### Related Formula The line passing through veca$\vec{a}$ and \parallel to direction vector vecv = lhatmathbfi + mhatmathbfj + nhatmathbfk$\vec{v} = l\hat{\mathbf{i}} + m\hat{\mathbf{j}} + n\hat{\mathbf{k}}$ is written in symmetric form as: fracx - x_al = fracy - y_am = fracz - z_an$\frac{x - x_a}{l} = \frac{y - y_a}{m} = \frac{z - z_a}{n}$ ### Core Logic Formulate the equation of the line passing through (-1, 2, 1)$(-1, 2, 1)$ with direction vector components (2, 3, 4)$(2, 3, 4)$: L_1: fracx + 12 = fracy - 23 = fracz - 14 = lambda quad dots (1)$L_1: \frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z - 1}{4} = \lambda \quad \dots (1)$

Any generic point on this line can be written as: P = (2lambda - 1, \, 3lambda + 2, \, 4lambda + 1)$P = (2\lambda - 1, \, 3\lambda + 2, \, 4\lambda + 1)$ The second given line equation is: L_2: fracx + 23 = fracy - 32 = fracz - 41 = mu quad dots (2)$L_2: \frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z - 4}{1} = \mu \quad \dots (2)$ Any generic point on line L_2$L_2$ is: P' = (3mu - 2, \, 2mu + 3, \, mu + 4)$P' = (3\mu - 2, \, 2\mu + 3, \, \mu + 4)$ ### Step 1: Compute the Point of Intersection At the point of intersection P$P$, equate the coordinates from both lines: 2lambda - 1 = 3mu - 2 implies 2lambda - 3mu = -1 quad dots (3)$2\lambda - 1 = 3\mu - 2 \implies 2\lambda - 3\mu = -1 \quad \dots (3)$ 3lambda + 2 = 2mu + 3 implies 3lambda - 2mu = 1 quad dots (4)$3\lambda + 2 = 2\mu + 3 \implies 3\lambda - 2\mu = 1 \quad \dots (4)$ 4lambda + 1 = mu + 4 implies 4lambda - mu = 3 quad dots (5)$4\lambda + 1 = \mu + 4 \implies 4\lambda - \mu = 3 \quad \dots (5)$ Solving equations (3) and (4) simultaneously gives: lambda = 1, quad mu = 1$\lambda = 1, \quad \mu = 1$ Verify these values using equation (5): 4(1) - 1 = 3 quad (textSatisfied)$4(1) - 1 = 3 \quad (\text{Satisfied})$ Thus, substituting lambda = 1$\lambda = 1$ gives the coordinates of point P$P$: P = (2(1) - 1, \, 3(1) + 2, \, 4(1) + 1) = (1, 5, 5)$P = (2(1) - 1, \, 3(1) + 2, \, 4(1) + 1) = (1, 5, 5)$ ### Step 2: Calculate Euclidean Distance PQ Find the distance between P(1, 5, 5)$P(1, 5, 5)$ and Q(4, -5, 1)$Q(4, -5, 1)$ using the 3D distance formula: PQ = sqrt(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2$PQ = \sqrt{(4 - 1)^2 + (-5 - 5)^2 + (1 - 5)^2}$ PQ = sqrt3^2 + (-10)^2 + (-4)^2 = sqrt9 + 100 + 16$PQ = \sqrt{3^2 + (-10)^2 + (-4)^2} = \sqrt{9 + 100 + 16}$ PQ = sqrt125 = 5sqrt5$PQ = \sqrt{125} = 5\sqrt{5}$ ### Pattern Recognition When finding the intersection point of two 3D lines, always use the third coordinate equation to verify the parameter values obtained from the first two equations to ensure consistency. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula Equation of a line passing through P(x_0, y_0, z_0)$P(x_0, y_0, z_0)$ along direction vector vecv = ahati+bhatj+chatk$\vec{v} = a\hat{i}+b\hat{j}+c\hat{k}$: fracx-x_0a = fracy-y_0b = fracz-z_0c = lambda$\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c} = \lambda$ ### Core Logic Let the given point be Pleft(frac157, frac327, 7right)$P\left(\frac{15}{7}, \frac{32}{7}, 7\right)$. We need to find the distance measured along the line passing through P$P$ parallel to the vector vecv = hati+4hatj+7hatk$\vec{v} = \hat{i}+4\hat{j}+7\hat{k}$. Equation of this line PQ$PQ$ is: fracx - frac1571 = fracy - frac3274 = fracz - 77 = lambda$\frac{x - \frac{15}{7}}{1} = \frac{y - \frac{32}{7}}{4} = \frac{z - 7}{7} = \lambda$ Any general point Q$Q$ on this line can be written as: Qleft(lambda + frac157, 4lambda + frac327, 7lambda + 7right)$Q\left(\lambda + \frac{15}{7}, 4\lambda + \frac{32}{7}, 7\lambda + 7\right)$ ### Step 1: Find Intersection Point Q with Given Line Point Q$Q$ must lie on the given target line L: fracx+13 = fracy+35 = fracz+57$L: \frac{x+1}{3} = \frac{y+3}{5} = \frac{z+5}{7}$. Substitute coordinates of Q$Q$ into the first and third fractions: fracleft(lambda + frac157right) + 13 = frac(7lambda + 7) + 57$\frac{\left(\lambda + \frac{15}{7}\right) + 1}{3} = \frac{(7\lambda + 7) + 5}{7}$ fraclambda + frac2273 = frac7lambda + 127$\frac{\lambda + \frac{22}{7}}{3} = \frac{7\lambda + 12}{7}$ frac7lambda + 2221 = frac7lambda + 127$\frac{7\lambda + 22}{21} = \frac{7\lambda + 12}{7}$ Multiplying by 21: 7lambda + 22 = 3(7lambda + 12)$7\lambda + 22 = 3(7\lambda + 12)$ 7lambda + 22 = 21lambda + 36$7\lambda + 22 = 21\lambda + 36$ 14lambda = -14 implies lambda = -1$14\lambda = -14 \implies \lambda = -1$ ### Step 2: Calculate Coordinates of Q and Distance squared Substituting lambda = -1$\lambda = -1$ into coordinates of Q$Q$: Qleft(-1 + frac157, -4 + frac327, -7 + 7right) = Qleft(frac87, frac47, 0right)$Q\left(-1 + \frac{15}{7}, -4 + \frac{32}{7}, -7 + 7\right) = Q\left(\frac{8}{7}, \frac{4}{7}, 0\right)$ Now, compute the distance squared (PQ)^2$(PQ)^2$: (PQ)^2 = left(frac157 - frac87right)^2 + left(frac327 - frac47right)^2 + (7 - 0)^2$(PQ)^2 = \left(\frac{15}{7} - \frac{8}{7}\right)^2 + \left(\frac{32}{7} - \frac{4}{7}\right)^2 + (7 - 0)^2$ (PQ)^2 = left(frac77right)^2 + left(frac287 ight)^2 + 7^2 = 1^2 + 4^2 + 49 = 1 + 16 + 49 = 66$(PQ)^2 = \left(\frac{7}{7}\right)^2 + \left(\frac{28}{7} ight)^2 + 7^2 = 1^2 + 4^2 + 49 = 1 + 16 + 49 = 66$ ### Pattern Recognition Instead of finding coordinates of Q$Q$, we could also use the vector form directly: vecPQ = lambda(hati + 4hatj + 7hatk)$\vec{PQ} = \lambda(\hat{i} + 4\hat{j} + 7\hat{k})$. Distance squared is PQ^2 = lambda^2(1^2 + 4^2 + 7^2) = (-1)^2(1 + 16 + 49) = 66$PQ^2 = \lambda^2(1^2 + 4^2 + 7^2) = (-1)^2(1 + 16 + 49) = 66$. This saves a lot of fractional coordinate subtraction arithmetic! ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula textEquation of line passing through vecr_0 text parallel to vecv: vecr = vecr_0 + alpha vecv$\text{Equation of line passing through } \vec{r}_0 \text{ \parallel to } \vec{v}: \vec{r} = \vec{r}_0 + \alpha \vec{v}$ ### Core Logic Express general points on L_1$L_1$ and L_2$L_2$ vector components: L_1: vecr = (lambda - 1)hati + 2(lambda + 1)hatj + (lambda + 1)hatk$L_1: \vec{r} = (\lambda - 1)\hat{i} + 2(\lambda + 1)\hat{j} + (\lambda + 1)\hat{k}$ L_2: vecr = 2muhati + (1 + 7mu)hatj + (1 + 3mu)hatk$L_2: \vec{r} = 2\mu\hat{i} + (1 + 7\mu)\hat{j} + (1 + 3\mu)\hat{k}$ ### Step 1: Equate components to find point of intersection lambda - 1 = 2mu quad dots (1)$\lambda - 1 = 2\mu \quad \dots (1)$ 2lambda + 2 = 1 + 7mu quad dots (2)$2\lambda + 2 = 1 + 7\mu \quad \dots (2)$ lambda + 1 = 1 + 3mu implies lambda = 3mu quad dots (3)$\lambda + 1 = 1 + 3\mu \implies \lambda = 3\mu \quad \dots (3)$ Substituting (3) into (1): 3mu - 1 = 2mu implies mu = 1 implies lambda = 3$3\mu - 1 = 2\mu \implies \mu = 1 \implies \lambda = 3$ Point of intersection vecr_0 = 2(1)hati + (1+7)hatj + (1+3)hatk = 2hati + 8hatj + 4hatk$\vec{r}_0 = 2(1)\hat{i} + (1+7)\hat{j} + (1+3)\hat{k} = 2\hat{i} + 8\hat{j} + 4\hat{k}$. ### Step 2: Formulate line L3 Direction vector vecv = veca + vecb = 3hati + 9hatj + 4hatk$\vec{v} = \vec{a} + \vec{b} = 3\hat{i} + 9\hat{j} + 4\hat{k}$. L_3: vecr = (2hati + 8hatj + 4hatk) + alpha(3hati + 9hatj + 4hatk)$L_3: \vec{r} = (2\hat{i} + 8\hat{j} + 4\hat{k}) + \alpha(3\hat{i} + 9\hat{j} + 4\hat{k})$ L_3(alpha) = (2 + 3alpha)hati + (8 + 9alpha)hatj + (4 + 4alpha)hatk$L_3(\alpha) = (2 + 3\alpha)\hat{i} + (8 + 9\alpha)\hat{j} + (4 + 4\alpha)\hat{k}$ ### Step 3: Match options For alpha = 2$\alpha = 2$: vecr = (2+6)hati + (8+18)hatj + (4+8)hatk = 8hati + 26hatj + 12hatk$\vec{r} = (2+6)\hat{i} + (8+18)\hat{j} + (4+8)\hat{k} = 8\hat{i} + 26\hat{j} + 12\hat{k}$ This matches coordinates (8, 26, 12)$(8, 26, 12)$. ### Pattern Recognition Instead of checking all lines simultaneously, match options via quick ratio checks: fracx - 23 = fracy - 89 = fracz - 44$\frac{x - 2}{3} = \frac{y - 8}{9} = \frac{z - 4}{4}$. This eliminates false choices instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Vector Algebra

### Related Formula textDirection ratios of a line perpendicular to both vectors: vecv = vecd_1 times vecd_2$\text{Direction ratios of a line perpendicular to both vectors: } \vec{v} = \vec{d}_1 \times \vec{d}_2$ ### Core Logic Compute direction ratios for L_3$L_3$ using cross product of direction vectors of L_1$L_1$ and L_2$L_2$: vecv = left| beginarrayccc hati & hatj & hatk \\ 1 & -1 & 2 \\ -1 & 2 & 1 endarray right| = -5hati - 3hatj + hatk$\vec{v} = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{array} \right| = -5\hat{i} - 3\hat{j} + \hat{k}$ ### Step 1: Establish intersection constraints Let A$A$ be a general point on L_3$L_3$ and B$B$ be a general point on L_1$L_1$: A = (alpha - 5lambda, beta - 3lambda, gamma + lambda)$A = (\alpha - 5\lambda, \beta - 3\lambda, \gamma + \lambda)$ B = (k+1, -k+2, 2k+1)$B = (k+1, -k+2, 2k+1)$ Since L_3$L_3$ intersects L_1$L_1$, at the point of intersection A equiv B$A \equiv B$: alpha - 5lambda = k + 1 implies alpha = 5lambda + k + 1$\alpha - 5\lambda = k + 1 \implies \alpha = 5\lambda + k + 1$ beta - 3lambda = -k + 2 implies beta = 3lambda - k + 2$\beta - 3\lambda = -k + 2 \implies \beta = 3\lambda - k + 2$ gamma + lambda = 2k + 1 implies gamma = -lambda + 2k + 1$\gamma + \lambda = 2k + 1 \implies \gamma = -\lambda + 2k + 1$ ### Step 2: Solve final algebraic target equation Substitute expressions into the target mod expression: 5alpha - 11beta - 8gamma = 5(5lambda + k + 1) - 11(3lambda - k + 2) - 8(-lambda + 2k + 1)$5\alpha - 11\beta - 8\gamma = 5(5\lambda + k + 1) - 11(3\lambda - k + 2) - 8(-\lambda + 2k + 1)$ = lambda(25 - 33 + 8) + k(5 + 11 - 16) + (5 - 22 - 8)$= \lambda(25 - 33 + 8) + k(5 + 11 - 16) + (5 - 22 - 8)$ = 0lambda + 0k - 25 = -25$= 0\lambda + 0k - 25 = -25$ Taking absolute value yields |-25| = 25$|-25| = 25$. ### Pattern Recognition The cancellation of internal variables (lambda$\lambda$ and k$k$) shows that the absolute expression describes a invariant geometric plane coordinate constraint, allowing simple baseline parameter values (like k=0, lambda=0$k=0, \lambda=0$) to evaluate the question instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry