- WELCOME TO A VIDEO ON VECTOR OPERATIONS.
THIS VIDEO WILL COVER SCALAR MULTIPLICATION,
ADDITION AND SUBTRACTION OF VECTORS,
AND ALSO HOW TO DETERMINE THE DOT PRODUCT OF TWO VECTORS.
THE SUM OF TWO OR MORE VECTORS IS CALLED THE RESULTANT
OF THE VECTORS.
TO ADD VECTORS YOU CAN USE THE TRIANGULAR METHOD
AND THE PARALLELOGRAM METHOD.
TO USE THE TRIANGULAR METHOD,
WE DRAW THE VECTORS ONE AFTER ANOTHER
AND PLACE THE INITIAL POINT OF THE SECOND VECTOR
AT THE TERMINAL POINT OF THE FIRST VECTOR.
THIS IS ALSO CALLED THE TAIL-TO-TIP METHOD.
SO THIS IS VECTOR "A" + VECTOR V.
SO HERE WE HAVE A SKETCH OF VECTOR "A" + VECTOR B.
NOTICE THE INITIAL POINT OF VECTOR B
IS AT THE TERMINAL POINT OF VECTOR "A,"
AND THE RESULTANT VECTOR IS THIS DASH VECTOR HERE.
LET'S SAY WE WANT TO FIND THE SUM OF VECTOR V AND U.
WE'LL GO AHEAD AND GRAPH VECTOR V AS A POSITION VECTOR.
NOW WE'LL GRAPH VECTOR U STARTING AT THE TERMINAL POINT
OF VECTOR V.
SO THE RESULTANT VECTOR, OR THE SUM OF THESE VECTORS,
WOULD BE THIS VECTOR THAT HAS ITS INITIAL POINT
AT THE ORIGIN AND ITS TERMINAL POINT
AT THE TERMINAL POINT OF VECTOR U.
AND THIS IS THE TAIL-TO-TIP, OR TRIANGULAR METHOD.
IF WE WANTED TO KNOW
THE MAGNITUDE OF THE RESULTANT VECTOR
WE COULD MEASURE THIS WITH A RULER,
AND WE COULD ALSO FIND THE DIRECTION ANGLE
WITH A PROTRACTOR.
TO USE THE PARALLELOGRAM METHOD WE DRAW THE VECTORS
SO THAT THEIR INITIAL POINTS MEET.
THEN WE DRAW IN LINES TO FORM A PARALLELOGRAM,
AND THE RESULTANT IS THE DIAGONAL FROM THE INITIAL POINT
TO THE OPPOSITE VERTEX OF THE PARALLELOGRAM.
SO HERE IS VECTOR "A" + VECTOR B.
NOTICE THEY SHARE THE SAME INITIAL POINT.
THE DASHED VECTORS WOULD CONSTRUCT THE PARALLELOGRAM,
AND THE DIAGONAL IS THE RESULTANT,
OR THE SUM OF THOSE TWO VECTORS.
YOU CAN NOTICE THAT THE PARALLELOGRAM METHOD
IS REALLY A VERSION OF THE TRIANGULAR
OR TIP-TO-TAIL METHOD.
IF YOU LOOK AT THE TOP PORTION OF THE FIGURE ABOVE,
YOU CAN SEE THAT ONE SIDE OF THE PARALLELOGRAM
IS REALLY VECTOR B TRANS--
LET'S GO AHEAD AND ADD THESE VECTORS
USING THE TRIANGULAR METHOD.
SO WE'LL PLACE BOTH INITIAL POINTS AT THE ORIGIN.
NOW WE'LL FORM A PARALLELOGRAM
AND THE DIAGONAL OF THIS PARALLELOGRAM
IS THE RESULTANT, LIKE SO.
SO THIS VECTOR HERE IS VECTOR V + VECTOR U.
NOW LET'S GO AHEAD
AND TAKE A LOOK AT SUBTRACTION JUST FOR A MOMENT.
NOTICE ON THIS PROBLEM WE WANT TO HAVE 2 x VECTOR U - VECTOR V.
WELL, 2 x VECTOR U MEANS WE'RE GOING TO SKETCH THIS
AS A POSITION VECTOR, BUT WE'LL DOUBLE THE MAGNITUDE,
OR DOUBLE THE LENGTH.
SO IF THIS IS U, THIS WOULD BE 2U.
NOW, BECAUSE WE'RE SUBTRACTING V,
WE'LL ACTUALLY GRAPH V NOT IN THIS DIRECTION,
BUT IN THE OPPOSITE DIRECTION, LIKE SO.
SO THE RESULTANT, USING THE TRIANGULAR METHOD,
WOULD BE THE VECTOR THAT STARTS AT THE ORIGIN
AND HAS ITS TERMINAL POINT HERE, 2 x VECTOR U - VECTOR V.
NOW, THE ADDITION AND SUBTRACTION OF VECTORS
IN COMPONENT FORM
IS ACTUALLY VERY STRAIGHTFORWARD.
YOU CAN SEE WHAT WE HAVE TO DO TO ADD TWO VECTORS
IN COMPONENT FORM
IS JUST ADD HORIZONTAL COMPONENTS
AND ADD THE VERTICAL COMPONENTS.
AND FOR SUBTRACTION,
WE JUST SUBTRACT THE HORIZONTAL COMPONENTS
AND SUBTRACT THE VERTICAL COMPONENTS.
IF WE HAVE THESE TWO VECTORS AND WE WANT TO FIND THEIR SUM,
WE WOULD HAVE 8 + -7 AND 4 + 3.
SO WE WOULD HAVE THE COVECTOR (1, 7).
SO AS YOU CAN SEE, IT'S VERY STRAIGHTFORWARD.
AND THEN FOR SUBTRACTION IT'S VERY SIMILAR,
EXCEPT NOW WE JUST SUBTRACT.
SO WE'LL HAVE 8 - -7 AND 4 - 3 FOR VECTOR (15, 1).
NOW LET'S TAKE A LOOK AT SCALAR MULTIPLICATION.
IT'LL PROBABLY REMIND YOU OF DISTRIBUTION.
IF YOU HAVE K x THE VECTOR U,
YOU JUST MULTIPLY EACH COMPONENT BY THE SCALAR K.
SO 1/2 x VECTOR U IS GOING TO BE 1/2 x THE VECTOR (8, 4).
WELL, 1/2 x (8, 4), 1/2 x 4, THAT WOULD BE 2.
-2 x VECTOR V,
THAT WOULD JUST BE THE VECTOR
WITH THE HORIZONTAL COMPONENT 14,
AND THE VERTICAL COMPONENT -6.
IF WE WANTED TO FIND THE SUM OF THESE TWO VECTORS,
WE WOULD FIRST MULTIPLY VECTOR U x -3.
LET'S GO AHEAD AND DO THAT FIRST.
HERE'S VECTOR U. SO WE WOULD HAVE (-24, -12).
WE WANT ADD THIS VECTOR 2 x VECTOR V,
THAT WOULD BE -46,
SO THE RESULTANT WOULD BE (-38, - 6).
LET'S FINISH OFF BY TALKING ABOUT DOT PRODUCT.
THE DOT PRODUCT OF VECTOR U
AND VECTOR V IS DENOTED WITH THIS NOTATION,
AND IT'S READ, U.V, AND IS GIVEN BY "A" x C + B x D.
SO WE FIND THE PRODUCT OF THE HORIZONTAL
AND VERTICAL COMPONENTS, AND THEN WE SUM THEM.
AND THE DOT PRODUCT DOES TELL US SOME INTERESTING INFORMATION
ABOUT THE TWO VECTORS.
IF THE DOT PRODUCT IS POSITIVE,
THE ANGLE BETWEEN THE TWO VECTORS IS ACUTE.
IF THE DOT PRODUCT IS EQUAL TO ZERO,
THE VECTORS ARE PERPENDICULAR, OR ORTHOGONAL.
IF THE DOT PRODUCT IS NEGATIVE,
THE ANGLE BETWEEN THE TWO VECTORS IS OBTUSE.
LET'S SEE IF WE CAN FIND THE DOT PRODUCT FOR THESE VECTORS.
V.W = 2 x -3 + 4 x 1, IT WOULD BE -6 + (4, -2).
SINCE THE DOT PRODUCT IS NEGATIVE,
THE ANGLE BETWEEN THESE TWO VECTORS IS AN OBTUSE ANGLE.
X.Y = 5 x -3, THAT'S -15 + 3 x 15, THAT'S 15.
THAT'S EQUAL TO ZERO.
SO THAT MEANS THESE TWO VECTORS ARE PERPENDICULAR OR ORTHOGONAL.
AND THEN LASTLY, WE MAY WANT TO REWRITE THESE VECTORS
IN COMPONENT FORM.
THIS WOULD BE (2, -3) AND THIS WOULD BE (0, -2).
SO THAT MEANS THE DOT PRODUCT WOULD EQUAL 2 x 0,
OF COURSE, THAT'S 0 + -3 x -2, THAT WOULD BE 6.
SO NOW OUR DOT PRODUCT IS POSITIVE,
SO THE ANGLE BETWEEN THESE TWO VECTORS WOULD BE ACUTE.
LET'S TAKE A LOOK AT THE GEOMETRIC INTERPRETATION
OF THE DOT PRODUCT.
IF THETA IS THE ANGLE BETWEEN THE TWO NONZERO VECTORS U AND V,
WHERE THETA IS IN THE CLOSED INTERVAL
FROM ZERO TO 180 DEGREES,
THEN U.V IS EQUAL TO THE MAGNITUDE OF U
x THE MAGNITUDE OF V x COSINE THETA.
OR IF WE SOLVE THIS FOR COSINE THETA,
WE WOULD HAVE THE EQUATION IN THIS FORM.
NOW, WE CAN DERIVE THIS FORMULA
BY USING A SKETCH SIMILAR TO THIS
AND USING THE LAW OF COSINES.
BUT RIGHT NOW WE DON'T HAVE TIME FOR THAT.
LET'S GO AHEAD AND TAKE A LOOK AT ONE PROBLEM.
WE WANT TO DETERMINE THE ANGLE BETWEEN THESE TWO VECTORS.
SO USING THIS FORMULA,
WE'RE GOING TO HAVE COSINE THETA IS EQUAL TO THE DOT PRODUCT.
SO 2 x -3 WOULD BE -6 + 1 x 1 IS 1
MAGNITUDE OF U x THE MAGNITUDE OF V.
WELL, OUR VECTORS ARE V AND W, BUT IT'S THE SAME IDEA.
SO THE MAGNITUDE OF V
IS EQUAL TO THE SQUARE ROOT OF "A" SQUARED + B SQUARED,
2 SQUARED + 1 SQUARED,
THAT WOULD BE THE SQUARE ROOT OF 5.
THE MAGNITUDE OF W WOULD BE -3 SQUARED + 1 SQUARED,
THAT WOULD BE 9 + 1.
SO WE HAVE THE SQUARE ROOT OF 5 x THE SQUARE ROOT OF 10.
SO COSINE THETA = -5/SQUARE ROOT OF 50.
IT DOES SIMPLIFY TO -- SQUARE ROOT 2,
SO WE HAVE COSINE THETA = -1/SQUARE ROOT OF 2.
REMEMBER, SINCE THIS VALUE IS NEGATIVE,
WE WILL HAVE AN OBTUSE ANGLE BETWEEN THESE TWO VECTORS.
THIS SHOULD REMIND YOU OF A REFERENCE ANGLE OF 45 DEGREES.
BUT IN THE SECOND QUADRANT IT WOULD BE 135 DEGREES.
LET'S GO AHEAD AND JUST VERIFY THAT THOUGH.
SO LET'S GO AHEAD AND VERIFY THAT WITH THE CALCULATOR.
IF WE PRESS 2nd, COSINE, WE'RE BRINGING UP INVERSE COSINE.
WE HAVE -1 DIVIDED BY THE SQUARE ROOT OF 2.
AGAIN, WE SHOULDN'T REALLY HAVE TO DO THIS,
BUT THERE'S OUR ANGLE.
THETA = 135 DEGREES.
AND, AGAIN, THIS SHOULD HAVE REMINDED YOU OF THIS TRIANGLE
WHERE THE REFERENCE ANGLE WOULD BE 45 DEGREES,
BUT IT WOULD BE IN THE SECOND QUADRANT,
GIVING US 135 DEGREES. OKAY.
I HOPE YOU'VE FOUND THIS VIDEO HELPFUL.
HAVE A GOOD DAY.