MHTCET 25 Apr 2025 Shift 2 Paper

Concept:A cross-product equation $\bar b \times \bar c = \bar a$ does not determine $\bar b$ completely; the component of $\bar b$ parallel to $\bar c$ remains arbitrary.Explanation:Let $\bar b = x\hat{i}+y\hat{j}+z\hat{k}$.Then$$\bar b \times \bar c = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ x & y & z\\ 5 & -3 & 2 \end{vmatrix} = (2y+3z)\hat{i} + (5z-2x)\hat{j} + (-3x-5y)\hat{k}.$$Since $\bar b \times \bar c = \hat{i}+\hat{j}-\hat{k}$,$2y+3z=1$,$5z-2x=1$,$-3x-5y=-1$.Solving the first two equations gives$x=\frac{5z-1}{2}$ and $y=\frac{1-3z}{2}$.These values also satisfy the third equation for every real value of $z$, so $z$ is a free parameter.Hence infinitely many vectors $\bar b$ are possible.For these vectors,$$|\bar b|^2 = \left(\frac{5z-1}{2}\right)^2+\left(\frac{1-3z}{2}\right)^2+z^2 = \frac{19z^2-8z+1}{2}.$$This expression changes with $z$; for example, $z=0$ gives $|\bar b|=\frac{1}{\sqrt{2}}$, while $z=1$ gives $|\bar b|=\sqrt{6}$.Answer:$|\bar b|$ cannot be determined uniquely from the given condition.Therefore, none of the given options A-D is correct; the data are insufficient.