Suppose ( aru=langle-5,0,2 angle ) and ( arv=langle...

Suppose ( ar{u}=langle-5,0,2 angle ) and ( ar{v}=langle 3,-4,0 angle ). Then: [ egin{array}{r} ar{u}+ar{v}= \ ar{u}-ar{v}= \ ar{v}-ar{u}= \ 8 ar{u}= \ -frac{1}{3} ar{v}= \ 8 ar{u}-3 ar{v}= end{array} ]

Transcript

00:01 Suppose u is a vector negative 5, 0, 2 and v is a vector 3, negative 4, 0.

00:08 We want to find the sum of u and v, u minus v, v minus u, a times u, negative one -third times v, and the linear combination 8u minus 3b.

00:22 So first let's calculate the sum of u and v, that is negative 5, 0, 2, plus 3, negative 4, 0.

00:37 And we know this operation is done component -wise and we get a new vector which is negative 5 plus 3, first component, second 0 plus negative 4 or 0 minus 4 and the third one is 2 plus 0.

00:56 And so we get negative 2, negative 4 and 2.

01:07 So u plus v is a vector negative 2, negative 4, 2.

01:20 Now let's calculate u minus v, that is negative 5, 0, 2, minus 3, negative 4, 0.

01:38 And that is negative 5 minus 3, 0 minus negative 4 is 0 plus 4 and 2 minus 0.

01:57 And that is equal to negative 8, 4 and 2.

02:11 And so we can say that u minus v is a vector negative 8, 4, 2.

02:20 Okay, now part 3, v minus u, that will be vector 3, negative 4, 0, minus vector negative 5, 0, 2.

02:47 And then that gives us 3 minus negative 5 is 3 plus 5, negative 4 minus 0 and 0 minus 2.

03:04 Finally this is equal to vector 8, negative 4 and negative 2.

03:22 So v minus u is the vector 8, negative 4, negative 2.

03:34 Which is the opposite or negative 1 times the vector u minus v found in part 2.

03:41 Good, so in part 4 we're going to calculate 8 times vector u, that is 8 times vector negative 5, 0, 2.

04:01 And the scalar times the vector is a new vector such that each component is the component, the corresponding component of the given vector times the scalar.

04:11 So we get 8 times negative 5, first component, then 8 times 0, second component and the third component 8 times 2.

04:26 That is we multiply the scalar by each of the components of the vector and that's a new vector...

Question

Suppose ( ar{u}=langle-5,0,2 angle ) and ( ar{v}=langle 3,-4,0 angle ). Then: [ egin{array}{r} ar{u}+ar{v}= \ ar{u}-ar{v}= \ ar{v}-ar{u}= \ 8 ar{u}= \ -frac{1}{3} ar{v}= \ 8 ar{u}-3 ar{v}= end{array} ]

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Suppose ( ar{u}=langle-5,0,2 angle ) and ( ar{v}=langle 3,-4,0 angle ). Then: [ egin{array}{r} ar{u}+ar{v}= \ ar{u}-ar{v}= \ ar{v}-ar{u}= \ 8 ar{u}= \ -frac{1}{3} ar{v}= \ 8 ar{u}-3 ar{v}= end{array} ]