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 textParametric form of a line: vecr = veca + lambdavecb$$\text{Parametric form of a line: } \vec{r} = \vec{a} + \lambda\vec{b}$$ ### Core Logic Let the given point be P(-1, 2, 3)$P(-1, 2, 3)$. A line passing through P$P$ intersects L_1$L_1$ at M$M$ and L_2$L_2$ at N$N$. Write general coordinates for M$M$ on L_1$L_1$ and N$N$ on L_2$L_2$: L_1: fracx-13 = fracy-22 = fracz+1-2 = lambda Rightarrow M(3lambda + 1, 2lambda + 2, -2lambda - 1)$L_1: \frac{x-1}{3} = \frac{y-2}{2} = \frac{z+1}{-2} = \lambda \Rightarrow M(3\lambda + 1, 2\lambda + 2, -2\lambda - 1)$ L_2: fracx+2-3 = fracy-2-2 = fracz-14 = mu Rightarrow N(-3mu - 2, -2mu + 2, 4mu + 1)$L_2: \frac{x+2}{-3} = \frac{y-2}{-2} = \frac{z-1}{4} = \mu \Rightarrow N(-3\mu - 2, -2\mu + 2, 4\mu + 1)$ ### Step 1: Collinearity Condition Since P, M, N$P, M, N$ lie on the same straight line, the direction vectors vecPM$\vec{PM}$ and vecPN$\vec{PN}$ must be proportional. vecPM = (3lambda + 1 - (-1))hati + (2lambda + 2 - 2)hatj + (-2lambda - 1 - 3)hatk = (3lambda + 2)hati + (2lambda)hatj + (-2lambda - 4)hatk$\vec{PM} = (3\lambda + 1 - (-1))\hat{i} + (2\lambda + 2 - 2)\hat{j} + (-2\lambda - 1 - 3)\hat{k} = (3\lambda + 2)\hat{i} + (2\lambda)\hat{j} + (-2\lambda - 4)\hat{k}$ vecPN = (-3mu - 2 - (-1))hati + (-2mu + 2 - 2)hatj + (4mu + 1 - 3)hatk = (-3mu - 1)hati + (-2mu)hatj + (4mu - 2)hatk$\vec{PN} = (-3\mu - 2 - (-1))\hat{i} + (-2\mu + 2 - 2)\hat{j} + (4\mu + 1 - 3)\hat{k} = (-3\mu - 1)\hat{i} + (-2\mu)\hat{j} + (4\mu - 2)\hat{k}$ Proportionality yields: frac3lambda + 2-3mu - 1 = frac2lambda-2mu = frac-2lambda - 44mu - 2$$\frac{3\lambda + 2}{-3\mu - 1} = \frac{2\lambda}{-2\mu} = \frac{-2\lambda - 4}{4\mu - 2}$$ ### Step 2: Solving for Lambda and Mu

From the first pair: frac3lambda + 2-3mu - 1 = frac-lambdamu$$\frac{3\lambda + 2}{-3\mu - 1} = \frac{-\lambda}{\mu}$$ mu(3lambda + 2) = -lambda(-3mu - 1)$$\mu(3\lambda + 2) = -\lambda(-3\mu - 1)$$ 3lambdamu + 2mu = 3lambdamu + lambda Rightarrow 2mu = lambda$$3\lambda\mu + 2\mu = 3\lambda\mu + \lambda \Rightarrow 2\mu = \lambda$$ From the second pair (using 2mu = lambda$2\mu = \lambda$): frac-lambdamu = frac-lambda - 22mu - 1$$\frac{-\lambda}{\mu} = \frac{-\lambda - 2}{2\mu - 1}$$ -2mu = frac-2mu - 22mu - 1$$-2\mu = \frac{-2\mu - 2}{2\mu - 1}$$ -2mu(2mu - 1) = -2mu - 2$$-2\mu(2\mu - 1) = -2\mu - 2$$ -4mu^2 + 2mu = -2mu - 2$$-4\mu^2 + 2\mu = -2\mu - 2$$ -4mu^2 + 4mu + 2 = 0 quad text(Wait, alternative grouping):$$-4\mu^2 + 4\mu + 2 = 0 \quad \text{(Wait, alternative grouping)}:$$ Re-checking the second pair with 2mu = lambda$2\mu = \lambda$: frac2lambda-2mu = frac-lambdamu$$\frac{2\lambda}{-2\mu} = \frac{-\lambda}{\mu}$$ frac-lambdamu = frac-2lambda - 44mu - 2$$\frac{-\lambda}{\mu} = \frac{-2\lambda - 4}{4\mu - 2}$$ Substitute lambda = 2mu$\lambda = 2\mu$: frac-2mumu = frac-4mu - 44mu - 2$$\frac{-2\mu}{\mu} = \frac{-4\mu - 4}{4\mu - 2}$$ -2(4mu - 2) = -4mu - 4$$-2(4\mu - 2) = -4\mu - 4$$ -8mu + 4 = -4mu - 4$$-8\mu + 4 = -4\mu - 4$$ 8 = 4mu Rightarrow mu = 2$$8 = 4\mu \Rightarrow \mu = 2$$ Then lambda = 2(2) = 4$\lambda = 2(2) = 4$. ### Step 3: Finalizing Target Coordinates For lambda = 4$\lambda = 4$: alpha + beta + gamma = (3(4)+1) + (2(4)+2) + (-2(4)-1) = 13 + 10 - 9 = 14$\alpha + \beta + \gamma = (3(4)+1) + (2(4)+2) + (-2(4)-1) = 13 + 10 - 9 = 14$ For mu = 2$\mu = 2$: a + b + c = (-3(2)-2) + (-2(2)+2) + (4(2)+1) = -8 - 2 + 9 = -1$a + b + c = (-3(2)-2) + (-2(2)+2) + (4(2)+1) = -8 - 2 + 9 = -1$ Finally, compute the target fraction: frac(alpha + beta + gamma)^2(a + b + c)^2 = frac(14)^2(-1)^2 = frac1961 = 196$$\frac{(\alpha + \beta + \gamma)^2}{(a + b + c)^2} = \frac{(14)^2}{(-1)^2} = \frac{196}{1} = 196$$ ### Pattern Recognition A line crossing two given lines acts as a transversal holding 3 collinear points. Setting vecPM = k cdot vecPN$\vec{PM} = k \cdot \vec{PN}$ rapidly locks the parametric constants lambda$\lambda$ and mu$\mu$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry

### Related Formula a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 quad text(Condition for perpendicular vectors)$$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \quad \text{(Condition for perpendicular vectors)}$$ ### Core Logic

Let the given point be A(1, 2, 3)$A(1, 2, 3)$. Let the foot of the perpendicular on the line be P(alpha, beta, gamma)$P(\alpha, \beta, \gamma)$. The equation of the line is fracx + 35 = fracy - 12 = fracz + 43 = k$\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} = k$. So, any point on the line can be written as (5k - 3, 2k + 1, 3k - 4)$(5k - 3, 2k + 1, 3k - 4)$. Let this point be P$P$. Thus, P(alpha, beta, gamma) equiv (5k - 3, 2k + 1, 3k - 4)$P(\alpha, \beta, \gamma) \equiv (5k - 3, 2k + 1, 3k - 4)$. ### Step 1: Finding Direction Ratios Direction ratios (DR's) of the line segment AP$AP$ are: ((5k - 3) - 1, (2k + 1) - 2, (3k - 4) - 3) equiv (5k - 4, 2k - 1, 3k - 7)$$((5k - 3) - 1, (2k + 1) - 2, (3k - 4) - 3) \equiv (5k - 4, 2k - 1, 3k - 7)$$ Direction ratios of the given line are (5, 2, 3)$(5, 2, 3)$. ### Step 2: Using Perpendicularity Condition Since AP$AP$ is perpendicular to the given line, the dot product of their direction ratios is zero: 5(5k - 4) + 2(2k - 1) + 3(3k - 7) = 0$$5(5k - 4) + 2(2k - 1) + 3(3k - 7) = 0$$ (25k - 20) + (4k - 2) + (9k - 21) = 0$$(25k - 20) + (4k - 2) + (9k - 21) = 0$$ 38k - 43 = 0$38k - 43 = 0$ k = frac4338$$k = \frac{43}{38}$$ ### Step 3: Calculating target expression We need to find 19(alpha + beta + gamma)$19(\alpha + \beta + \gamma)$. alpha + beta + gamma = (5k - 3) + (2k + 1) + (3k - 4)$$\alpha + \beta + \gamma = (5k - 3) + (2k + 1) + (3k - 4)$$ alpha + beta + gamma = 10k - 6$$\alpha + \beta + \gamma = 10k - 6$$ Substituting k = frac4338$k = \frac{43}{38}$: 19(alpha + beta + gamma) = 19 left(10left(frac4338right) - 6right)$$19(\alpha + \beta + \gamma) = 19 \left(10\left(\frac{43}{38}\right) - 6\right)$$ = 19 left( frac43038 - 6 right) = frac4302 - 114 = 215 - 114 = 101$$= 19 \left( \frac{430}{38} - 6 \right) = \frac{430}{2} - 114 = 215 - 114 = 101$$ ### Pattern Recognition To find coordinates of a foot of perpendicular, define a general point on the line using parameter k$k$, create the direction vector, and dot it with the line's direction vector set to zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry

### Related Formula textShortest distance between vecr = veca_1 + lambda vecb_1 text and vecr = veca_2 + mu vecb_2 text is d = frac|(veca_2 - veca_1) cdot (vecb_1 times vecb_2)||vecb_1 times vecb_2|$$\text{Shortest distance between } \vec{r} = \vec{a}_1 + \lambda \vec{b}_1 \text{ and } \vec{r} = \vec{a}_2 + \mu \vec{b}_2 \text{ is } d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$$ ### Core Logic For d_1$d_1$, rewrite L_1$L_1$ and L_2$L_2$ into standard form: L_1: fracx + 11 = fracy1/2 = fracz-1/12$L_1: \frac{x + 1}{1} = \frac{y}{1/2} = \frac{z}{-1/12}$ veca_1 = (-1, 0, 0)$\vec{a}_1 = (-1, 0, 0)$, vecb_1 = left(1, frac12, -frac112right) propto (12, 6, -1)$\vec{b}_1 = \left(1, \frac{1}{2}, -\frac{1}{12}\right) \propto (12, 6, -1)$ L_2: x = y + 2 = 6(z - 1) Rightarrow fracx1 = fracy + 21 = fracz - 11/6$L_2: x = y + 2 = 6(z - 1) \Rightarrow \frac{x}{1} = \frac{y + 2}{1} = \frac{z - 1}{1/6}$ veca_2 = (0, -2, 1)$\vec{a}_2 = (0, -2, 1)$, vecb_2 = left(1, 1, frac16right) propto (6, 6, 1)$\vec{b}_2 = \left(1, 1, \frac{1}{6}\right) \propto (6, 6, 1)$ ### Step 1: Computing d1 veca_2 - veca_1 = (1, -2, 1)$\vec{a}_2 - \vec{a}_1 = (1, -2, 1)$ vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ 12 & 6 & -1 \\ 6 & 6 & 1 endvmatrix = hati(6 + 6) - hatj(12 + 6) + hatk(72 - 36) = 12hati - 18hatj + 36hatk propto 2hati - 3hatj + 6hatk$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 12 & 6 & -1 \\ 6 & 6 & 1 \end{vmatrix} = \hat{i}(6 + 6) - \hat{j}(12 + 6) + \hat{k}(72 - 36) = 12\hat{i} - 18\hat{j} + 36\hat{k} \propto 2\hat{i} - 3\hat{j} + 6\hat{k}$ |vecb_1 times vecb_2| = sqrt4 + 9 + 36 = sqrt49 = 7$|\vec{b}_1 \times \vec{b}_2| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$ d_1 = frac|(1, -2, 1) cdot (2, -3, 6)|7 = frac|2 + 6 + 6|7 = frac147 = 2$$d_1 = \frac{|(1, -2, 1) \cdot (2, -3, 6)|}{7} = \frac{|2 + 6 + 6|}{7} = \frac{14}{7} = 2$$ ### Step 2: Computing d2 L_3: fracx - 12 = fracy + 8-7 = fracz - 45$L_3: \frac{x - 1}{2} = \frac{y + 8}{-7} = \frac{z - 4}{5}$ veca_3 = (1, -8, 4)$\vec{a}_3 = (1, -8, 4)$, vecb_3 = (2, -7, 5)$\vec{b}_3 = (2, -7, 5)$ L_4: fracx - 12 = fracy - 21 = fracz - 6-3$L_4: \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{-3}$ veca_4 = (1, 2, 6)$\vec{a}_4 = (1, 2, 6)$, vecb_4 = (2, 1, -3)$\vec{b}_4 = (2, 1, -3)$ veca_4 - veca_3 = (0, 10, 2)$\vec{a}_4 - \vec{a}_3 = (0, 10, 2)$ vecb_3 times vecb_4 = beginvmatrix hati & hatj & hatk \\ 2 & -7 & 5 \\ 2 & 1 & -3 endvmatrix = hati(21 - 5) - hatj(-6 - 10) + hatk(2 + 14) = 16hati + 16hatj + 16hatk propto hati + hatj + hatk$\vec{b}_3 \times \vec{b}_4 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & 5 \\ 2 & 1 & -3 \end{vmatrix} = \hat{i}(21 - 5) - \hat{j}(-6 - 10) + \hat{k}(2 + 14) = 16\hat{i} + 16\hat{j} + 16\hat{k} \propto \hat{i} + \hat{j} + \hat{k}$ |vecb_3 times vecb_4| = sqrt1 + 1 + 1 = sqrt3$|\vec{b}_3 \times \vec{b}_4| = \sqrt{1 + 1 + 1} = \sqrt{3}$ d_2 = frac|(0, 10, 2) cdot (1, 1, 1)|sqrt3 = frac|0 + 10 + 2|sqrt3 = frac12sqrt3$$d_2 = \frac{|(0, 10, 2) \cdot (1, 1, 1)|}{\sqrt{3}} = \frac{|0 + 10 + 2|}{\sqrt{3}} = \frac{12}{\sqrt{3}}$$ ### Step 3: Evaluating final expression Target: frac32sqrt3d_1d_2$\frac{32\sqrt{3}d_1}{d_2}$ = frac32sqrt3 times 212 / sqrt3 = frac64sqrt3 cdot sqrt312 = frac64 times 312 = frac19212 = 16$$= \frac{32\sqrt{3} \times 2}{12 / \sqrt{3}} = \frac{64\sqrt{3} \cdot \sqrt{3}}{12} = \frac{64 \times 3}{12} = \frac{192}{12} = 16$$ ### Pattern Recognition Extracting direction ratios efficiently by normalizing the denominator scaling is crucial to avoid fraction arithmetic errors in cross products. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry

### Related Formula textDot product constraint: vecPR cdot vecd = 0 text where vecd text is direction of the line.$$\text{Dot product constraint: } \vec{PR} \cdot \vec{d} = 0 \text{ where } \vec{d} \text{ is direction of the line.}$$ ### Core Logic

Let P = (2, 3, 5)$P = (2, 3, 5)$ and its mirror image be R(alpha, beta, gamma)$R(\alpha, \beta, \gamma)$. The vector connecting the point and its mirror image, vecPR$\vec{PR}$, is strictly perpendicular to the given line. The direction ratios of the line are vecd = (2, 3, 4)$\vec{d} = (2, 3, 4)$. Therefore, the dot product must be zero: vecPR cdot vecd = 0$$\vec{PR} \cdot \vec{d} = 0$$ (alpha - 2, beta - 3, gamma - 5) cdot (2, 3, 4) = 0$$(\alpha - 2, \beta - 3, \gamma - 5) \cdot (2, 3, 4) = 0$$ 2(alpha - 2) + 3(beta - 3) + 4(gamma - 5) = 0$$2(\alpha - 2) + 3(\beta - 3) + 4(\gamma - 5) = 0$$ 2alpha - 4 + 3beta - 9 + 4gamma - 20 = 0$$2\alpha - 4 + 3\beta - 9 + 4\gamma - 20 = 0$$ 2alpha + 3beta + 4gamma = 33$$2\alpha + 3\beta + 4\gamma = 33$$ ### Pattern Recognition When asked for a linear combination of image coordinates matching the direction ratios of the line, skip finding the exact foot of perpendicular. Just use the orthogonality condition directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry

### Related Formula textShortest distance = frac|(veca_2 - veca_1) cdot (vecn_1 times vecn_2)||vecn_1 times vecn_2|$$\text{Shortest distance } = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{n}_1 \times \vec{n}_2)|}{|\vec{n}_1 \times \vec{n}_2|}$$ ### Core Logic For L_1$L_1$: point a_1(1, -1, -4)$a_1(1, -1, -4)$ and direction vecn_1 = (2, -3, 2)$\vec{n}_1 = (2, -3, 2)$. For L_2$L_2$: passes through A(-4, 4, 3)$A(-4, 4, 3)$ and B(-1, 6, 3)$B(-1, 6, 3)$. Direction vecn_2 = vecAB = (-1 - (-4), 6 - 4, 3 - 3) = (3, 2, 0)$\vec{n}_2 = \vec{AB} = (-1 - (-4), 6 - 4, 3 - 3) = (3, 2, 0)$. The condition "perpendicular to..." is extra confirming information, since (3,2,0) cdot (-2,3,1) = 0$(3,2,0) \cdot (-2,3,1) = 0$. Cross product of directions: vecn_1 times vecn_2 = beginvmatrix hati & hatj & hatk \\ 2 & -3 & 2 \\ 3 & 2 & 0 endvmatrix = (-4, 6, 13)$$\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{vmatrix} = (-4, 6, 13)$$ Magnitude: |vecn_1 times vecn_2| = sqrt(-4)^2 + 6^2 + 13^2 = sqrt16 + 36 + 169 = sqrt221$|\vec{n}_1 \times \vec{n}_2| = \sqrt{(-4)^2 + 6^2 + 13^2} = \sqrt{16 + 36 + 169} = \sqrt{221}$. Vector connecting lines: veca_2 - veca_1 = (-4 - 1, 4 - (-1), 3 - (-4)) = (-5, 5, 7)$$\vec{a}_2 - \vec{a}_1 = (-4 - 1, 4 - (-1), 3 - (-4)) = (-5, 5, 7)$$ Numerator of distance formula: |(veca_2 - veca_1) cdot (vecn_1 times vecn_2)| = |-5(-4) + 5(6) + 7(13)|$$|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{n}_1 \times \vec{n}_2)| = |-5(-4) + 5(6) + 7(13)|$$ = |20 + 30 + 91| = 141$$= |20 + 30 + 91| = 141$$ Shortest Distance = frac141sqrt221$= \frac{141}{\sqrt{221}}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry