRUCK ONE IS PULLING WITH A FORCE OF 635 POUNDS
AT A 51 DEGREE ANGLE FROM THE HORIZONTAL,
WHILE TOW TRUCK TWO IS PULLING WITH A FORCE OF 592 POUNDS
AT 39 DEGREES FROM THE HORIZONTAL.
WHAT IS THE MAGNITUDE
AND DIRECTION OF THE RESULTANT FORCE?
LET'S GO AHEAD AND SKETCH THESE POSITION VECTORS.
THE FIRST DIRECTION ANGLE IS 51 DEGREES WITH A MAGNITUDE OF 635.
THE SECOND VECTOR HAS A DIRECTION OF 39 DEGREES
WITH A MAGNITUDE OF 592.
SO TO FIND THE RESULTANT VECTOR
WE'RE GOING TO HAVE TO ADD THESE TOGETHER.
SO WHAT WE'LL DO IS WE'LL USE THE TRIANGULAR METHOD.
SO I'LL SKETCH A HORIZONTAL LINE PARALLEL TO THE X AXIS
THAT PASSES THROUGH THE TERMINAL POINT OF THIS FIRST VECTOR.
AND NOW WE'LL MOVE THE INITIAL POINT OF THIS BLUE VECTOR
UP TO THE TERMINAL POINT OF THE BLACK VECTOR.
SO IT MIGHT LOOK LIKE THIS.
SO THE RESULTANT VECTOR
WOULD HAVE ITS INITIAL POINT AT THE ORIGIN,
AND ITS TERMINAL POINT AT THE TERMINAL POINT
OF THE BLUE VECTOR.
SO WE HAVE A TRIANGLE THAT HAS A VERY LARGE OBTUSE ANGLE.
LET'S SEE IF WE CAN GATHER SOME ADDITIONAL INFORMATION
FROM OUR SKETCH.
IF WE KNOW THIS ANGLE HERE IS 39 DEGREES,
SINCE THESE TWO BLUE VECTORS ARE PARALLEL
AND THESE HORIZONTAL LINES ARE ALSO PARALLEL,
IF THIS IS 39 DEGREES,
THEN THE ANGLE BETWEEN THIS DASHED RED LINE
AND THIS BLUE VECTOR WOULD ALSO BE 39 DEGREES.
NEXT, IF THE ANGLE BETWEEN THIS BLACK VECTOR
AND THE X AXIS IS 51 DEGREES,
THIS ANGLE AND THIS ANGLE HERE WILL BE SUPPLEMENTARY.
SO 180 - 51 WILL GIVE US 129 DEGREES FOR THIS ANGLE.
AGAIN, 39 DEGREES WAS THIS SMALL ANGLE HERE.
SO THIS LARGE OBTUSE ANGLE WILL HAVE A MEASURE OF 129 DEGREES
+ 39 DEGREES, WHICH IS EQUAL TO 168 DEGREES.
SO THIS LARGE OBTUSE ANGLE HERE HAS A MEASURE OF 168 DEGREES.
NOW WE CAN USE THE LAW OF COSINES TO FIND THE LENGTH
OF THIS RED VECTOR OR ITS MAGNITUDE.
LET'S GO AHEAD AND LABEL THIS THE RESULTANT VECTOR.
AND, OF COURSE, THIS BLUE VECTOR HAS A LENGTH OF 592,
THE SAME AS THE BLUE VECTOR DOWN HERE.
NOW, ONCE WE KNOW THIS ANGLE IS 168 DEGREES,
THIS IS 592, THIS IS 635,
WE CAN FIND THE LENGTH OF THIS, OR ITS MAGNITUDE,
BY USING THE LAW OF COSINES.
SO THE SQUARE OF THE MAGNITUDE OF VECTOR R = 135 SQUARED
+ 592 SQUARED - 2 x 635,
THEN IT'S 592 x THE COSINE OF 168 DEGREES.
SO THIS ANGLE HERE IS REALLY THE KEY TO SETTING THIS UP
AND SOLVING IT.
AND, AGAIN, FOR THE SAKE OF TIME I'VE ALREADY CALCULATED THIS.
SO THE SQUARE OF THE MAGNITUDE OF R = 1,489,099.49,
OR APPROXIMATELY.
SO NOW WE CAN TAKE THE SQUARE ROOT OF BOTH SIDES.
SO THE MAGNITUDE OF R IS APPROXIMATELY 1,220.
SO 1,220 POUNDS IS THE FORCE BEING APPLIED TO THE TRUCK STUCK
IN THE MUD.
NEXT, THEY WANT US TO DETERMINE THE DIRECTION OF THAT FORCE.
SO WHAT WE'LL DO NOW IS USE THE LAW OF SINES
TO DETERMINE THIS ACUTE ANGLE IN THIS RIGHT TRIANGLE.
AND THEN FROM THAT WE'LL BE ABLE TO FIGURE OUT
WHAT THIS ANGLE HERE WOULD BE,
51 DEGREES - WHATEVER THAT --.
THE SINE OF 168 DEGREES DIVIDED BY 1,220
MUST EQUAL THE SINE OF THETA DIVIDED BY 592.
LET'S GO AHEAD AND TAKE THIS OVER TO THE NEXT SLIDE.
CROSS-MULTIPLY,
DIVIDE BY 1,220, IT'S APPROXIMATELY .1009.
SO WE'LL GO TO THE CALCULATOR, INVERSE, SINE, .1009,
SO OUR ANGLE IS APPROXIMATELY 5.8 DEGREES.
LET'S GO BACK TO OUR SKETCH.
AGAIN, THIS ANGLE RIGHT HERE IS APPROXIMATELY 5.8 DEGREES.
WELL, REMEMBER, THIS BLACK VECTOR FORMED A 51 DEGREE ANGLE
WITH THE POSITIVE X AXIS.
SO IF THAT'S 51 DEGREES,
AND THEN WE SUBTRACT 5.8 DEGREES,
THAT WOULD GIVE US OUR DIRECTION ANGLE.
THIS IS GOING TO COME OUT TO APPROXIMATELY 45 DEGREES
IF WE--TWO SIGNIFICANT DIGITS AS WE HAVE HERE.
SO OUR DIRECTION HERE TO HERE IS APPROXIMATELY 45 DEGREES.
HOPEFULLY, YOU CAN READ THIS IF YOU VIEW THIS IN FULL SCREEN.
AND I HOPE YOU'VE FOUND THIS EXPLANATION HELPFUL.
HAVE A GOOD DAY.