MHTCET 11 Apr 2026 Shift 1 Paper

Concept:When two vectors are perpendicular, their dot product is zero. Use both given conditions to form equations, then solve for the angle between the vectors.Explanation:Let $\theta$ be the angle between $\bar{a}$ and $\bar{b}$.Since $\bar{a}+3\bar{b}$ is perpendicular to $7\bar{a}-5\bar{b}$, their dot product is zero:$$(\bar{a}+3\bar{b})\cdot(7\bar{a}-5\bar{b})=0$$Expanding using dot product properties:$$7|\bar{a}|^{2}-5\bar{a}\cdot\bar{b}+21\bar{a}\cdot\bar{b}-15|\bar{b}|^{2}=0$$$$7|\bar{a}|^{2}+16\bar{a}\cdot\bar{b}-15|\bar{b}|^{2}=0 \quad \ldots (1)$$Since $\bar{a}-4\bar{b}$ is perpendicular to $7\bar{a}-2\bar{b}$, their dot product is zero:$$(\bar{a}-4\bar{b})\cdot(7\bar{a}-2\bar{b})=0$$Expanding:$$7|\bar{a}|^{2}-2\bar{a}\cdot\bar{b}-28\bar{a}\cdot\bar{b}+8|\bar{b}|^{2}=0$$$$7|\bar{a}|^{2}-30\bar{a}\cdot\bar{b}+8|\bar{b}|^{2}=0 \quad \ldots (2)$$Subtracting equation (2) from equation (1):$$46\bar{a}\cdot\bar{b}-23|\bar{b}|^{2}=0$$$$2\bar{a}\cdot\bar{b}=|\bar{b}|^{2}$$$$\bar{a}\cdot\bar{b}=\frac{|\bar{b}|^{2}}{2} \quad \ldots (3)$$Substituting equation (3) into equation (2):$$7|\bar{a}|^{2}-30\left(\frac{|\bar{b}|^{2}}{2}\right)+8|\bar{b}|^{2}=0$$$$7|\bar{a}|^{2}-15|\bar{b}|^{2}+8|\bar{b}|^{2}=0$$$$7|\bar{a}|^{2}=7|\bar{b}|^{2}$$$$|\bar{a}|=|\bar{b}|$$Using the dot product formula $\bar{a}\cdot\bar{b}=|\bar{a}||\bar{b}|\cos\theta$:$$\frac{|\bar{b}|^{2}}{2}=|\bar{b}|^{2}\cos\theta$$$$\cos\theta=\frac{1}{2}$$$$\theta=\frac{\pi}{3}$$Answer:The angle between $\bar{a}$ and $\bar{b}$ is $\frac{\pi}{3}$.Correct option: D. $\frac{\pi}{3}$