Question 132: Motion in 2D & Vectors

Question

The resultant of $\mathbf{A}$ and $\mathbf{B}$ is $\mathbf{R}_1$. On reversing the vector $\mathbf{B}$, the resultant becomes $\mathbf{R}_2$. What is the value of $R_1^2+R_2^2$ ?

Accepted Answer

Let's analyze the problem step by step.
When the vectors are added normally, we have:
$$\mathbf{R}_1 = \mathbf{A} + \mathbf{B}$$
The square of its magnitude is:
$$R_1^2 = A^2 + B^2 + 2\,\mathbf{A}\cdot\mathbf{B} = A^2 + B^2 + 2AB\cos	heta$$
where $$	heta$$ is the angle between vectors $$\mathbf{A}$$ and $$\mathbf{B}$$.
When vector $$\mathbf{B}$$ is reversed, the new resultant becomes:
$$\mathbf{R}_2 = \mathbf{A} - \mathbf{B}$$
Its squared magnitude is:
$$R_2^2 = A^2 + B^2 - 2\,\mathbf{A}\cdot\mathbf{B} = A^2 + B^2 - 2AB\cos	heta$$
To find $$R_1^2 + R_2^2$$, add the two expressions:
$    \begin{aligned}    R_1^2 + R_2^2 &= \left(A^2 + B^2 + 2AB\cos	hetaight) + \left(A^2 + B^2 - 2AB\cos	hetaight) \    &= 2A^2 + 2B^2 \    &= 2\left(A^2+B^2ight)    \end{aligned}    $
Thus, the value of $$R_1^2 + R_2^2$$ is:
$$\boxed{2\left(A^2+B^2ight)}$$
This corresponds to Option C.

Answer

$A^2+B^2$

Answer

$A^2-B^2$

Answer

$2\left(A^2+B^2\right)$

Answer

$2\left(A^2-B^2\right)$

Motion in 2D & Vectors

The resultant of \mathbf{A} and \mathbf{B} is \mathbf{R}_1. On reversing the vector \mathbf{B}, the resultant becomes \mathbf{R}_2. What is the value of R_1^2+R_2^2 ?

Choose the correct answer: