### Related Formula The vector connecting the shortest distance points P$P$ and Q$Q$ on two skew lines must be simultaneously perpendicular to the direction vectors vecb_1$\vec{b}_1$ and vecb_2$\vec{b}_2$ of both lines: vecPQ parallel (vecb_1 times vecb_2)$\vec{PQ} \parallel (\vec{b}_1 \times \vec{b}_2)$ ### Core Logic Let's define general points on both lines: - Point P$P$ on L_1$L_1$: (1+lambda, \, 2-lambda, \, 3+lambda)$(1+\lambda, \, 2-\lambda, \, 3+\lambda)$ - Point Q$Q$ on L_2$L_2$: (4+mu, \, 5+mu, \, 6-mu)$(4+\mu, \, 5+\mu, \, 6-\mu)$ The direction ratios of vector vecPQ$\vec{PQ}$ are: vecPQ = (3+mu-lambda)hati + (3+mu+lambda)hatj + (3-mu-lambda)hatk$\vec{PQ} = (3+\mu-\lambda)\hat{i} + (3+\mu+\lambda)\hat{j} + (3-\mu-\lambda)\hat{k}$ ### Step 1: Compute Perpendicular Direction Vector Calculate the cross product of the directions of lines L_1$L_1$ and L_2$L_2$: vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ 1 & -1 & 1 \\ 1 & 1 & -1 endvmatrix = 0hati + 2hatj + 2hatk$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = 0\hat{i} + 2\hat{j} + 2\hat{k}$ Since vecPQ$\vec{PQ}$ is parallel to (0, 2, 2)$(0, 2, 2)$, we compare the coordinate ratios: 3+mu-lambda = 0 implies lambda - mu = 3 quad implies (1)$3+\mu-\lambda = 0 \implies \lambda - \mu = 3 \quad \implies (1)$ frac3+mu+lambda2 = frac3-mu-lambda2 implies 2mu + 2lambda = 0 implies lambda + mu = 0 quad implies (2)$\frac{3+\mu+\lambda}{2} = \frac{3-\mu-\lambda}{2} \implies 2\mu + 2\lambda = 0 \implies \lambda + \mu = 0 \quad \implies (2)$

### Step 2: Solve for Parameters and Midpoint Solving linear equations (1) and (2) simultaneously: lambda = frac32, quad mu = -frac32$\lambda = \frac{3}{2}, \quad \mu = -\frac{3}{2}$ Substitute these values back to find the specific coordinates of points P$P$ and Q$Q$: - P = left(frac52, \, frac12, \, frac92 ight)$P = \left(\frac{5}{2}, \, \frac{1}{2}, \, \frac{9}{2} ight)$ - Q = left(frac52, \, frac72, \, frac152 ight)$Q = \left(\frac{5}{2}, \, \frac{7}{2}, \, \frac{15}{2} ight)$ The midpoint coordinates (alpha, beta, gamma)$(\alpha, \beta, \gamma)$ are: (alpha, beta, gamma) = left( frac5/2 + 5/22, \, frac1/2 + 7/22, \, frac9/2 + 15/22 right) = left(frac52, \, 2, \, 6 ight)$(\alpha, \beta, \gamma) = \left( \frac{5/2 + 5/2}{2}, \, \frac{1/2 + 7/2}{2}, \, \frac{9/2 + 15/2}{2} \right) = \left(\frac{5}{2}, \, 2, \, 6 ight)$ ### Step 3: Final Computation Calculate the required terms: 2(alpha+beta+gamma) = 2left(frac52 + 2 + 6right) = 5 + 4 + 12 = 21$2(\alpha+\beta+\gamma) = 2\left(\frac{5}{2} + 2 + 6\right) = 5 + 4 + 12 = 21$ ### Pattern Recognition Sees: Explicit endpoints of the shortest distance line vector segment. Shortcut: Since the cross product component along hati$\hat{i}$ is 0, the x-coordinates of both line points are identical, providing a massive shortcut to check algebraic equations immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula cos theta = fracveca cdot vecb|veca||vecb|$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$ ### Core Logic Let us compute the direction ratios of the vectors overrightarrowPQ$\overrightarrow{PQ}$ and overrightarrowPR$\overrightarrow{PR}$ originating from vertex P$P$: Direction ratios of overrightarrowPQ = (4 - 3, 6 - 2, 2 - 3) = (1, 4, -1)$\overrightarrow{PQ} = (4 - 3, 6 - 2, 2 - 3) = (1, 4, -1)$ Direction ratios of overrightarrowPR = (7 - 3, 3 - 2, 2 - 3) = (4, 1, -1)$\overrightarrow{PR} = (7 - 3, 3 - 2, 2 - 3) = (4, 1, -1)$ ### Step 1: Evaluating the Angle Using the dot product formula for the angle theta = angle QPR$\theta = \angle QPR$: cos theta = frac1(4) + 4(1) + (-1)(-1)sqrt1^2 + 4^2 + (-1)^2 cdot sqrt4^2 + 1^2 + (-1)^2$\cos \theta = \frac{1(4) + 4(1) + (-1)(-1)}{\sqrt{1^2 + 4^2 + (-1)^2} \cdot \sqrt{4^2 + 1^2 + (-1)^2}}$ cos theta = frac4 + 4 + 1sqrt18 cdot sqrt18 = frac918 = frac12$\cos \theta = \frac{4 + 4 + 1}{\sqrt{18} \cdot \sqrt{18}} = \frac{9}{18} = \frac{1}{2}$

Since cos theta = frac12$\cos \theta = \frac{1}{2}$, we have: theta = fracpi3$\theta = \frac{\pi}{3}$ ### Pattern Recognition When asked for an angle like angle QPR$\angle QPR$, always ensure both vectors diverge from the common vertex P$P$ (i.e., use overrightarrowPQ$\overrightarrow{PQ}$ and overrightarrowPR$\overrightarrow{PR}$) to avoid sign errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula The line segment of shortest distance MN$MN$ is perpendicular to both directing vectors vecb_1$\vec{b}_1$ and vecb_2$\vec{b}_2$. ### Core Logic Let general point coordinates be expressions of parameters lambda$\lambda$ and mu$\mu$: M = (4lambda + 5, lambda + 4, 3lambda + 5)$M = (4\lambda + 5, \lambda + 4, 3\lambda + 5)$ N = (12mu - 8, 5mu - 2, 9mu - 11)$N = (12\mu - 8, 5\mu - 2, 9\mu - 11)$ Vector direction ratios of overrightarrowMN$\overrightarrow{MN}$: overrightarrowMN = (4lambda - 12mu + 13, lambda - 5mu + 6, 3lambda - 9mu + 16)$\overrightarrow{MN} = (4\lambda - 12\mu + 13, \lambda - 5\mu + 6, 3\lambda - 9\mu + 16)$ The cross product vector perpendicular to both lines is vecb_1 times vecb_2 = (-6, 0, 8)$\vec{b}_1 \times \vec{b}_2 = (-6, 0, 8)$. ### Step 1: Finding Parameters via Perpendicular Conditions Equating directional proportional factors: frac4lambda - 12mu + 13-6 = fraclambda - 5mu + 60 = frac3lambda - 9mu + 168$\frac{4\lambda - 12\mu + 13}{-6} = \frac{\lambda - 5\mu + 6}{0} = \frac{3\lambda - 9\mu + 16}{8}$ From the zero denominator constraint: lambda - 5mu + 6 = 0 quad dots (iii)$\lambda - 5\mu + 6 = 0 \quad \dots (iii)$ From the first and third components: 8(4lambda - 12mu + 13) = -6(3lambda - 9mu + 16) implies 32lambda - 96mu + 104 = -18lambda + 54mu - 96$8(4\lambda - 12\mu + 13) = -6(3\lambda - 9\mu + 16) \implies 32\lambda - 96\mu + 104 = -18\lambda + 54\mu - 96$ 50lambda - 150mu + 200 = 0 implies lambda - 3mu + 4 = 0 quad dots (iv)$50\lambda - 150\mu + 200 = 0 \implies \lambda - 3\mu + 4 = 0 \quad \dots (iv)$ Solving equations (iii) and (iv) yields lambda = -1$\lambda = -1$ and mu = 1$\mu = 1$. ### Step 2: Vector Coordinate Evaluation Substituting parameter roots into point layout vectors: M = (4(-1)+5, -1+4, 3(-1)+5) = (1, 3, 2)$M = (4(-1)+5, -1+4, 3(-1)+5) = (1, 3, 2)$ N = (12(1)-8, 5(1)-2, 9(1)-11) = (4, 3, -2)$N = (12(1)-8, 5(1)-2, 9(1)-11) = (4, 3, -2)$ Evaluating target scalar products: overlinemathrmOM cdot overlinemathrmON = 1(4) + 3(3) + 2(-2) = 4 + 9 - 4 = 9$\overline{\mathrm{OM}} \cdot \overline{\mathrm{ON}} = 1(4) + 3(3) + 2(-2) = 4 + 9 - 4 = 9$ ### Pattern Recognition Shortest distance points constrain line segments to align with the direct cross product vector. Setting proportional components yields quick parameters. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula fracx-x_1a = fracy-y_1b = fracz-z_1c = lambda$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \lambda$ ### Core Logic Let point A = (7, -2, 11)$A = (7, -2, 11)$. We want the distance from A$A$ to the line L_1: fracx-61=fracy-40=fracz-83$L_1: \frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ measured *along* a line parallel to L_2: fracx-52=fracy-1-3=fracz-56$L_2: \frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$. The line passing through A$A$ parallel to L_2$L_2$ will have the equation: fracx-72 = fracy+2-3 = fracz-116 = lambda$\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6} = \lambda$ Any general point B$B$ on this new line can be represented as: B equiv (2lambda + 7, -3lambda - 2, 6lambda + 11)$B \equiv (2\lambda + 7, -3\lambda - 2, 6\lambda + 11)$ ### Step 1: Finding Intersection Point B Since B$B$ lies on the given line L_1$L_1$, its coordinates must satisfy the equation of L_1$L_1$: frac(2lambda + 7) - 61 = frac(-3lambda - 2) - 40 = frac(6lambda + 11) - 83$\frac{(2\lambda + 7) - 6}{1} = \frac{(-3\lambda - 2) - 4}{0} = \frac{(6\lambda + 11) - 8}{3}$ Focus on the middle term (since denominator is 0, numerator must equal 0 for intersection): -3lambda - 6 = 0 Rightarrow lambda = -2$-3\lambda - 6 = 0 \Rightarrow \lambda = -2$ ### Step 2: Coordinates of B and Distance Calculation Substitute lambda = -2$\lambda = -2$ back into the coordinates of point B$B$: B = (2(-2)+7, -3(-2)-2, 6(-2)+11) = (3, 4, -1)$B = (2(-2)+7, -3(-2)-2, 6(-2)+11) = (3, 4, -1)$ Now, calculate the distance AB$AB$ using the 3D distance formula: AB = sqrt(7-3)^2 + (-2-4)^2 + (11 - (-1))^2$AB = \sqrt{(7-3)^2 + (-2-4)^2 + (11 - (-1))^2}$ AB = sqrt4^2 + (-6)^2 + (12)^2$AB = \sqrt{4^2 + (-6)^2 + (12)^2}$ AB = sqrt16 + 36 + 144$AB = \sqrt{16 + 36 + 144}$ AB = sqrt196 = 14$AB = \sqrt{196} = 14$ ### Pattern Recognition Distance of a point from a line *along* another direction means finding the intersection of the given line and a new line passing through the point parallel to the direction vector. The zero in the direction ratio is a massive shortcut—just equate the corresponding numerator to zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry

### Related Formula d = frac|(veca_2 - veca_1) cdot (vecb_1 times vecb_2)||vecb_1 times vecb_2|$d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ ### Core Logic Identify the positional vectors and direction ratios for both lines: Line 1: veca_1 = (4, -1, 0)$\vec{a}_1 = (4, -1, 0)$, direction vecb_1 = (1, 2, -3)$\vec{b}_1 = (1, 2, -3)$ Line 2: veca_2 = (lambda, -1, 2)$\vec{a}_2 = (\lambda, -1, 2)$, direction vecb_2 = (2, 4, -5)$\vec{b}_2 = (2, 4, -5)$ Vector connecting lines: (veca_2 - veca_1) = (lambda - 4, 0, 2)$(\vec{a}_2 - \vec{a}_1) = (\lambda - 4, 0, 2)$ ### Step 1: Cross Product and Magnitude Find the normal vector vecn = vecb_1 times vecb_2$\vec{n} = \vec{b}_1 \times \vec{b}_2$: vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ 1 & 2 & -3 \\ 2 & 4 & -5 endvmatrix$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{vmatrix}$ = hati(-10 - (-12)) - hatj(-5 - (-6)) + hatk(4 - 4)$= \hat{i}(-10 - (-12)) - \hat{j}(-5 - (-6)) + \hat{k}(4 - 4)$ = 2hati - 1hatj + 0hatk = (2, -1, 0)$= 2\hat{i} - 1\hat{j} + 0\hat{k} = (2, -1, 0)$ Magnitude of the normal vector: |vecb_1 times vecb_2| = sqrt2^2 + (-1)^2 + 0^2 = sqrt5$|\vec{b}_1 \times \vec{b}_2| = \sqrt{2^2 + (-1)^2 + 0^2} = \sqrt{5}$ ### Step 2: Application of Shortest Distance Formula Dot product of normal vector and positional difference vector: (veca_2 - veca_1) cdot (vecb_1 times vecb_2) = (lambda - 4)(2) + (0)(-1) + (2)(0) = 2(lambda - 4)$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (\lambda - 4)(2) + (0)(-1) + (2)(0) = 2(\lambda - 4)$ Using the shortest distance formula given as frac6sqrt5$\frac{6}{\sqrt{5}}$: frac|2(lambda - 4)|sqrt5 = frac6sqrt5$\frac{|2(\lambda - 4)|}{\sqrt{5}} = \frac{6}{\sqrt{5}}$ |2(lambda - 4)| = 6$|2(\lambda - 4)| = 6$ |lambda - 4| = 3$|\lambda - 4| = 3$ ### Step 3: Finding Unknown values Solve the absolute value relation: lambda - 4 = 3 Rightarrow lambda = 7$\lambda - 4 = 3 \Rightarrow \lambda = 7$ lambda - 4 = -3 Rightarrow lambda = 1$\lambda - 4 = -3 \Rightarrow \lambda = 1$ Sum of possible values = 7 + 1 = 8$7 + 1 = 8$. ### Pattern Recognition Standard Shortest Distance methodology between skew lines. Cross product of direction vectors forms the perpendicular frame normal, and dot-producting the difference of positional anchor points yields the direct orthogonal projection. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry Class 12 Maths: Vector Algebra