Let's solve each problem step-by-step. --- ### **Problem 1** #### **Given:** - Particles \( A \) and \( B \) start at the origin. - Particle \( A \) velocity: \( \vec{v}_A = \hat{i} - 3\hat{j} \) - Particle \( B \) moves perpendicular to \( \vec{v}_A \) and is in the first quadrant after 1 unit time. - After 1 unit time, they are \( \sqrt{26} \) units apart. #### **Step 1: Find velocity direction for \( B \)** - \( \vec{v}_B \) is perpendicular to \( \vec{v}_A \). - If \( \vec{v}_B = a\hat{i} + b\hat{j} \), then \( \vec{v}_A \cdot \vec{v}_B = 0 \): \[ (1)(a) + (-3)(b) = 0 \implies a - 3b = 0 \implies a = 3b \] - Since B is in the first quadrant, both \( a > 0 \) and \( b > 0 \). So, \( \vec{v}_B = 3b\hat{i} + b\hat{j} \). #### **Step 2: After 1 unit time, position vectors:** - \( \vec{r}_A = \vec{v}_A \times 1 = \hat{i} - 3\hat{j} \) - \( \vec{r}_B = \vec{v}_B \times 1 = 3b\hat{i} + b\hat{j} \) #### **Step 3: Distance between A and B after 1 unit time:** \[ |\vec{r}_A - \vec{r}_B| = \sqrt{[(1 - 3b)]^2 + [(-3) - b]^2} \] Set this equal to \( \sqrt{26} \): \[ [(1 - 3b)]^2 + [(-3) - b]^2 = 26 \] Expand: \[ (1 - 3b)^2 + (-3 - b)^2 = 26 \] \[ (1 - 6b + 9b^2) + (9 + 6b + b^2) = 26 \] \[ (1 + 9) + (-6b + 6b) + (9b^2 + b^2) = 26 \] \[ 10 + 0 + 10b^2 = 26 \] \[ 10b^2 = 16 \implies b^2 = \frac{8}{5} \implies b = \frac{2\sqrt{10}}{5} \] Remember \( a = 3b \): \[ a = 3 \times \frac{2\sqrt{10}}{5} = \frac{6\sqrt{10}}{5} \] #### **Step 4: Velocity of Particle B** \[ \vec{v}_B = a\hat{i} + b\hat{j} = \frac{6\sqrt{10}}{5}\hat{i} + \frac{2\sqrt{10}}{5}\hat{j} \] **Final Answer (boxed):** \[ \boxed{\vec{v}_B = \frac{6\sqrt{10}}{5}\hat{i} + \frac{2\sqrt{10}}{5}\hat{j}} \] --- ### **Problem 2** #### **Given:** - Tangent plane to \( x^2 + \frac{y^2}{4} - \frac{z^2}{3} = -1 \) at \( (1, 2, 3) \). - Three forces: \[ \vec{F}_1 = -\hat{i} + 3\hat{j} + 2\hat{k} \] \[ \vec{F}_2 = 2\hat{i} + 5\hat{j} + \hat{k} \] \[ \vec{F}_3 = 4\hat{i} - 2\hat{j} + a\hat{k} \] Object remains on the tangent plane if the total force is perpendicular to the normal vector of the tangent plane. #### **Step 1: Find the normal vector to the surface at (1, 2, 3)** Given surface: \[ F(x, y, z) = x^2 + \frac{y^2}{4} - \frac{z^2}{3} \] Gradient: \[ \nabla F = \left(2x, \frac{y}{2}, -\frac{2z}{3}\right) \] At \( (1, 2, 3) \): \[ \nabla F = \left(2, 1, -2\right) \] #### **Step 2: Net force** \[ \vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 \] Sum components: - \( \hat{i} \): \( -1 + 2 + 4 = 5 \) - \( \hat{j} \): \( 3 + 5 - 2 = 6 \) - \( \hat{k} \): \( 2 + 1 + a = a + 3 \) So, \[ \vec{F}_{\text{net}} = 5\hat{i} + 6\hat{j} + (a+3)\hat{k} \] #### **Step 3: Force perpendicular to normal** Condition: \( \vec{F}_{\text{net}} \cdot \nabla F = 0 \): \[ (5)(2) + (6)(1) + (a+3)(-2) = 0 \] \[ 10 + 6 - 2(a+3) = 0 \] \[ 16 - 2a - 6 = 0 \] \[ 10 - 2a = 0 \implies 2a = 10 \implies a = 5 \] **Final Answer (boxed):** \[ \boxed{a = 5} \] --- **Summary of Answers:** 1. \(\boxed{\vec{v}_B = \frac{6\sqrt{10}}{5}\hat{i} + \frac{2\sqrt{10}}{5}\hat{j}}\) 2. \(\boxed{a = 5}\)