# Math Homework Help Vectors Grade 12?

I'm studying for my math test, but I can't seem to understand these two questions:

1)The angle between the vectors A and B is 120 degrees. The magnitude of A is 2 and the magnitude of B is 4. Determine the scalar projection of 3A+4B on B

2)The vector R is twice as long as the vector S. The angle between the vectors is 110 degrees.The vector projection of S on R is (2, -1, 7). Determine R

Please show ALL steps/work.

### 2 Answers

- az_lenderLv 710 months agoFavorite Answer
(1) will be easier if we take a specific case. Let B = 4i, and then A could be -i+j*sqrt(3). Then 3A + 4B would be 13i + 3j*sqrt(3). The projection of the latter vector onto B is simply its projection on the x-axis, which has scalar magnitude 13.

(2) If the vector projection of S on R is (2,-1,7), then R has the same direction as (2,-1,7).

Formally, (2,-1,7) = (S "dot" R/ ||R||)(R / ||R||).The quantity in the last parentheses is the unit vector in the direction (2,-1,7), which would be (2,-1,7)/sqrt(54).

Therefore, sqrt(54) = (S "dot" R)/ ||R||, and

||R|| sqrt(54) = ||R|| ||S|| cos(110) =>

||S|| = sqrt(54)/cos(110).

As the cos(110) is negative, I guess I can use the negative square root of 54, and get

||S|| = 11.603... .

And so ||R|| = 23.206...

and R = (23.206)*(2,-1,7)/sqrt(54)

= 3.158*(2,-1,7) = (6.316,-3.158, 22 point something, use calculator)

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- ted sLv 710 months ago
1)....you know that W dot S = | W || S | cos <(W,S)....and scalar projection of W on S is | W | cos <(W , S )....

let W = 3 A + 4 B & S = B......( 3 A + 4 B ) dot( B ) = 3 ( A dot B) + 4 | B |²....

projection is { 3 | A | | B | cos < (A,B)+ 4 | B |² } / | B | = 3 (2) ( -1 / 2) + 4 (4) =13

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