If # bb(ul u) = << −1,0,2 >> #, # bb(ul v) = << 3,1,2 >> # and # bb(ul w) = << 1,−2,−2 >> # then find?

(a) # 3bb(ul v)−2bb(ul u) #

(b) # kbb(ul u) + bb(ul v) + kbb(ul w) #

(c) # || bb(vec(PQ)) || # where # bb(vec(OP)) = -3bb(ul u) # and # bb(vec(OQ)) = bb(ul v) + 5bb(ul w)#

(d) Area of the parallelogram with sides # bb(ul u)# and #2bb(ul v) #

1 Answer
Aug 24, 2017

We have:

# bb(ul u) = << −1,0,2 >> #
# bb(ul v) = << 3,1,2 >> #
# bb(ul w) = << 1,−2,−2 >> #

Then:

Part (a):

# 3bb(ul v)−2bb(ul u) = 3 << 3,1,2 >> - 2<< −1,0,2 >> #
# " " = << 9,3,6 >> - << −2,0,4 >> #
# " " = << 11,3,2 >> #

Part (b):

# kbb(ul u) + bb(ul v) + kbb(ul w) = k<< −1,0,2 >> + << 3,1,2 >> + k<< 1,−2,−2 >> #

# " " = << −k,0,2k >> + << 3,1,2 >> + << k,−2k,−2k >> #
# " " = << −k+3+k,0+1-2k,2k+2-2k >> #
# " " = << 3,1-2k,2 >> #

Part (c):

Let:

# bb(vec(OP)) = -3bb(ul u) #
# " " = -3<< −1,0,2 >> #
# " " = << 3,0,-6 >> #

# bb(vec(OQ)) = bb(ul v) + 5bb(ul w)#
# " " = << 3,1,2 >> + 5<< 1,−2,−2 >> #
# " " = << 3,1,2 >> + << 5,−10,−10 >> #
# " " = << 8,-9,-8 >> #

Then:

# bb(vec(PQ)) = bb(vec(OQ)) - bb(vec(OP)) #
# " " = << 8,-9,-8 >> - << 3,0,-6 >> #
# " " = << 5,-9,-2 >> #

Finally:

# || bb(vec(PQ)) || = || << 5,-9,-2 >> || #
# " " = sqrt( 5^2 + (-9)^2 +(-2)^2 ) #
# " " = sqrt( 25 + 81 + 4 ) #
# " " = sqrt( 110 ) #

Part (c):

The area is given by the magnitude of the cross product:

# A = || bb(ul u) xx bb(ul v) || #

we calculate the cross product using:

# bb(ul u) xx bb(ul v) = << −1,0,2 >> xx << 3,1,2 >> #
# \ \ \ = | ( bb(ul(hat i)),bb(ul(hat j)),bb(ul(hat k))), (−1,0,2), (3,1,2) | #

# \ \ \ = | (0,2),(1,2)|bb(ul(hat i)) - | (-1,2),(3,2)|bb(ul(hat j)) + | (-1,0),(3,1)|bb(ul(hat k)) #

# \ \ \ = (0-2) bb(ul(hat i)) - (-2-6)bb(ul(hat j)) + (-1-0)bb(ul(hat k)) #

# \ \ \ = -2 bb(ul(hat i)) +8bb(ul(hat j)) -bb(ul(hat k)) #

Hence,

# A = || bb(ul u) xx bb(ul v) || #
# \ \ \ = sqrt((-2)^2 + 8^2+(-1)^2) #
# \ \ \ = sqrt(4+64+1) || #
# \ \ \ = sqrt(69) #