### Related Formula Dot product of vectors: overrightarrowOA cdot overrightarrowOB = x_A x_B + y_A y_B + z_A z_B$\overrightarrow{OA} \cdot \overrightarrow{OB} = x_A x_B + y_A y_B + z_A z_B$ ### Core Logic Let any general point Q$Q$ on line L$L$ be (3lambda + 6, 2lambda + 7, -2lambda + 7)$(3\lambda + 6, 2\lambda + 7, -2\lambda + 7)$. Vector from P(1,2,3)$P(1,2,3)$ to Q$Q$ is overrightarrowPQ = (3lambda + 5)hati + (2lambda + 5)hatj + (-2lambda + 4)hatk$\overrightarrow{PQ} = (3\lambda + 5)\hat{i} + (2\lambda + 5)\hat{j} + (-2\lambda + 4)\hat{k}$. Since Q$Q$ is the foot of perpendicular, overrightarrowPQ cdot vecb = 0$\overrightarrow{PQ} \cdot \vec{b} = 0$, where vecb = 3hati + 2hatj - 2hatk$\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}$: 3(3lambda + 5) + 2(2lambda + 5) - 2(-2lambda + 4) = 0$3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0$ 9lambda + 15 + 4lambda + 10 + 4lambda - 8 = 0 implies 17lambda = -17 implies lambda = -1$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0 \implies 17\lambda = -17 \implies \lambda = -1$ Thus, the foot of perpendicular is Q(3, 5, 9)$Q(3, 5, 9)$. ### Step 1: Locate Points A and B Points A$A$ and B$B$ are on line L$L$, expressed via a distance parameter mu$\mu$ away from Q$Q$: General point expression from Q$Q$: A, B = (3mu + 3, 2mu + 5, -2mu + 9)$A, B = (3\mu + 3, 2\mu + 5, -2\mu + 9)$ [relative tracking points]. Distance squared = 68$= 68$: (3mu)^2 + (2mu)^2 + (-2mu)^2 = 68 implies 17mu^2 = 68 implies mu = pm 2$(3\mu)^2 + (2\mu)^2 + (-2\mu)^2 = 68 \implies 17\mu^2 = 68 \implies \mu = \pm 2$ For mu = 2$\mu = 2$: A = (3(2)+3, 2(2)+5, -2(2)+9) = (9, 9, 5)$A = (3(2)+3, 2(2)+5, -2(2)+9) = (9, 9, 5)$ For mu = -2$\mu = -2$: B = (3(-2)+3, 2(-2)+5, -2(-2)+9) = (-3, 1, 13)$B = (3(-2)+3, 2(-2)+5, -2(-2)+9) = (-3, 1, 13)$

### Step 2: Vector Dot Product Evaluation $overrightarrowOA cdot overrightarrowOB = 9(-3) + 9(1) + 5(13) = -27 + 9 + 65 = 47$$\overrightarrow{OA} \cdot \overrightarrow{OB} = 9(-3) + 9(1) + 5(13) = -27 + 9 + 65 = 47$ ### Pattern Recognition When symmetrical points on a line are equidistant from a central foot position, utilizing standard parametric tracking simplifies vector resolution instantly. Double check calculations sequentially via parallel distance components. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula The direction vector of a line perpendicular to two given lines with direction vectors vecv_1$\vec{v}_1$ and \vec{v}_2 is determined by their cross product: vecv = vecv_1 times vecv_2$\vec{v} = \vec{v}_1 \times \vec{v}_2$ ### Core Logic The given point left(0, -frac12, 0right)$\left(0, -\frac{1}{2}, 0\right)$ lies on the required line: fracx - 1-2 = fracy + 4d = fracz - c-4$\frac{x - 1}{-2} = \frac{y + 4}{d} = \frac{z - c}{-4}$ Substituting the point coordinates into the equation: frac0 - 1-2 = frac-frac12 + 4d = frac0 - c-4$\frac{0 - 1}{-2} = \frac{-\frac{1}{2} + 4}{d} = \frac{0 - c}{-4}$ frac12 = frac72d = fracc4 implies d = 7, quad c = 2$\frac{1}{2} = \frac{7}{2d} = \frac{c}{4} \implies d = 7, \quad c = 2$ ### Step 1: Cross Product Direction Ratios The direction vectors of the lines are vecv_1 = (1, a, b)$\vec{v}_1 = (1, a, b)$ and vecv_2 = (-b, a, 5)$\vec{v}_2 = (-b, a, 5)$. vecv = beginvmatrix hati & hatj & hatk \\ 1 & a & b \\ -b & a & 5 endvmatrix = hati(5a - ab) - hatj(5 + b^2) + hatk(a + ab)$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & a & b \\ -b & a & 5 \end{vmatrix} = \hat{i}(5a - ab) - \hat{j}(5 + b^2) + \hat{k}(a + ab)$ Thus, the direction ratios of the line are proportional to: frac5a - ab-2 = frac-(b^2 + 5)7 = fraca + ab-4 quad dots text(i)$\frac{5a - ab}{-2} = \frac{-(b^2 + 5)}{7} = \frac{a + ab}{-4} \quad \dots \text{(i)}$ ### Step 2: Solve for a and b From the first and third components of equation (i): frac5a - ab-2 = fraca + ab-4 implies 2(5a - ab) = a + ab$\frac{5a - ab}{-2} = \frac{a + ab}{-4} \implies 2(5a - ab) = a + ab$ 10a - 2ab = a + ab implies 9a = 3ab implies b = 3$10a - 2ab = a + ab \implies 9a = 3ab \implies b = 3$ Now use the second component ratio with b = 3$b = 3$ and d = 7$d = 7$: frac-(3^2 + 5)7 = fraca + a(3)-4 implies frac-147 = frac4a-4 implies -2 = -a implies a = 2$\frac{-(3^2 + 5)}{7} = \frac{a + a(3)}{-4} \implies \frac{-14}{7} = \frac{4a}{-4} \implies -2 = -a \implies a = 2$ ### Step 3: Sum the Variables Summing up a, b, c, d$a, b, c, d$: a + b + c + d = 2 + 3 + 2 + 7 = 14$a + b + c + d = 2 + 3 + 2 + 7 = 14$ ### Pattern Recognition Substituting known point values into symmetric equations immediately determines structural values like c$c$ and d$d$ before running cross product systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula The vector area of a triangle given two adjacent position vectors vecu$\vec{u}$ and vecv$\vec{v}$ is calculated as: textArea = frac12 |vecu times vecv|$\text{Area} = \frac{1}{2} |\vec{u} \times \vec{v}|$ ### Core Logic For line L_1$L_1$: fracx-11 = fracy-21 = fracz-01$\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-0}{1}$. Let a general point be Q(lambda+1, lambda+2, lambda)$Q(\lambda+1, \lambda+2, \lambda)$. vecPQ = (lambda-4, lambda+1, lambda+3)$\vec{PQ} = (\lambda-4, \lambda+1, \lambda+3)$ Since vecPQ cdot vecm_1 = 0$\vec{PQ} \cdot \vec{m}_1 = 0$ (direction vector of L_1$L_1$ is (1,1,1)$(1,1,1)$): (lambda-4)(1) + (lambda+1)(1) + (lambda+3)(1) = 0 implies 3lambda = 0 implies lambda = 0$(\lambda-4)(1) + (\lambda+1)(1) + (\lambda+3)(1) = 0 \implies 3\lambda = 0 \implies \lambda = 0$ Thus, Q(1, 2, 0)$Q(1, 2, 0)$ and vecPQ = (-4, 1, 3)$\vec{PQ} = (-4, 1, 3)$.

### Step 1: Compute Foot R For line L_2$L_2$: fracx-21 = fracy1 = fracz-11$\frac{x-2}{1} = \frac{y}{1} = \frac{z-1}{1}$. Let a general point be R(mu+2, mu, mu+1)$R(\mu+2, \mu, \mu+1)$. vecPR = (mu-3, mu-1, mu+4)$\vec{PR} = (\mu-3, \mu-1, \mu+4)$ Since vecPR cdot vecm_2 = 0$\vec{PR} \cdot \vec{m}_2 = 0$ (direction vector of L_2$L_2$ is (1,1,1)$(1,1,1)$): (mu-3)(1) + (mu-1)(1) + (mu+4)(1) = 0 implies 3mu = 0 implies mu = 0$(\mu-3)(1) + (\mu-1)(1) + (\mu+4)(1) = 0 \implies 3\mu = 0 \implies \mu = 0$ Thus, R(2, 0, 1)$R(2, 0, 1)$ and vecPR = (-3, 1, 4)$\vec{PR} = (-3, 1, 4)$. ### Step 2: Area Vector Calculation The area A$A$ of Delta PQR$\Delta PQR$ is given by: A = frac12 |vecPQ times vecPR|$A = \frac{1}{2} |\vec{PQ} \times \vec{PR}|$ vecPQ times vecPR = beginvmatrix hati & hatj & hatk \\ -4 & 1 & 3 \\ -3 & 1 & 4 endvmatrix = hati(4-3) - hatj(-16+9) + hatk(-4+3) = hati + 7hatj - hatk$\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & 1 & 3 \\ -3 & 1 & 4 \end{vmatrix} = \hat{i}(4-3) - \hat{j}(-16+9) + \hat{k}(-4+3) = \hat{i} + 7\hat{j} - \hat{k}$ textMagnitude squared: |vecPQ times vecPR|^2 = 1^2 + 7^2 + (-1)^2 = 1 + 49 + 1 = 51$\text{Magnitude squared: } |\vec{PQ} \times \vec{PR}|^2 = 1^2 + 7^2 + (-1)^2 = 1 + 49 + 1 = 51$ Let's re-verify the matrix arithmetic layout: vecPQ = (-4, 1, 3), vecPR = (-3, 1, 4)$\vec{PQ} = (-4, 1, 3), \vec{PR} = (-3, 1, 4)$ = 7hati + 7hatj + 7hatk$= 7\hat{i} + 7\hat{j} + 7\hat{k}$ |7(hati + hatj + hatk)|^2 = 49 cdot 3 = 147$|7(\hat{i} + \hat{j} + \hat{k})|^2 = 49 \cdot 3 = 147$ ### Step 3: Evaluate 4A^2 Since A = frac12 sqrt147$A = \frac{1}{2} \sqrt{147}$: 4A^2 = 4 cdot left(frac14 cdot 147right) = 147$4A^2 = 4 \cdot \left(\frac{1}{4} \cdot 147\right) = 147$ ### Pattern Recognition Setting up dot products systematically with general parametric forms quickly locks in spatial feet indices without complex geometric drawings. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula 1. Area of a \triangle with perpendicular height h$h$ and base b$b$: textArea = frac12 times b times h$\text{Area} = \frac{1}{2} \times b \times h$ 2. Since P$P$ is the reflection image of Q$Q$ across line L$L$, the line acts as a perpendicular bisector. For any point R$R$ lying on the line, the height from R$R$ to the line is RT$RT$, and the base QP = 2QT$QP = 2QT$. ### Core Logic Determine parameters for point R$R$ lying directly on line L [cite: 3405, 4071]: frac5-12 = fracp+13 = fracq4 Rightarrow 2 = fracp+13 = fracq4$\frac{5-1}{2} = \frac{p+1}{3} = \frac{q}{4} \Rightarrow 2 = \frac{p+1}{3} = \frac{q}{4}$ p+1 = 6 Rightarrow p = 5, quad q = 8 Rightarrow R = (5, 5, 8)$p+1 = 6 \Rightarrow p = 5, \quad q = 8 \Rightarrow R = (5, 5, 8)$ [cite: 4071, 4073]

### Step 1: Locate Foot of Perpendicular (T$T$) Let the foot of the perpendicular from Q(7, -2, 5)$Q(7, -2, 5)$ on line L$L$ be T(2lambda+1, 3lambda-1, 4lambda)$T(2\lambda+1, 3\lambda-1, 4\lambda)$ . The directional direction of L$L$ is vecb = 2hati + 3hatj + 4hatk$\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ . Vector overrightarrowQT = (2lambda - 6)hati + (3lambda + 1)hatj + (4lambda - 5)hatk$\overrightarrow{QT} = (2\lambda - 6)\hat{i} + (3\lambda + 1)\hat{j} + (4\lambda - 5)\hat{k}$ . Apply orthogonality condition overrightarrowQT cdot vecb = 0$\overrightarrow{QT} \cdot \vec{b} = 0$ : 2(2lambda - 6) + 3(3lambda + 1) + 4(4lambda - 5) = 0$2(2\lambda - 6) + 3(3\lambda + 1) + 4(4\lambda - 5) = 0$ 4lambda - 12 + 9lambda + 3 + 16lambda - 20 = 0 Rightarrow 29lambda - 29 = 0 Rightarrow lambda = 1$4\lambda - 12 + 9\lambda + 3 + 16\lambda - 20 = 0 \Rightarrow 29\lambda - 29 = 0 \Rightarrow \lambda = 1$ [cite: 4077, 4078] Thus, T = (3, 2, 4)$T = (3, 2, 4)$. ### Step 2: Measure Geometric Distances Compute length QT$QT$ using distance metrics : QT = sqrt(3-7)^2 + (2 - (-2))^2 + (4-5)^2 = sqrt16 + 16 + 1 = sqrt33$QT = \sqrt{(3-7)^2 + (2 - (-2))^2 + (4-5)^2} = \sqrt{16 + 16 + 1} = \sqrt{33}$ Since P$P$ is the symmetrical image, base QP = 2QT = 2sqrt33$QP = 2QT = 2\sqrt{33}$. Compute length RT$RT$ representing height from vertex R(5, 5, 8)$R(5, 5, 8)$ to base line at T(3, 2, 4)$T(3, 2, 4)$ : RT = sqrt(5-3)^2 + (5-2)^2 + (8-4)^2 = sqrt4 + 9 + 16 = sqrt29$RT = \sqrt{(5-3)^2 + (5-2)^2 + (8-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$ ### Step 3: Calculate Squared Area Compute the \triangle area squared value [cite: 3406, 4081]: textArea = frac12 times QP times RT = frac12 times left(2sqrt33right) times sqrt29 = sqrt957$\text{Area} = \frac{1}{2} \times QP \times RT = \frac{1}{2} \times \left(2\sqrt{33}\right) \times \sqrt{29} = \sqrt{957}$ left(textArearight)^2 = 957$\left(\text{Area}\right)^2 = 957$ [cite: 4081, 4083] ### Pattern Recognition Because the image geometry creates an isosceles pairing from any point on the mirror line to the object and image point, the area reduces beautifully to 2 times textArea(triangle QTR) = QT times RT$2 \times \text{Area}(\triangle QTR) = QT \times RT$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula The area of a triangle given base length b$b$ and altitude perpendicular height h$h$ is evaluated as: textArea = frac12 cdot b cdot h$\text{Area} = \frac{1}{2} \cdot b \cdot h$ ### Core Logic The base side AC$AC$ lies entirely along the line equation. Let M$M$ be the foot of the perpendicular dropped from vertex B(1,2,3)$B(1,2,3)$ to the line segment AC$AC$:

Any coordinate point on the line can be represented parametrically by setting the line fractions equal to lambda$\lambda$: M = (3lambda + 6, \, 2lambda + 7, \, -2lambda + 7)$M = (3\lambda + 6, \, 2\lambda + 7, \, -2\lambda + 7)$ ### Step 1: Compute Foot of Perpendicular Construct the vector direction representing line segment BM$BM$: vecBM = (3lambda + 6 - 1)hatmathbfi + (2lambda + 7 - 2)hatmathbfj + (-2lambda + 7 - 3)hatmathbfk$\vec{BM} = (3\lambda + 6 - 1)\hat{\mathbf{i}} + (2\lambda + 7 - 2)\hat{\mathbf{j}} + (-2\lambda + 7 - 3)\hat{\mathbf{k}}$ vecBM = (3lambda + 5)hatmathbfi + (2lambda + 5)hatmathbfj + (-2lambda + 4)hatmathbfk$\vec{BM} = (3\lambda + 5)\hat{\mathbf{i}} + (2\lambda + 5)\hat{\mathbf{j}} + (-2\lambda + 4)\hat{\mathbf{k}}$ Since BM$BM$ is perpendicular to the base line segment direction vector vecv = 3hatmathbfi + 2hatmathbfj - 2hatmathbfk$\vec{v} = 3\hat{\mathbf{i}} + 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}}$, their dot product must equal zero: vecBM cdot vecv = 3(3lambda + 5) + 2(2lambda + 5) - 2(-2lambda + 4) = 0$\vec{BM} \cdot \vec{v} = 3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0$ 9lambda + 15 + 4lambda + 10 + 4lambda - 8 = 0$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$ 17lambda + 17 = 0 implies lambda = -1$17\lambda + 17 = 0 \implies \lambda = -1$ ### Step 2: Find Perpendicular Length and Calculate Area Substitute lambda = -1$\lambda = -1$ back into the vector equation to find the altitude magnitude: vecBM = (3(-1) + 5)hatmathbfi + (2(-1) + 5)hatmathbfj + (-2(-1) + 4)hatmathbfk = 2hatmathbfi + 3hatmathbfj + 6hatmathbfk$\vec{BM} = (3(-1) + 5)\hat{\mathbf{i}} + (2(-1) + 5)\hat{\mathbf{j}} + (-2(-1) + 4)\hat{\mathbf{k}} = 2\hat{\mathbf{i}} + 3\hat{\mathbf{j}} + 6\hat{\mathbf{k}}$ h = \|vecBM\| = sqrt2^2 + 3^2 + 6^2 = sqrt4 + 9 + 36 = sqrt49 = 7$h = \|\vec{BM}\| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ Now, plug the base length AC = 6$AC = 6$ and altitude h = 7$h = 7$ into the standard area formula: textArea = frac12 cdot 6 cdot 7 = 21 text sq. units$\text{Area} = \frac{1}{2} \cdot 6 \cdot 7 = 21 \text{ sq. units}$ ### Pattern Recognition Instead of determining the absolute coordinates for individual triangle vertices A$A$ and C$C$, treating the problem via altitude minimization relative to the given parametric vector direction saves substantial calculation steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry