### Related Formula The shortest distance between two skew lines vecr = veca_1 + svecb_1$\vec{r} = \vec{a}_1 + s\vec{b}_1$ and vecr = veca_2 + tvecb_2$\vec{r} = \vec{a}_2 + t\vec{b}_2$ is given by: d = frac|(veca_2 - veca_1) cdot (vecb_1 times vecb_2)||vecb_1 times vecb_2|$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ ### Core Logic Identify the vectors from the equations of the lines: - Line 1 passes through veca_1 = lambdahati + 2hatj + hatk$\vec{a}_1 = \lambda\hat{i} + 2\hat{j} + \hat{k}$ with direction vecb_1 = -2hati + hatj + hatk$\vec{b}_1 = -2\hat{i} + \hat{j} + \hat{k}$. - Line 2 passes through veca_2 = sqrt3hati + hatj + 2hatk$\vec{a}_2 = \sqrt{3}\hat{i} + \hat{j} + 2\hat{k}$ with direction vecb_2 = hati - 2hatj + hatk$\vec{b}_2 = \hat{i} - 2\hat{j} + \hat{k}$. Calculate the coordinate difference vector: veca_2 - veca_1 = (sqrt3 - lambda)hati - hatj + hatk$\vec{a}_2 - \vec{a}_1 = (\sqrt{3} - \lambda)\hat{i} - \hat{j} + \hat{k}$ ### Step 1: Compute the Cross Product of the Direction Vectors Find the cross product matrix: vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ -2 & 1 & 1 \\ 1 & -2 & 1 endvmatrix$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix}$ vecb_1 times vecb_2 = hati(1 - (-2)) - hatj(-2 - 1) + hatk(4 - 1) = 3hati + 3hatj + 3hatk$\vec{b}_1 \times \vec{b}_2 = \hat{i}(1 - (-2)) - \hat{j}(-2 - 1) + \hat{k}(4 - 1) = 3\hat{i} + 3\hat{j} + 3\hat{k}$ Compute its magnitude: |vecb_1 times vecb_2| = sqrt3^2 + 3^2 + 3^2 = sqrt27 = 3sqrt3$|\vec{b}_1 \times \vec{b}_2| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$

### Step 2: Solve the Absolute Distance Equation for lambda Substitute these values into the shortest distance formula: 1 = frac|((sqrt3 - lambda)hati - hatj + hatk) cdot (3hati + 3hatj + 3hatk)|3sqrt3$1 = \frac{|((\sqrt{3} - \lambda)\hat{i} - \hat{j} + \hat{k}) \cdot (3\hat{i} + 3\hat{j} + 3\hat{k})|}{3\sqrt{3}}$ 1 = frac|3(sqrt3 - lambda) - 3 + 3|3sqrt3 = frac3|sqrt3 - lambda|3sqrt3 = frac|sqrt3 - lambda|sqrt3$1 = \frac{|3(\sqrt{3} - \lambda) - 3 + 3|}{3\sqrt{3}} = \frac{3|\sqrt{3} - \lambda|}{3\sqrt{3}} = \frac{|\sqrt{3} - \lambda|}{\sqrt{3}}$ This gives: |sqrt3 - lambda| = sqrt3$|\sqrt{3} - \lambda| = \sqrt{3}$ Unfolding the absolute value gives two cases: - **Case A:** sqrt3 - lambda = sqrt3 implies lambda = 0$\sqrt{3} - \lambda = \sqrt{3} \implies \lambda = 0$ - **Case B:** sqrt3 - lambda = -sqrt3 implies lambda = 2sqrt3$\sqrt{3} - \lambda = -\sqrt{3} \implies \lambda = 2\sqrt{3}$ ### Step 3: Calculate the Final Sum The sum of all possible values of lambda$\lambda$ is: textSum = 0 + 2sqrt3 = 2sqrt3$\text{Sum} = 0 + 2\sqrt{3} = 2\sqrt{3}$ ### Pattern Recognition Sees: Shortest distance setup between vector paths containing free variables. Shortcut: In equations like |c - lambda| = d$|c - \lambda| = d$, the sum of the roots is simply equal to 2c$2c$, because the roots are symmetrically balanced around the center point c$c$. Thus, textSum = 2 times sqrt3 = 2sqrt3$\text{Sum} = 2 \times \sqrt{3} = 2\sqrt{3}$ holds instantly without calculating individual values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Vector Algebra