physicsvector analysis
I was able to determine a resultant vector based on the sum of two vectors and told to express them in vector units. Here is my answer that was correct:
$\overrightarrow{C}=12.01\hat{i}+14.62\hat{j}$
I was told to then determine the magnitude of $\overrightarrow{C}$ but have no idea what to do.
Any tips are greatly appreciated.
The vector $T$ is the component of $V$ perpendicular to $CD$. Assuming the center of the circle $C$ is the origin, this is just $$T = V - \frac{D(D\cdot V)}{\lVert D\rVert^2}.$$
If you want the signed magnitude of the vector, with positive being anticlockwise and negative being clockwise, you should take the cross product of $D/\lVert D\rVert$ and $V$ instead. In two dimensions, the cross product of two vectors is a scalar, $$ u \times v = \begin{vmatrix}u_x & v_x \\ u_y & v_y\end{vmatrix} = u_xv_y - u_yv_x. $$
If the circle is not centered at the origin, just replace $D$ with $D - C$ in all of the above expressions.
To the problem with vectors, we have to break the trip up into two parts, I'll call them $A$ (airport to change heading) and $B$ (change heading to emergency landing) like you did. From the picture we can see that the vector $A$ uses an angle of $68^{\circ}$ from the vertical. So $A = \langle 140\sin(68^{\circ}), 140\cos(68^{\circ})\rangle \approx \langle 129.806, 52.445\rangle$. You could also have done $A = \langle 140\cos(22^{\circ}), 140\sin(22^{\circ})\rangle$ if you wanted to work with the angle formed between $A$ and the horizontal. For vector $B$, we have the angle with the horizontal but it is below the horizontal. So the vertical component of $B$ is negative. $B = \langle 255\cos(48^{\circ}), -255\sin(48^{\circ})\rangle \approx \langle 170.628, -189.502\rangle$. The resultant is $R = A+B = \langle 300.434, -137.507\rangle$. $|R| = \sqrt{(300.434)^2+(-137.507)^2} \approx 330.220$ km. The angle they need to travel at is $\tan^{-1}\left(\frac{137.507}{300.434}\right) \approx 27.24^{\circ}$ below the horizontal or south of east.
Best Answer
In general, if $\vec{x} = a\vec{i}+b\vec{j}$, then the magnitude of $\vec{x}$, denoted $\lVert \vec{x}\rVert,$ is given by $\lVert \vec{x}\rVert = \sqrt{a^2+b^2}$. Does this help?