### Related Formula By orthogonal vector decomposition (Pythagorean property): |vecv|^2 = |vecv_1|^2 + |vecv_2|^2 quad textwhen vecv_1 cdot vecv_2 = 0$|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2 \quad \text{when } \vec{v}_1 \cdot \vec{v}_2 = 0$ ### Core Logic Compute lambda$\lambda$ using dot product formula: costheta = fracvecu cdot vecv|vecu||vecv| implies fracsqrt52sqrt7 = frac3(2) + (-1)(1)sqrt3^2 + (-1)^2 sqrt2^2 + 1^2 + (-lambda)^2$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \implies \frac{\sqrt{5}}{2\sqrt{7}} = \frac{3(2) + (-1)(1)}{\sqrt{3^2 + (-1)^2} \sqrt{2^2 + 1^2 + (-\lambda)^2}}$ fracsqrt52sqrt7 = frac5sqrt10sqrt5 + lambda^2 implies frac12sqrt7 = fracsqrt5sqrt10sqrt5 + lambda^2 = frac1sqrt2sqrt5 + lambda^2$\frac{\sqrt{5}}{2\sqrt{7}} = \frac{5}{\sqrt{10}\sqrt{5 + \lambda^2}} \implies \frac{1}{2\sqrt{7}} = \frac{\sqrt{5}}{\sqrt{10}\sqrt{5 + \lambda^2}} = \frac{1}{\sqrt{2}\sqrt{5 + \lambda^2}}$ ### Step 1: Solve for lambda Square both sides of equation: frac128 = frac12(5 + lambda^2) implies 2(5 + lambda^2) = 28 implies 5 + lambda^2 = 14 implies lambda^2 = 9 implies lambda = 3$\frac{1}{28} = \frac{1}{2(5 + \lambda^2)} \implies 2(5 + \lambda^2) = 28 \implies 5 + \lambda^2 = 14 \implies \lambda^2 = 9 \implies \lambda = 3$ Since vecv = 2hati + hatj - 3hatk$\vec{v} = 2\hat{i} + \hat{j} - 3\hat{k}$. ### Step 2: Apply Identity Since components are orthogonal, direct magnitude squared holds: |vecv_1|^2 + |vecv_2|^2 = |vecv|^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14$|\vec{v}_1|^2 + |\vec{v}_2|^2 = |\vec{v}|^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14$ ### Pattern Recognition Do not waste time explicitly projecting components vecv_1$\vec{v}_1$ and vecv_2$\vec{v}_2$ if only the sum of their squared magnitudes is requested. The scalar length matches the total vector length invariant under any orthogonal basis change. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra