Vector Algebra Part 3

Concept:Projection of a vector on another vector is given by the dot product divided by the magnitude of the base vector.Explanation:In parallelogram $ABCD$, diagonal $AC = AB + AD$.$AC = (2+1)\hat{i} + (4+2)\hat{j} + (-5+\lambda)\hat{k} = 3\hat{i} + 6\hat{j} + (\lambda-5)\hat{k}$.Given $\vec{v} = \hat{i} + \hat{j} + \hat{k}$.Dot product: $\vec{v} \cdot \vec{AC} = 3 + 6 + (\lambda-5) = \lambda + 4$.Magnitude: $|\vec{AC}| = \sqrt{3^2 + 6^2 + (\lambda-5)^2} = \sqrt{\lambda^2 - 10\lambda + 70}$.Projection length = $\frac{\vec{v} \cdot \vec{AC}}{|\vec{AC}|} = \frac{\lambda+4}{\sqrt{\lambda^2 - 10\lambda + 70}} = 1$.Square both sides: $(\lambda+4)^2 = \lambda^2 - 10\lambda + 70$.Expand: $\lambda^2 + 8\lambda + 16 = \lambda^2 - 10\lambda + 70$.Simplify: $8\lambda + 16 = -10\lambda + 70$.Thus $18\lambda = 54$, so $\lambda = 3$.Substitute $\lambda = 3$ into the quadratic: $\lambda^2 x^2 - 6\lambda x + 5 = 0$ becomes $9x^2 - 18x + 5 = 0$.Solve: $x = \frac{18 \pm \sqrt{324 - 180}}{18} = \frac{18 \pm 12}{18}$.Roots: $x = \frac{30}{18} = \frac{5}{3}$ and $x = \frac{6}{18} = \frac{1}{3}$.Given $\alpha > \beta$, so $\alpha = \frac{5}{3}$, $\beta = \frac{1}{3}$.Compute $2\alpha - \beta = 2 \cdot \frac{5}{3} - \frac{1}{3} = \frac{10}{3} - \frac{1}{3} = \frac{9}{3} = 3$.Answer:$3$ (Option A)