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Let v = −1 , −1 , 1 and w = 2 , 0 , 1 . Find a unit vector in the direction of 5 v + 3 w .

1 3 10 , 5 3 10 , 8 3 10

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Throwing a forward pass

A quarterback is standing on the football field preparing to throw a pass. His receiver is standing 20 yd down the field and 15 yd to the quarterback’s left. The quarterback throws the ball at a velocity of 60 mph toward the receiver at an upward angle of 30 ° (see the following figure). Write the initial velocity vector of the ball, v , in component form.

This figure is an image of two football players with the first one throwing the football to the second one. There is a line segment from each player to the bottom of the image. The distance from the first player to the bottom of the image is 20 yards. The distance from the second player to the same point on the bottom of the image is 15 yards. The two line segments are perpendicular. There is a broken line segment from the first player to the second player. There is a vector from the first player. The angle between the broken line and the vector is 30 degrees.

The first thing we want to do is find a vector in the same direction as the velocity vector of the ball. We then scale the vector appropriately so that it has the right magnitude. Consider the vector w extending from the quarterback’s arm to a point directly above the receiver’s head at an angle of 30 ° (see the following figure). This vector would have the same direction as v , but it may not have the right magnitude.

This figure is the image of two football players with the first player throwing the football to the second player. The distance between the two players is represented with a broken line segment. There is a vector from the first player. The angle between the vector and the broken line segment is 30 degrees. There is a vertical broken line segment from the second player. Also, there is a right triangle formed from the two broken line segments and the vector from the first player is labeled “w” and is the hypotenuse.

The receiver is 20 yd down the field and 15 yd to the quarterback’s left. Therefore, the straight-line distance from the quarterback to the receiver is

Dist from QB to receiver = 15 2 + 20 2 = 225 + 400 = 625 = 25 yd .

We have 25 w = cos 30 ° . Then the magnitude of w is given by

w = 25 cos 30 ° = 25 · 2 3 = 50 3 yd

and the vertical distance from the receiver to the terminal point of w is

Vert dist from receiver to terminal point of w = w sin 30 ° = 50 3 · 1 2 = 25 3 yd .

Then w = 20 , 15 , 25 3 , and has the same direction as v .

Recall, though, that we calculated the magnitude of w to be w = 50 3 , and v has magnitude 60 mph. So, we need to multiply vector w by an appropriate constant, k . We want to find a value of k so that k w = 60 mph. We have

k w = k w = k 50 3 mph,

so we want

k 50 3 = 60 k = 60 3 50 k = 6 3 5 .

Then

v = k w = k 20 , 15 , 25 3 = 6 3 5 20 , 15 , 25 3 = 24 3 , 18 3 , 30 .

Let’s double-check that v = 60 . We have

v = ( 24 3 ) 2 + ( 18 3 ) 2 + ( 30 ) 2 = 1728 + 972 + 900 = 3600 = 60 mph .

So, we have found the correct components for v .

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Assume the quarterback and the receiver are in the same place as in the previous example. This time, however, the quarterback throws the ball at velocity of 40 mph and an angle of 45 ° . Write the initial velocity vector of the ball, v , in component form.

v = 16 2 , 12 2 , 20 2

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Key concepts

  • The three-dimensional coordinate system is built around a set of three axes that intersect at right angles at a single point, the origin. Ordered triples ( x , y , z ) are used to describe the location of a point in space.
  • The distance d between points ( x 1 , y 1 , z 1 ) and ( x 2 , y 2 , z 2 ) is given by the formula
    d = ( x 2 x 1 ) 2 + ( y 2 y 1 ) 2 + ( z 2 z 1 ) 2 .
  • In three dimensions, the equations x = a , y = b , and z = c describe planes that are parallel to the coordinate planes.
  • The standard equation of a sphere with center ( a , b , c ) and radius r is
    ( x a ) 2 + ( y b ) 2 + ( z c ) 2 = r 2 .
  • In three dimensions, as in two, vectors are commonly expressed in component form, v = x , y , z , or in terms of the standard unit vectors, x i + y j + z k .
  • Properties of vectors in space are a natural extension of the properties for vectors in a plane. Let v = x 1 , y 1 , z 1 and w = x 2 , y 2 , z 2 be vectors, and let k be a scalar.
    • Scalar multiplication: k v = k x 1 , k y 1 , k z 1
    • Vector addition: v + w = x 1 , y 1 , z 1 + x 2 , y 2 , z 2 = x 1 + x 2 , y 1 + y 2 , z 1 + z 2
    • Vector subtraction: v w = x 1 , y 1 , z 1 x 2 , y 2 , z 2 = x 1 x 2 , y 1 y 2 , z 1 z 2
    • Vector magnitude: v = x 1 2 + y 1 2 + z 1 2
    • Unit vector in the direction of v: v v = 1 v x 1 , y 1 , z 1 = x 1 v , y 1 v , z 1 v , v 0

Questions & Answers

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Mostly, they use nano carbon for electronics and for materials to be strengthened.
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carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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