### Related Formula |vecA| = sqrtA_x^2 + A_y^2$|\vec{A}| = \sqrt{A_x^2 + A_y^2}$ vecN = |vecA| hatB$\vec{N} = |\vec{A}| \hat{B}$ ### Core Logic We need to find a new vector vecN$\vec{N}$ that has the magnitude of vecA$\vec{A}$ and the direction of vecB$\vec{B}$. Magnitude of vecA$\vec{A}$: |vecA| = sqrt3^2 + 4^2 = sqrt25 = 5$|\vec{A}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$. Unit vector in the direction of vecB$\vec{B}$: hatB = fracvecB|vecB| = frac4hati + 3hatjsqrt4^2 + 3^2 = frac4hati + 3hatj5$\hat{B} = \frac{\vec{B}}{|\vec{B}|} = \frac{4\hat{i} + 3\hat{j}}{\sqrt{4^2 + 3^2}} = \frac{4\hat{i} + 3\hat{j}}{5}$. ### Step 1: Construct the Vector vecN = |vecA| hatB = 5 left( frac4hati + 3hatj5 right)$\vec{N} = |\vec{A}| \hat{B} = 5 \left( \frac{4\hat{i} + 3\hat{j}}{5} \right)$ vecN = 4hati + 3hatj$\vec{N} = 4\hat{i} + 3\hat{j}$ ### Step 2: Match Components The x$x$ and y$y$ components are given as x$x$ and 3$3$. From vecN = 4hati + 3hatj$\vec{N} = 4\hat{i} + 3\hat{j}$, we see the x$x$-component is 4$4$. Therefore, x = 4$x = 4$. ### Pattern Recognition Constructing a vector matching magnitude and direction is a simple scalar multiplication of the desired magnitude by the target direction's unit vector. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Motion in a Plane