This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
To determine the unit vector in the y-z plane that is perpendicular to A = i + j + k, let the required unit vector be u. Step 1: Define the vector u based on the given conditions. Since the vector is in the y-z plane, its x-component must be 0. Let u = yj + zk. Step 2: Use the condition that u is a unit vector. The magnitude of a unit vector is 1. |u| = sqrt(y^2 + z^2) = 1 Squaring both sides gives: y^2 + z^2 = 1 (*) Step 3: Use the condition that u is perpendicular to A. If two vectors are perpendicular, their dot product is 0. u · A = 0 (0i + yj + zk) · (i + j + k) = 0 (0)(1) + (y)(1) + (z)(1) = 0 y + z = 0 This implies: z = -y (**) Step 4: Solve the system of equations. Substitute equation (*) into equation (): y^2 + (-y)^2 = 1 y^2 + y^2 = 1 2y^2 = 1 y^2 = (1)/(2) y = ± sqrt((1)/(2)) = ± (1)/(sqrt(2)) = ± sqrt(2)2 Step 5: Find the corresponding values for z and the unit vectors. Case 1: If y = sqrt(2)2 Using z = -y, we get z = -sqrt(2)2. So, the unit vector is u_1 = sqrt(2)2j - sqrt(2)2k. Case 2: If y = -sqrt(2)2 Using z = -y, we get z = -(-sqrt(2)2) = sqrt(2)2. So, the unit vector is u_2 = -sqrt(2)2j + sqrt(2)2k. Both vectors satisfy the given conditions. The unit vectors are sqrt(2)2j - sqrt(2)2k and -sqrt(2)2j + sqrt(2)2k.