If the image of the point (4, 4, 3)$(4, 4, 3)$ in the line fracx - 12 = fracy - 21 = fracz - 13$\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}$ is (alpha, beta, gamma)$(\alpha, \beta, \gamma)$, then alpha + beta + gamma$\alpha + \beta + \gamma$ is equal to (1) 9 (2) 12 (3) 8 (4) 7

### Related Formula The shortest distance between two skew lines vecr = veca_1 + svecb_1$\vec{r} = \vec{a}_1 + s\vec{b}_1$ and vecr = veca_2 + tvecb_2$\vec{r} = \vec{a}_2 + t\vec{b}_2$ is given by: 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 vectors from the equations of the lines: - Line 1 passes through veca_1 = lambdahati + 2hatj + hatk$\vec{a}_1 = \lambda\hat{i} + 2\hat{j} + \hat{k}$ with direction vecb_1 = -2hati + hatj + hatk$\vec{b}_1 = -2\hat{i} + \hat{j} + \hat{k}$. - Line 2 passes through veca_2 = sqrt3hati + hatj + 2hatk$\vec{a}_2 = \sqrt{3}\hat{i} + \hat{j} + 2\hat{k}$ with direction vecb_2 = hati - 2hatj + hatk$\vec{b}_2 = \hat{i} - 2\hat{j} + \hat{k}$. Calculate the coordinate difference vector: veca_2 - veca_1 = (sqrt3 - lambda)hati - hatj + hatk$$\vec{a}_2 - \vec{a}_1 = (\sqrt{3} - \lambda)\hat{i} - \hat{j} + \hat{k}$$ ### Step 1: Compute the Cross Product of the Direction Vectors Find the cross product matrix: vecb_1 times vecb_2 = beginvmatrix hati & hatj & hatk \\ -2 & 1 & 1 \\ 1 & -2 & 1 endvmatrix$$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \end{vmatrix}$$ vecb_1 times vecb_2 = hati(1 - (-2)) - hatj(-2 - 1) + hatk(4 - 1) = 3hati + 3hatj + 3hatk$$\vec{b}_1 \times \vec{b}_2 = \hat{i}(1 - (-2)) - \hat{j}(-2 - 1) + \hat{k}(4 - 1) = 3\hat{i} + 3\hat{j} + 3\hat{k}$$ Compute its magnitude: |vecb_1 times vecb_2| = sqrt3^2 + 3^2 + 3^2 = sqrt27 = 3sqrt3$$|\vec{b}_1 \times \vec{b}_2| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt{3}$$

### Step 2: Solve the Absolute Distance Equation for lambda Substitute these values into the shortest distance formula: 1 = frac|((sqrt3 - lambda)hati - hatj + hatk) cdot (3hati + 3hatj + 3hatk)|3sqrt3$$1 = \frac{|((\sqrt{3} - \lambda)\hat{i} - \hat{j} + \hat{k}) \cdot (3\hat{i} + 3\hat{j} + 3\hat{k})|}{3\sqrt{3}}$$ 1 = frac|3(sqrt3 - lambda) - 3 + 3|3sqrt3 = frac3|sqrt3 - lambda|3sqrt3 = frac|sqrt3 - lambda|sqrt3$$1 = \frac{|3(\sqrt{3} - \lambda) - 3 + 3|}{3\sqrt{3}} = \frac{3|\sqrt{3} - \lambda|}{3\sqrt{3}} = \frac{|\sqrt{3} - \lambda|}{\sqrt{3}}$$ This gives: |sqrt3 - lambda| = sqrt3$$|\sqrt{3} - \lambda| = \sqrt{3}$$ Unfolding the absolute value gives two cases: - **Case A:** sqrt3 - lambda = sqrt3 implies lambda = 0$\sqrt{3} - \lambda = \sqrt{3} \implies \lambda = 0$ - **Case B:** sqrt3 - lambda = -sqrt3 implies lambda = 2sqrt3$\sqrt{3} - \lambda = -\sqrt{3} \implies \lambda = 2\sqrt{3}$ ### Step 3: Calculate the Final Sum The sum of all possible values of lambda$\lambda$ is: textSum = 0 + 2sqrt3 = 2sqrt3$$\text{Sum} = 0 + 2\sqrt{3} = 2\sqrt{3}$$ ### Pattern Recognition Sees: Shortest distance setup between vector paths containing free variables. Shortcut: In equations like |c - lambda| = d$|c - \lambda| = d$, the sum of the roots is simply equal to 2c$2c$, because the roots are symmetrically balanced around the center point c$c$. Thus, textSum = 2 times sqrt3 = 2sqrt3$\text{Sum} = 2 \times \sqrt{3} = 2\sqrt{3}$ holds instantly without calculating individual values. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Vector Algebra

### 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