### Related Formula textArea A = frac12 |vecAB times vecAC|$\text{Area } A = \frac{1}{2} |\vec{AB} \times \vec{AC}|$ ### Core Logic Determine the three intersection vertex positions for the matching coordinate line segments, then calculate vector cross expansions to determine face boundaries. ### Step 1: Locate Intersection Vertices Solving line pairs intersection matrices: * L_1 cap L_2 implies A(-2, 1, 0)$L_1 \cap L_2 \implies A(-2, 1, 0)$ * L_2 cap L_3 implies B(3, 0, 1)$L_2 \cap L_3 \implies B(3, 0, 1)$ * L_3 cap L_1 implies C(0, 3, 2)$L_3 \cap L_1 \implies C(0, 3, 2)$ ### Step 2: Construct Vectors Cross Matrix Using vertex values to form component arrays: vecAB = -5hati + hatj - hatk, quad vecAC = -3hati + 3hatj + hatk$\vec{AB} = -5\hat{i} + \hat{j} - \hat{k}, \quad \vec{AC} = -3\hat{i} + 3\hat{j} + \hat{k}$ vecAB times vecAC = beginvmatrix hati & hatj & hatk \\ -5 & 1 & -1 \\ -3 & 3 & 1 endvmatrix = 4hati + 8hatj - 12hatk$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -1 \\ -3 & 3 & 1 \end{vmatrix} = 4\hat{i} + 8\hat{j} - 12\hat{k}$ ### Step 3: Final Area Squared Derivation A = frac12sqrt16 + 64 + 144 = frac12sqrt224 = sqrt56$A = \frac{1}{2}\sqrt{16 + 64 + 144} = \frac{1}{2}\sqrt{224} = \sqrt{56}$ A^2 = 56$A^2 = 56$ {{SOL_IMG_73}} ### Pattern Recognition Finding the area of a triangle formed by intersecting lines involves grouping directional cross vectors once coordinates are solved. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula The direction vector of a line perpendicular to two vectors vecu$\vec{u}$ and vecv$\vec{v}$ is obtained via the cross product: vecn = vecu times vecv$\vec{n} = \vec{u} \times \vec{v}$ ### Core Logic Extract direction vectors of the given lines: vecu = 2hati + hatj - 2hatk$\vec{u} = 2\hat{i} + \hat{j} - 2\hat{k}$ vecv = hati + 3hatj + 4hatk$\vec{v} = \hat{i} + 3\hat{j} + 4\hat{k}$ Compute the cross product: vecn = beginvmatrix hati & hatj & hatk \\ 2 & 1 & -2 \\ 1 & 3 & 4 endvmatrix$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix}$ = hati(4 - (-6)) - hatj(8 - (-2)) + hatk(6 - 1) = 10hati - 10hatj + 5hatk$= \hat{i}(4 - (-6)) - \hat{j}(8 - (-2)) + \hat{k}(6 - 1) = 10\hat{i} - 10\hat{j} + 5\hat{k}$ = 5(2hati - 2hatj + hatk)$= 5(2\hat{i} - 2\hat{j} + \hat{k})$ ### Step 1: Write Line Equation and Intersect with Plane Equation of line L$L$ through P(2, -1, 3)$P(2, -1, 3)$ with direction (2, -2, 1)$(2, -2, 1)$: fracx - 22 = fracy + 1-2 = fracz - 31 = lambda$\frac{x - 2}{2} = \frac{y + 1}{-2} = \frac{z - 3}{1} = \lambda$ Any random point on this line is Q(2lambda + 2, -2lambda - 1, lambda + 3)$Q(2\lambda + 2, -2\lambda - 1, \lambda + 3)$. For intersection with the yz$yz$-plane, set x = 0$x = 0$: 2lambda + 2 = 0 implies lambda = -1$2\lambda + 2 = 0 \implies \lambda = -1$ ### Step 2: Find Distance Substituting lambda = -1$\lambda = -1$ into the coordinate matrix of Q$Q$ gives: Q(0, 1, 2)$Q(0, 1, 2)$ Calculate distance d(P, Q)$d(P, Q)$: d = sqrt(2 - 0)^2 + (-1 - 1)^2 + (3 - 2)^2 = sqrt4 + 4 + 1 = 3$d = \sqrt{(2 - 0)^2 + (-1 - 1)^2 + (3 - 2)^2} = \sqrt{4 + 4 + 1} = 3$ ### Pattern Recognition Perpendicularity to two lines always indicates using the cross-product to lock down the direction ratios. Intersection with the yz$yz$-plane simply forces x = 0$x = 0$ immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula Dot product of vector projection matching orthogonal axes equals zero: vecAP cdot vecd = 0$\vec{AP} \cdot \vec{d} = 0$ ### Core Logic Let the target source coordinates tracking point match A(1, 2, 2)$A(1, 2, 2)$. General parameter points on line L$L$ are defined by parameter mu$\mu$: P(mu + 1, -mu - 1, 2mu + 2)$P(\mu + 1, -\mu - 1, 2\mu + 2)$

vecAP = muhati - (mu + 3)hatj + 2muhatk$\vec{AP} = \mu\hat{i} - (\mu + 3)\hat{j} + 2\mu\hat{k}$ Line direction vector vecd = hati - hatj + 2hatk$\vec{d} = \hat{i} - \hat{j} + 2\hat{k}$. ### Step 1: Isolate Foot and Intersection Positions (mu)cdot 1 - (-mu - 3)cdot 1 + (2mu)cdot 2 = 0 implies 6mu + 3 = 0 implies mu = -frac12$(\mu)\cdot 1 - (-\mu - 3)\cdot 1 + (2\mu)\cdot 2 = 0 \implies 6\mu + 3 = 0 \implies \mu = -\frac{1}{2}$ Substituting back yields coordinate positions for foot P$P$: Pleft(frac12, -frac12, 1right)$P\left(\frac{1}{2}, -\frac{1}{2}, 1\right)$ Equating general vectors between standard linear constraints tracks intersection point Q$Q$ at mu = -2$\mu = -2$: Q(-1, 1, -2)$Q(-1, 1, -2)$ ### Step 2: Distance Formulation Compute length of line segment squared: PQ^2 = left(frac12 - (-1)right)^2 + left(-frac12 - 1right)^2 + (1 - (-2))^2$PQ^2 = \left(\frac{1}{2} - (-1)\right)^2 + \left(-\frac{1}{2} - 1\right)^2 + (1 - (-2))^2$ = frac94 + frac94 + 9 = frac544$= \frac{9}{4} + \frac{9}{4} + 9 = \frac{54}{4}$ 2(PQ)^2 = 2 left(frac544right) = 27$2(PQ)^2 = 2 \left(\frac{54}{4}\right) = 27$ ### Pattern Recognition Always separate foot evaluations from line-intersection parameter updates to ensure you do not mix up variables tracking linear metrics. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula 3D Cartesian distance formula: d = sqrt(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$ ### Core Logic Since A(x,y,z)$A(x,y,z)$ lies in the xy-plane, its z-coordinate must be zero (z = 0$z = 0$). Let the reference targets be P(0,3,2)$P(0,3,2)$, Q(2,0,3)$Q(2,0,3)$, and R(0,0,1)$R(0,0,1)$. Setting AP^2 = AR^2$AP^2 = AR^2$: x^2 + (y-3)^2 + (0-2)^2 = x^2 + y^2 + (0-1)^2 implies y = 2$x^2 + (y-3)^2 + (0-2)^2 = x^2 + y^2 + (0-1)^2 \implies y = 2$ ### Step 1: Locating Coordinate Dimensions Setting AQ^2 = AR^2$AQ^2 = AR^2$ with y=2$y=2$: (x-2)^2 + 2^2 + 3^2 = x^2 + 2^2 + 1^2 implies x = 3$(x-2)^2 + 2^2 + 3^2 = x^2 + 2^2 + 1^2 \implies x = 3$ Thus, A$A$ is precisely located at (3,2,0)$(3,2,0)$. ### Step 2: Triangle Side and Area Assessment Calculate the lengths between A(3,2,0)$A(3,2,0)$, B(1,4,-1)$B(1,4,-1)$, and C(2,0,-2)$C(2,0,-2)$: AB = sqrt(3-1)^2 + (2-4)^2 + (0+1)^2 = 3$AB = \sqrt{(3-1)^2 + (2-4)^2 + (0+1)^2} = 3$ AC = sqrt(3-2)^2 + (2-0)^2 + (0+2)^2 = 3$AC = \sqrt{(3-2)^2 + (2-0)^2 + (0+2)^2} = 3$ BC = sqrt(1-2)^2 + (4-0)^2 + (-1+2)^2 = sqrt18$BC = \sqrt{(1-2)^2 + (4-0)^2 + (-1+2)^2} = \sqrt{18}$ Since AB = AC = 3$AB = AC = 3$ and AB^2 + AC^2 = BC^2$AB^2 + AC^2 = BC^2$, it forms an isosceles right-angled triangle. Thus, (S1) is true. textArea = frac12 times 3 times 3 = frac92$\text{Area} = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$ Therefore, (S2) is false. ### Pattern Recognition Planar locations instantly zero out specific coordinate dimensions (z=0$z=0$ for xy-planes), simplifying system matrices down rapidly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Three Dimensional Geometry

### Related Formula Perpendicularity condition for vectors: vecu cdot vecv = 0$\vec{u} \cdot \vec{v} = 0$ ### Core Logic Let Q$Q$ be the projection point on the given line parameterized by lambda$\lambda$: Q(2lambda + 1, lambda + 2, 3lambda + 1)$Q(2\lambda + 1, \lambda + 2, 3\lambda + 1)$. The vector overrightarrowPQ$\overrightarrow{PQ}$ from P(4,4,3)$P(4,4,3)$ is:

overrightarrowPQ = (2lambda - 3)hati + (lambda - 2)hatj + (3lambda - 2)hatk$\overrightarrow{PQ} = (2\lambda - 3)\hat{i} + (\lambda - 2)\hat{j} + (3\lambda - 2)\hat{k}$. ### Step 1: Solving for Projected Intersection Points Since overrightarrowPQ$\overrightarrow{PQ}$ is perpendicular to the line's direction vector (2, 1, 3)$(2, 1, 3)$: 2(2lambda - 3) + 1(lambda - 2) + 3(3lambda - 2) = 0 implies 14lambda - 14 = 0 implies lambda = 1$2(2\lambda - 3) + 1(\lambda - 2) + 3(3\lambda - 2) = 0 \implies 14\lambda - 14 = 0 \implies \lambda = 1$ Thus, Q$Q$ is located at (3,3,4)$(3,3,4)$. ### Step 2: Transforming using Midpoint Mappings The projection point Q$Q$ acts as the midpoint between original point P$P$ and its target image R(alpha, beta, gamma)$R(\alpha, \beta, \gamma)$: fracalpha + 42 = 3, quad fracbeta + 42 = 3, quad fracgamma + 32 = 4$\frac{\alpha + 4}{2} = 3, \quad \frac{\beta + 4}{2} = 3, \quad \frac{\gamma + 3}{2} = 4$ Evaluating this gives (alpha, beta, gamma) = (2, 2, 5)$(\alpha, \beta, \gamma) = (2, 2, 5)$. textSum = 2 + 2 + 5 = 9$\text{Sum} = 2 + 2 + 5 = 9$ Wait, checking the options from the paper layout: option (2) represents the correct numerical matrix sum choice value 12? Let's verify the options mapping sequence matching. Ah, let's look at the calculation value carefully: 2+2+5=9$2+2+5=9$, which corresponds to choice (1). ### Pattern Recognition Midpoint properties safely speed up spatial image transitions once you locate the perpendicular projection foot. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry