### Related Formula textEquation of a line through (x_1, y_1, z_1) text with direction ratios (a, b, c): quad fracx-x_1a = fracy-y_1b = fracz-z_1c$\text{Equation of a line through } (x_1, y_1, z_1) \text{ with direction ratios } (a, b, c): \quad \frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ textCondition for Perpendicular Lines: quad a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$\text{Condition for Perpendicular Lines:} \quad a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$ ### Core Logic Let R$R$ be the foot of the perpendicular drawn from point P(1,0,3)$P(1,0,3)$ to the line joining A(4,7,1)$A(4,7,1)$ and B(3,5,3)$B(3,5,3)$. Since Q(alpha,beta,gamma)$Q(\alpha,\beta,\gamma)$ is the reflection (image) of P$P$ across the line, point R$R$ serves as the midpoint of the line segment PQ$PQ$. ### Step 1: Find the Equation of the Line AB The direction ratios of the line AB are: vecd = (3 - 4, 5 - 7, 3 - 1) = (-1, -2, 2) equiv (1, 2, -2)$\vec{d} = (3 - 4, 5 - 7, 3 - 1) = (-1, -2, 2) \equiv (1, 2, -2)$ Using point B(3,5,3)$B(3,5,3)$, the symmetric equation of the line AB is: fracx - 31 = fracy - 52 = fracz - 3-2 = lambda$\frac{x - 3}{1} = \frac{y - 5}{2} = \frac{z - 3}{-2} = \lambda$ Any general point R$R$ on this line can be written in terms of parameter lambda$\lambda$: R equiv (lambda + 3, 2lambda + 5, -2lambda + 3)$R \equiv (\lambda + 3, 2\lambda + 5, -2\lambda + 3)$ ### Step 2: Find the Foot of the Perpendicular R The direction ratios of the line segment PR are: vecmathrmPR = (lambda + 3 - 1, \, 2lambda + 5 - 0, \, -2lambda + 3 - 3) = (lambda + 2, \, 2lambda + 5, \, -2lambda)$\vec{\mathrm{PR}} = (\lambda + 3 - 1, \, 2\lambda + 5 - 0, \, -2\lambda + 3 - 3) = (\lambda + 2, \, 2\lambda + 5, \, -2\lambda)$ Since PR is perpendicular to the line AB, the dot product of their direction vectors must equal zero: 1(lambda + 2) + 2(2lambda + 5) - 2(-2lambda) = 0$1(\lambda + 2) + 2(2\lambda + 5) - 2(-2\lambda) = 0$ lambda + 2 + 4lambda + 10 + 4lambda = 0$\lambda + 2 + 4\lambda + 10 + 4\lambda = 0$ lambda = -frac43$\lambda = -\frac{4}{3}$ ### Step 3: Coordinates of Foot of Perpendicular Substitute lambda = -frac43$\lambda = -\frac{4}{3}$ into the general coordinates of R$R$: R equiv left(-frac43 + 3, \, 2left(-frac43right) + 5, \, -2left(-frac43right) + 3right)$R \equiv \left(-\frac{4}{3} + 3, \, 2\left(-\frac{4}{3}\right) + 5, \, -2\left(-\frac{4}{3}\right) + 3\right)$ R equiv left(frac53, \, frac73, \, frac173right)$R \equiv \left(\frac{5}{3}, \, \frac{7}{3}, \, \frac{17}{3}\right)$ ### Step 4: Solve for the Image Coordinates and Sum Since R$R$ is the midpoint of PQ$PQ$, where P = (1, 0, 3)$P = (1, 0, 3)$ and Q = (alpha, beta, gamma)$Q = (\alpha, \beta, \gamma)$: - fracalpha + 12 = frac53 implies alpha = frac103 - 1 = frac73$\frac{\alpha + 1}{2} = \frac{5}{3} \implies \alpha = \frac{10}{3} - 1 = \frac{7}{3}$ - fracbeta + 02 = frac73 implies beta = frac143$\frac{\beta + 0}{2} = \frac{7}{3} \implies \beta = \frac{14}{3}$ - fracgamma + 32 = frac173 implies gamma = frac343 - 3 = frac253$\frac{\gamma + 3}{2} = \frac{17}{3} \implies \gamma = \frac{34}{3} - 3 = \frac{25}{3}$ Now, let us calculate the sum: alpha + beta + gamma = frac73 + frac143 + frac253 = frac463$\alpha + \beta + \gamma = \frac{7}{3} + \frac{14}{3} + \frac{25}{3} = \frac{46}{3}$ ### Pattern Recognition Standard Midpoint reflection: The image coordinates are given directly by x_textimage = 2 x_textfoot - x_textpoint$x_{\text{image}} = 2 x_{\text{foot}} - x_{\text{point}}$, y_textimage = 2 y_textfoot - y_textpoint$y_{\text{image}} = 2 y_{\text{foot}} - y_{\text{point}}$, and z_textimage = 2 z_textfoot - z_textpoint$z_{\text{image}} = 2 z_{\text{foot}} - z_{\text{point}}$. Finding the parameter lambda$\lambda$ by using the perpendicular vector dot-product rule is the fastest and most robust method. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula textShortest Distance d = fracleft| (vecr_2 - vecr_1) cdot (veca times vecb) right||veca times vecb|$\text{Shortest Distance } d = \frac{\left| (\vec{r}_2 - \vec{r}_1) \cdot (\vec{a} \times \vec{b}) \right|}{|\vec{a} \times \vec{b}|}$ ### Core Logic Shortest distance between two skew lines is the projection of the vector joining any two points of the lines onto their common normal. ### Step 1: Find the vector joining the two points Let the points be P_1(7, 6, 2)$P_1(7, 6, 2)$ on L_1$L_1$ and P_2(5, 3, 4)$P_2(5, 3, 4)$ on L_2$L_2$: vecr_2 - vecr_1 = (5 - 7)hatmathrmi + (3 - 6)hatmathrmj + (4 - 2)hatmathrmk = -2hatmathrmi - 3hatmathrmj + 2hatmathrmk$\vec{r}_2 - \vec{r}_1 = (5 - 7)\hat{\mathrm{i}} + (3 - 6)\hat{\mathrm{j}} + (4 - 2)\hat{\mathrm{k}} = -2\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 2\hat{\mathrm{k}}$ ### Step 2: Find the common normal vector The direction is given by the cross product of the direction vectors: veca times vecb = beginvmatrix hatmathrmi & hatmathrmj & hatmathrmk \\ -3 & 2 & 4 \\ 2 & 1 & 3 endvmatrix$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix}$ veca times vecb = hatmathrmi(6 - 4) - hatmathrmj(-9 - 8) + hatmathrmk(-3 - 4) = 2hatmathrmi + 17hatmathrmj - 7hatmathrmk$\vec{a} \times \vec{b} = \hat{\mathrm{i}}(6 - 4) - \hat{\mathrm{j}}(-9 - 8) + \hat{\mathrm{k}}(-3 - 4) = 2\hat{\mathrm{i}} + 17\hat{\mathrm{j}} - 7\hat{\mathrm{k}}$ ### Step 3: Calculate the distance Calculate the dot product of the vectors: (vecr_2 - vecr_1) cdot (veca times vecb) = (-2)(2) + (-3)(17) + (2)(-7) = -4 - 51 - 14 = -69$(\vec{r}_2 - \vec{r}_1) \cdot (\vec{a} \times \vec{b}) = (-2)(2) + (-3)(17) + (2)(-7) = -4 - 51 - 14 = -69$ Calculate the magnitude of the cross product: |veca times vecb| = sqrt2^2 + 17^2 + (-7)^2 = sqrt4 + 289 + 49 = sqrt342 = 3sqrt38$|\vec{a} \times \vec{b}| = \sqrt{2^2 + 17^2 + (-7)^2} = \sqrt{4 + 289 + 49} = \sqrt{342} = 3\sqrt{38}$ Now, compute the shortest distance: d = frac|-69|3sqrt38 = frac23sqrt38$d = \frac{|-69|}{3\sqrt{38}} = \frac{23}{\sqrt{38}}$ ### Pattern Recognition Matrix determinant check: In skew lines problems, the numerator can also be computed as the determinant of the 3x3 matrix composed of (vecr_2-vecr_1)$(\vec{r}_2-\vec{r}_1)$, veca$\vec{a}$, and vecb$\vec{b}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula textArea of triangle ABC = frac12 left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) right|$\text{Area of triangle ABC} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$ textMaximum area of an inscribed parallelogram = frac12 times textArea of triangle$\text{Maximum area of an inscribed parallelogram} = \frac{1}{2} \times \text{Area of triangle}$ ### Core Logic First, we compute the total area of the triangle ABC from the given coordinate points. We then apply the geometric maximum area property for inscribed parallelograms. ### Step 1: Calculate the area of triangle ABC The coordinate points are A(4, -2)$A(4, -2)$, B(1, 1)$B(1, 1)$, and C(9, -3)$C(9, -3)$: textArea(Delta textABC) = frac12 beginvmatrix 4 & -2 & 1 \\ 1 & 1 & 1 \\ 9 & -3 & 1 endvmatrix$\text{Area}(\Delta \text{ABC}) = \frac{1}{2} \begin{vmatrix} 4 & -2 & 1 \\ 1 & 1 & 1 \\ 9 & -3 & 1 \end{vmatrix}$ textArea(Delta textABC) = frac12 left| 4(1 - (-3)) - (-2)(1 - 9) + 1(-3 - 9) right|$\text{Area}(\Delta \text{ABC}) = \frac{1}{2} \left| 4(1 - (-3)) - (-2)(1 - 9) + 1(-3 - 9) \right|$ textArea(Delta textABC) = frac12 left| 4(4) + 2(-8) + 1(-12) right|$\text{Area}(\Delta \text{ABC}) = \frac{1}{2} \left| 4(4) + 2(-8) + 1(-12) \right|$ textArea(Delta textABC) = frac12 left| 16 - 16 - 12 right| = 6 text square units$\text{Area}(\Delta \text{ABC}) = \frac{1}{2} \left| 16 - 16 - 12 \right| = 6 \text{ square units}$ ### Step 2: Apply the maximum area theorem The maximum area of a parallelogram inscribed in a triangle of area Delta$\Delta$ is always exactly half the area of the triangle: textMaximum Area = frac12 times textArea(Delta textABC) = frac12 times 6 = 3 text square units$\text{Maximum Area} = \frac{1}{2} \times \text{Area}(\Delta \text{ABC}) = \frac{1}{2} \times 6 = 3 \text{ square units}$ ### Pattern Recognition Inscribed shapes maximization: The maximum area of any inscribed parallelogram AFDE$AFDE$ on the sides of a triangle ABC$ABC$ occurs when the vertices D, E, F$D, E, F$ are exactly the midpoints of the respective sides of the triangle. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines

### Related Formula Area of a triangle given base length b$b$ and perpendicular height h$h$: textArea = frac12 cdot b cdot h$\text{Area} = \frac{1}{2} \cdot b \cdot h$ ### Core Logic Since Q$Q$ and R$R$ lie on the line, the length QR=5$QR=5$ forms the base of the triangle. The perpendicular distance from point P$P$ to the line represents the height h$h$.

### Step 1: Define Perpendicular Foot coordinates Let M$M$ be the foot of the perpendicular from P(0,2,3)$P(0,2,3)$ to the line. Parametric form of any point on the line: M(5lambda - 3, 2lambda + 1, 3lambda - 4)$M(5\lambda - 3, 2\lambda + 1, 3\lambda - 4)$ Direction ratios of line segment PM$PM$: textDRs = (5lambda - 3 - 0, 2lambda + 1 - 2, 3lambda - 4 - 3) = (5lambda - 3, 2lambda - 1, 3lambda - 7)$\text{DRs} = (5\lambda - 3 - 0, 2\lambda + 1 - 2, 3\lambda - 4 - 3) = (5\lambda - 3, 2\lambda - 1, 3\lambda - 7)$ ### Step 2: Solve for Parameter using Perpendicularity Since PM$PM$ is perpendicular to the given line (DRs: 5, 2, 3$5, 2, 3$): 5(5lambda - 3) + 2(2lambda - 1) + 3(3lambda - 7) = 0$5(5\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 7) = 0$ 25lambda - 15 + 4lambda - 2 + 9lambda - 21 = 0 implies 38lambda = 38 implies lambda = 1$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0 \implies 38\lambda = 38 \implies \lambda = 1$ ### Step 3: Compute Perpendicular Distance For lambda = 1$\lambda = 1$, the coordinates of M$M$ are (2, 3, -1)$(2, 3, -1)$. Calculate length PM$PM$: PM = sqrt(2-0)^2 + (3-2)^2 + (-1-3)^2 = sqrt4 + 1 + 16 = sqrt21$PM = \sqrt{(2-0)^2 + (3-2)^2 + (-1-3)^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$ ### Step 4: Formulate the Area Equation textArea = frac12 cdot QR cdot PM = frac12 cdot 5 cdot sqrt21 = fracmn$\text{Area} = \frac{1}{2} \cdot QR \cdot PM = \frac{1}{2} \cdot 5 \cdot \sqrt{21} = \frac{m}{n}$ frac5sqrt212 = fracmn implies 2m = 5sqrt21n implies 2m - 5sqrt21n = 0$\frac{5\sqrt{21}}{2} = \frac{m}{n} \implies 2m = 5\sqrt{21}n \implies 2m - 5\sqrt{21}n = 0$ ### Pattern Recognition Finding the foot of a perpendicular via parametric variables is a guaranteed shortcut for 3D area problems instead of cross-product calculations, keeping computation times minimal. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula De Gua's Theorem (3D extension of Pythagorean theorem) for right corner tetrahedron states: [textArea(Delta BCD)]^2 = [textArea(Delta ABC)]^2 + [textArea(Delta ACD)]^2 + [textArea(Delta ADB)]^2$[\text{Area}(\Delta BCD)]^2 = [\text{Area}(\Delta ABC)]^2 + [\text{Area}(\Delta ACD)]^2 + [\text{Area}(\Delta ADB)]^2$ ### Core Logic Since the three edges meeting at vertex A$A$ are mutually perpendicular, we can align them directly with orthogonal coordinate axes to compute the slanted face area. ### Step 1: Apply Square Area Summation Let the target area be Delta$\Delta$. Delta^2 = 5^2 + 6^2 + 7^2$\Delta^2 = 5^2 + 6^2 + 7^2$ Delta^2 = 25 + 36 + 49 = 110$\Delta^2 = 25 + 36 + 49 = 110$ Delta = sqrt110$\Delta = \sqrt{110}$ ### Pattern Recognition This is exactly analogous to finding the length of a vector given three perpendicular component lengths. Squaring, adding, and taking the square root yields the value immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Vector Algebra