1.) Let P = (0,-1), Q = (-1,0), R = (1,1), S = (2,0). Are the vectors PQ and SR

Question

1.) Let P = (0,-1), Q = (-1,0), R = (1,1), S = (2,0). Are the vectors PQ and SR equivalent?

2.) Suppose v = <2,5> and u = <1,2>.
a. Calculate 2v - 9u.

b. Calculate 3(v+u).

c. Are there numbers a and b so that av + bu = <0,1>? If so, find the values. Otherwise, justify why not.

3. Find a vector of length 3 that is in the same direction as <2,-1,1>.

4.) Find two vectors of length 1 that are parallel to 2i - j +1k.

5.) Find a velocity factor that is 45 degrees relative to the x-axis, which has speed 1.  

Explanation / Answer

1)

Two vectors are said to be equivalent if they have the same magnitude and direction.

here given P = (0,-1), Q = (-1,0), R = (1,1), S = (2,0).

PQ = Q-P = (-1,0) - (0,-1) = (-1 , 1 )

SR = R-S = (1,1) - (2,0) = (-1 , 1 )

we cansee that they have same direction but different magnitude .. hence they are equivalent

2)

v = <2,5> and u = <1,2>.

a)
2v - 9u = 2 *<2,5> - 9*<1,2>.

= <4,10> - <9,18>

=< -5 , -8 >

b)

3(v+u) = 3*( <2,5> + <1,2>)

=3*(<3,7>)

=< 9 , 21 >

c)

av + bu

= a *<2,5> + b*<1,2>.

=< 2a , 5a > + < 1b , 2b >

=< 2a + b , 5a + 2b>

equating it to <0,1>

we have

2a + b = 0 --------------(1)

5a + 2b = 1 --------------(2)

we can solve this as follows

multiply equation (1) with 2 and then subtract it from equation (2)

so we have

4a + 2b = 0

- (5a + 2b = 1)

=> -a = -1

a = 1

which gives b = -2

3)

since it is given that it will have the same direction as <2,-1,1>.

then the vector will have the same direction cosines as <2,-1,1>.

so we can write the required vetor as <2a , -1a , a >

given that its magnitude is 3

so sqrt((2a)^2 + (-1a)^2 + (a)^2) = 3

6a^2 = 9

a = sqrt(1.5) = 1.225

so the required vector = <2*1.225 ,-1*1.225 ,1*1.225>.

= < 2.45 , -1.225 , 1.225 >

4)

given that the vectors are parallel so we need to find the two vectors that are aligned with the given 2i - j + k

writing it in dirction cosines form < 2 ,-1 ,1 >

then the vector will have the same direction cosines as <2,-1,1>.

so we can write the required vetor as <2a , -1a , a >

given that its magnitude is 1

so sqrt((2a)^2 + (-1a)^2 + (a)^2) = 1

6a^2 = 1

a = +/- 0.41

so the required vectors = +/- <2*0.41 ,-1*0.41 ,1*0.41>.

= < 0.82 , - 0.41 , 0.41 > and < - 0.82 , 0.41 ,- 0.41 >

5)

taking counter clock wise as positive

the vectors which are at 45 degees to X axis are i+j

so its unit vectors are <1 / sqrt(1+1) , 1/ sqrt(1+1) >

so the rwquired vector is 1/sqrt(2) i + 1/ sqrt(2) j

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