Question

Physics Vectors Help!!! Am I supposed to somehow use the Pythagorean theorem to answer this? I've tried multiple times to answer, but what is the best way to solve this sort of problem???

Answer 1

Consider that,

`vec M` is the sum of `vec P` and `vec Q`. Moreover, consider that,

`vec P` and `vec Q` have been resolved into their components in the question. With this data, we may find the magnitude via,

`V = sqrt(V_x^2+V_y^2)`

Thus,

`abs(P) = sqrt(21^2+6^2) approx 21cm`, and

`abs(Q) = sqrt(25^2+9^2) approx 27cm`

Recall that `vec M` is the sum of the preceding vectors,

`M_x = P_x+Q_x = 4cm`, and
`M_y = P_x+Q_x = 15cm`

Therefore, the magnitude of the resultant vector `vec M` is,

`abs(M) = sqrt(4^2+15^2) approx 16cm`

Hence, I would wager that,

is a fairly accurate sketch of the vectors, color coded per the question's specifications.

I'm open to feedback if I made a mistake!

Answer 2

`"please review the details below."`

Explanation:

• The P vector and its components are shown on the drawing.

• You can find the magnitude of the P vector with the Pythagorean theorem.

`P=sqrt((-21)^2+6^2)=sqrt(441+36)=21.84" cm"`

• The Q vector and its components are shown on the drawing.

• Magnitude of Q Vector :

`P=sqrt((25)^2+9^2)=sqrt(625+81)=26.57 " cm"`

• View of the P and Q vectors.

• Let's draw the vector M = P + Q.

`vec M_x=vec P_x+vec Q_x=-21+25=+4" cm"`

`vec M_y=vec P_y+vec Q_y=+6+9=+15" cm"`

• The magnitude of M=P+Q:

`M=sqrt(4^2+15^2)=sqrt(16+225)=sqrt(241)=15.52 " cm"`