Consider the lines mathrmL_1: mathrmx - 1 = mathrmy - 2 = mathrmz$\mathrm{L}_1: \mathrm{x} - 1 = \mathrm{y} - 2 = \mathrm{z}$ and mathrmL_2: mathrmx - 2 = mathrmy = mathrmz - 1$\mathrm{L}_2: \mathrm{x} - 2 = \mathrm{y} = \mathrm{z} - 1$. Let the feet of the perpendiculars from the point mathrmP(5,1,-3)$\mathrm{P}(5,1,-3)$ on the lines mathrmL_1$\mathrm{L}_1$ and mathrmL_2$\mathrm{L}_2$ be mathrmQ$\mathrm{Q}$ and mathrmR$\mathrm{R}$ respectively. If the area of the triangle PQR is mathrmA$\mathrm{A}$, then 4mathrmA^2$4\mathrm{A}^2$ is equal to:

### Related Formula textShortest Distance = left| fracvecAB cdot (vecp times vecq)|vecp times vecq| right|$$\text{Shortest Distance} = \left| \frac{\vec{AB} \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right|$$ ### Core Logic Identify points A(1, 2, 3)$A(1, 2, 3)$ and B(lambda, 4, 5)$B(\lambda, 4, 5)$ on the lines with directions vecp = 2hati + 3hatj + 4hatk$\vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and vecq = 3hati + 4hatj + 5hatk$\vec{q} = 3\hat{i} + 4\hat{j} + 5\hat{k}$ respectively. Use the shortest distance formula to determine lambda_1$\lambda_1$ and lambda_2$\lambda_2$. ### Step 1: Calculate Cross Product and Direction Vector vecp times vecq = beginvmatrix hati & hatj & hatk \\ 2 & 3 & 4 \\ 3 & 4 & 5 endvmatrix = -hati + 2hatj - hatk$$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = -\hat{i} + 2\hat{j} - \hat{k}$$ |vecp times vecq| = sqrt(-1)^2 + 2^2 + (-1)^2 = sqrt6$$|\vec{p} \times \vec{q}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}$$ vecAB = (lambda - 1)hati + 2hatj + 2hatk$$\vec{AB} = (\lambda - 1)\hat{i} + 2\hat{j} + 2\hat{k}$$ ### Step 2: Solve for Lambda frac1sqrt6 = left| frac((lambda - 1)hati + 2hatj + 2hatk) cdot (-hati + 2hatj - hatk)sqrt6 right|$$\frac{1}{\sqrt{6}} = \left| \frac{((\lambda - 1)\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-\hat{i} + 2\hat{j} - \hat{k})}{\sqrt{6}} \right|$$ implies |-lambda + 1 + 4 - 2| = 1 implies |lambda - 3| = 1$$\implies |-\lambda + 1 + 4 - 2| = 1 \implies |\lambda - 3| = 1$$ implies lambda = 4 text or 2$$\implies \lambda = 4 \text{ or } 2$$ ### Step 3: Radius of the Passing Circle The circle passes through (0,0)$(0,0)$, (4,2)$(4,2)$ and (2,4)$(2,4)$. Using the circumradius formula R = fracabc4Delta$R = \frac{abc}{4\Delta}$: a = sqrt20, quad b = sqrt20, quad c = sqrt8$$a = \sqrt{20}, \quad b = \sqrt{20}, \quad c = \sqrt{8}$$ Delta = frac12 beginvmatrix 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 endvmatrix = 6$$\Delta = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 \end{vmatrix} = 6$$ R = fracsqrt20 times sqrt20 times sqrt84 times 6 = frac40sqrt224 = frac5sqrt23$$R = \frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times 6} = \frac{40\sqrt{2}}{24} = \frac{5\sqrt{2}}{3}$$ ### Pattern Recognition Shortest distance values create symmetric configurations. When finding a circle passing through (0,0)$(0,0)$, (x,y)$(x,y)$, and (y,x)$(y,x)$, the symmetry about y=x$y=x$ simplifies radius calculations immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Circles

### 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