Find parallel vector component

[blackandyello]

Homework Statement

Resolve the position vector A of the car (measured from fixed point O)
into components parallel to OB and OC.

Figure: http://screensnapr.com/e/4D2OAz.png

Homework Equations

Projection of vector A to OC = |A| * |unit vector OC| * unit vector

The Attempt at a Solution

my solution for finding the component parallel to OC is by finding the angle of triangle AOC, then use that angle to compute for parallel component at OC. my answer ended up to be the same 3500m. but the book says, 2570m along OC.

 

[humanist rho]

A.B = ABCosθ is the projection of vector A along B.

 

[JHamm]

A.B = ABCosθ is the projection of vector A along B.

But if you take $B = A$ then you find the projection of $A$ along $A$ is $A^2$

 

[humanist rho]

But if you take $B = A$ then you find the projection of $A$ along $A$ is $A^2$

oh sorry. I mean B cosθ is the projection of B along A.

 

[asteriatic]

To find the parallel component of the position vector A, we can use the projection formula: projection of vector A to OC = |A| * |unit vector OC| * cosθ, where θ is the angle between vector A and OC.

To find the angle θ, we can use the dot product formula: A · OC = |A| * |OC| * cosθ. Rearranging this formula, we get cosθ = (A · OC) / (|A| * |OC|).

Substituting the values given in the figure, we get cosθ = (3000 * 1000 + 4000 * (-3000)) / (5000 * 4000) = -0.2. Therefore, θ = arccos(-0.2) = 101.54°.

Now, plugging this value of θ into the projection formula, we get projection of A to OC = 5000 * 1000 * cos(101.54°) = 2570m.

Therefore, the parallel component of A along OC is 2570m. This differs from the book's answer of 3500m, which may be due to rounding errors or a different method used to find the angle θ.