1.) Find a ⋅ b. a = 7i - 4j, b = 4i + 3j <28, -12> 16 -40 <11, -1> 2.)Let u = <-4, 1>, v = <-1, 6>. Find -2u + 4v. <4, 22> <4, 7> <12, -26> <10, -14>

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1.) Find a ⋅ b. a = 7i - 4j, b = 4i + 3j <28, -12> 16 -40 <11, -1> 2.)Let u = <-4, 1>, v = <-1, 6>. Find -2u + 4v. <4, 22> <4, 7> <12, -26> <10, -14>

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is that a-b
@princeharryyy a times b
the product @em2000

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Other answers:

yes
multiply i by i and j by j
i got <11, -1> for the final answer, is that what you got? @princeharryyy
u added but we need to multiply
try A
what happened @em2000
7*4 is 28 and -4*3 is -12
@princeharryyy gotcha that makes sense. i got <12, -26> for #2 is that wrong?
but before i tell you about the first and second answer, are you doing scalar vectors.
then it's my bad. the first ones answer won't be A part
@princeharryyy actually im not sure, the instructions dont specify
< and > means vectors that's why i asked
let me tell how they work
i.i =1, j.j =1 i.j =0 (iand j are at 90 degrees)
|dw:1435096143456:dw|
i, j, k are all at 90 degrees any two quantities when multiplied at 90 degrees gives 0 as i.j.cos90 =0, j.k.cos90 =0, k.i.cos90 =0
are you getting it @em2000
So, for the first question a = 7i - 4j, b = 4i + 3j a.b is written as a.b = (7i - 4j).(4i - 3j)
so, a.b = 7*4*i.i -7*3*i.j -4*4*j.j-4*3j.j so, a.b = 28-0-0-12 =16 answer @em2000
for the first question
now for the second
Let u = <-4, 1>, v = <-1, 6>. Find -2u + 4v. => u = -4i+1j, v = -1i+6j -2u+4v = -2(-4i+1j) +4(-1i+6j) = 8i-2j -4i+24j = 4i+22j => <4,22> answer @em2000
which is option A @em2000 . Did you get it.
Scalar multiplication of vectors: \(\bf{A} = \sf A_x + B_y\) \(\bf B = \sf B_x + B_y\) \(\bf A \cdot B \sf = A_xB_x + A_yB_y\) Answer is scalar.
@princeharryyy thank u so much
Do you know the answers now @em2000
for first one it is 16 and for second one i t is <4,22> @em2000
:)
\(\bf A \cdot B \sf = A_xB_x + A_yB_y = (7)(4) + (-4)(3) = 28 - 12 = 16\)
|dw:1435096919904:dw|
thanks @mathstudent55. Sorry, for the medal she patiently tried to understand the concept. That's why I gave her the medal. Sorry, that I can't offer them to two people on one question. @em2000 could you please give a medal to @mathstudent55
@mathstudent55 tried to help you out as well.
no not to me @em2000 , but to @mathstudent55
@princeharryyy Don't worry about the medal. I gave you one because I thought your explanation was great, and you deserved it!
I never worry about medals @mathstudent55 but most people at open study help others for medals.

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