Question

Two vectors, A = 10i + 15j + 20k and B = 12i – 6j + k are given.
What is the magnitude of C = 4A - 5B
What is the scalar product AB?
What is the angle between A and B?
What is the vector product AxB?

Answer 1

What are you having troubles with?

Answer 2

see i know how to the magnitude thing
but can you help with th scalar product

Answer 3

The scalar product, also known as the dot product, is the product defined by multiplying the corresponding components of each vector and adding them together.

Answer 4

So take vectors i and j. i * j is zero as 1*0 + 0*1 + 0*0 = 0

Answer 5

?

Answer 6

Ok. Let's say we have two vectors \[x = \left(\begin{matrix}1 \\ 2 \\ 3\end{matrix}\right) y = \left(\begin{matrix}0 \\ 1 \\ -1\end{matrix}\right)\]

Answer 7

ok

Answer 8

Then the scalar product, x*y, is \[x * y = 1*0 + 2*1 + 1*(-1) = \]

Answer 9

oops

Answer 10

I mean x*y = 1*0 + 2*1 + 3 *(-1) = -1

Answer 11

ok lets start will the magnitude one
\[A= 50i + 25j + 40k\]
\[B = 42i -16j + 34k\]
\[C = 2A - B\]

Answer 12

find the magnitude of A and B

Answer 13

Do you need help with calculating the magnitude?

Answer 14

can you help with calculating magnitude of A and B
and for C i got 58i+66j+46k

Answer 15

C should be 58i + 36j + 46k

Answer 16

And the magnitude of a vector is the sqrt of the sum of its components squared. so for A, |A| = sqrt( 50^2 + 25^2 + 40^2 ) = sqrt(4752)

Answer 17

so the answer would be 68.738

Answer 18

for C this square root will be our next method

Answer 19

Nextmethod?

Answer 20

since we had found C now we will find |C|

Answer 21

Ok. So we just do what we did with A.

Answer 22

squart(50^2 + 25^ 2 + 40^2 )

Answer 23

We wouldn't use the components of A. We use the components of C.