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 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

### Related Formula For a right-angled triangle, the circumcenter is exactly the midpoint of the hypotenuse, and the circumradius is half the hypotenuse. textArea = frac12 times textbase times textheight$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ textPerimeter = textsum of all side lengths$$\text{Perimeter} = \text{sum of all side lengths}$$ ### Core Logic Observe the coordinates of the vertices: A = (a, -2)$A = (a, -2)$ B = (a, 6)$B = (a, 6)$ C = (a/4, -2)$C = (a/4, -2)$ Vertices A$A$ and B$B$ share the same x-coordinate (x=a$x=a$), meaning side AB$AB$ is perfectly vertical. Vertices A$A$ and C$C$ share the same y-coordinate (y=-2$y=-2$), meaning side AC$AC$ is perfectly horizontal. Since a vertical and horizontal line meet at a right angle, Delta ABC$\Delta ABC$ is a right-angled triangle, with the right angle residing at vertex A$A$. In a right-angled triangle, the hypotenuse is the side opposite the right angle, which is BC$BC$. The circumcenter O$O$ lies exactly at the midpoint of hypotenuse BC$BC$. ### Step 1: Evaluate parameter a The midpoint of BC$BC$ is given by: M = left(fraca + a/42, frac6 - 22right) = left(frac5a8, 2right)$$M = \left(\frac{a + a/4}{2}, \frac{6 - 2}{2}\right) = \left(\frac{5a}{8}, 2\right)$$ We are given the circumcenter O$O$ is (5, fraca4)$(5, \frac{a}{4})$. Equating M$M$ and O$O$: For the x-coordinate: frac5a8 = 5 Rightarrow 5a = 40 Rightarrow a = 8$$\frac{5a}{8} = 5 \Rightarrow 5a = 40 \Rightarrow a = 8$$ Let's verify with the y-coordinate: fraca4 = frac84 = 2 quad text(Consistent)$$\frac{a}{4} = \frac{8}{4} = 2 \quad \text{(Consistent)}$$ So, the vertices are A(8, -2)$A(8, -2)$, B(8, 6)$B(8, 6)$, and C(2, -2)$C(2, -2)$. ### Step 2: Calculate Geometric Quantities Calculate the side lengths: AB = |6 - (-2)| = 8$AB = |6 - (-2)| = 8$ AC = |8 - 2| = 6$AC = |8 - 2| = 6$ BC = sqrt8^2 + 6^2 = sqrt64 + 36 = sqrt100 = 10$BC = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10$ Calculate the requested parameters: 1. Circumradius alpha = fractextHypotenuse2 = fracBC2 = frac102 = 5$\alpha = \frac{\text{Hypotenuse}}{2} = \frac{BC}{2} = \frac{10}{2} = 5$ 2. Area beta = frac12 cdot AB cdot AC = frac12 cdot 8 cdot 6 = 24$\beta = \frac{1}{2} \cdot AB \cdot AC = \frac{1}{2} \cdot 8 \cdot 6 = 24$ 3. Perimeter gamma = AB + AC + BC = 8 + 6 + 10 = 24$\gamma = AB + AC + BC = 8 + 6 + 10 = 24$ ### Step 3: Final Calculation Sum the components: alpha + beta + gamma = 5 + 24 + 24 = 53$$\alpha + \beta + \gamma = 5 + 24 + 24 = 53$$ ### Pattern Recognition Whenever you see two coordinates sharing a common x-value and a common y-value amongst three vertices, immediately flag the triangle as right-angled. This skips the arduous circumcenter distance formulas and jumps straight to hypotenuse midpoint rules. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Coordinate Geometry

### Related Formula The centroid G$G$ divides the median from any vertex to the midpoint of the opposite side in a 2:1$2:1$ ratio. G = frac2 cdot M + 1 cdot R3$$G = \frac{2 \cdot M + 1 \cdot R}{3}$$ d = sqrt(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$ ### Core Logic Given R(-1,4,2)$R(-1,4,2)$ and the midpoint of the opposite side PQ$PQ$ is M(2,1,2)$M(2,1,2)$. The centroid G$G$ lies on the median RM$RM$, dividing it internally in the ratio 2:1$2:1$ from vertex R$R$. G = left(frac2(2) + 1(-1)3, frac2(1) + 1(4)3, frac2(2) + 1(2)3right)$$G = \left(\frac{2(2) + 1(-1)}{3}, \frac{2(1) + 1(4)}{3}, \frac{2(2) + 1(2)}{3}\right)$$ G = left(frac4-13, frac2+43, frac4+23right) = (1, 2, 2)$$G = \left(\frac{4-1}{3}, \frac{2+4}{3}, \frac{4+2}{3}\right) = (1, 2, 2)$$ ### Step 1: Find Line Intersection Let the first line be L_1$L_1$: fracx-20 = fracy2 = fracz+3-1 = t$\frac{x-2}{0} = \frac{y}{2} = \frac{z+3}{-1} = t$. Any general point on L_1$L_1$ is A(2, 2t, -t-3)$A(2, 2t, -t-3)$. To find the intersection, substitute point A$A$ into the equation of the second line L_2$L_2$: fracx-11 = fracy+3-3 = fracz+11$\frac{x-1}{1} = \frac{y+3}{-3} = \frac{z+1}{1}$. Using the x-coordinate: frac2-11 = 1$$\frac{2-1}{1} = 1$$ This means the ratio must equal 1$1$ for all coordinates at the intersection. Checking the y-coordinate: frac2t+3-3 = 1 Rightarrow 2t+3 = -3 Rightarrow 2t = -6 Rightarrow t = -3$$\frac{2t+3}{-3} = 1 \Rightarrow 2t+3 = -3 \Rightarrow 2t = -6 \Rightarrow t = -3$$ Checking the z-coordinate with t = -3$t = -3$ to verify intersection: frac-(-3)-3+11 = frac3-3+11 = 1$$\frac{-(-3)-3+1}{1} = \frac{3-3+1}{1} = 1$$. It matches perfectly. Substitute t = -3$t = -3$ back into the coordinates of A$A$: A = (2, 2(-3), -(-3)-3) = (2, -6, 0)$A = (2, 2(-3), -(-3)-3) = (2, -6, 0)$. ### Step 2: Calculate Final Distance Now compute the distance between the centroid G(1,2,2)$G(1,2,2)$ and the intersection point A(2,-6,0)$A(2,-6,0)$: AG = sqrt(1 - 2)^2 + (2 - (-6))^2 + (2 - 0)^2$$AG = \sqrt{(1 - 2)^2 + (2 - (-6))^2 + (2 - 0)^2}$$ AG = sqrt(-1)^2 + (8)^2 + (2)^2$$AG = \sqrt{(-1)^2 + (8)^2 + (2)^2}$$ AG = sqrt1 + 64 + 4$$AG = \sqrt{1 + 64 + 4}$$ AG = sqrt69$AG = \sqrt{69}$ ### Pattern Recognition When given a vertex and the midpoint of the opposite side, bypass the full coordinate sum (x_1+x_2+x_3)/3 $ (x_1+x_2+x_3)/3 $ and go straight to the section formula (2M + R)/3 $ (2M + R)/3 $. For line intersections, equating one fixed coordinate (like the x$x$-component divided by 0) instantly locks the parametric variable. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula textDot Product for Orthogonality: vecA cdot vecB = 0 implies a_1b_1 + a_2b_2 + a_3b_3 = 0$$\text{Dot Product for Orthogonality: } \vec{A} \cdot \vec{B} = 0 \implies a_1b_1 + a_2b_2 + a_3b_3 = 0$$ textDistance between 3D points: d = sqrt(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2$$\text{Distance between 3D points: } d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$ ### Core Logic Let the first line be L_1: x = y+2 = z = t$L_1: x = y+2 = z = t$. Any point P$P$ on L_1$L_1$ has coordinates (t, t-2, t)$(t, t-2, t)$. Let the second line be L_2: fracx+22 = y = z = s$L_2: \frac{x+2}{2} = y = z = s$. Any point Q$Q$ on L_2$L_2$ has coordinates (2s-2, s, s)$(2s-2, s, s)$. The line segment PQ$PQ$ has direction ratios given by the difference of coordinates: DR_PQ = (2s-2-t, s-(t-2), s-t) = (2s-t-2, s-t+2, s-t)$$DR_{PQ} = (2s-2-t, s-(t-2), s-t) = (2s-t-2, s-t+2, s-t)$$ We are given the fixed direction ratios of PQ$PQ$ as (2, 1, 2)$(2, 1, 2)$. Because direction ratios are proportional, we set up equivalence ratios: frac2s-t-22 = fracs-t+21 = fracs-t2$$\frac{2s-t-2}{2} = \frac{s-t+2}{1} = \frac{s-t}{2}$$ ### Step 1: Solve for Line PQ Using the 2nd and 3rd parts of the proportion: fracs-t+21 = fracs-t2$$\frac{s-t+2}{1} = \frac{s-t}{2}$$ 2s - 2t + 4 = s - t Rightarrow s - t = -4 Rightarrow t = s + 4$$2s - 2t + 4 = s - t \Rightarrow s - t = -4 \Rightarrow t = s + 4$$ Using the 1st and 3rd parts of the proportion: frac2s-t-22 = fracs-t2 Rightarrow 2s-t-2 = s-t Rightarrow s = 2$$\frac{2s-t-2}{2} = \frac{s-t}{2} \Rightarrow 2s-t-2 = s-t \Rightarrow s = 2$$ Substitute s=2$s=2$ to find t$t$: t = 2 + 4 = 6$t = 2 + 4 = 6$ Now, substitute these parameters back to find points P and Q: P = (6, 6-2, 6) = (6, 4, 6)$$P = (6, 6-2, 6) = (6, 4, 6)$$ Q = (2(2)-2, 2, 2) = (2, 2, 2)$$Q = (2(2)-2, 2, 2) = (2, 2, 2)$$ The equation of line PQ$PQ$ passing through Q(2,2,2)$Q(2,2,2)$ with direction ratios (2,1,2)$(2,1,2)$ is: fracx-22 = fracy-21 = fracz-22 = lambda$$\frac{x-2}{2} = \frac{y-2}{1} = \frac{z-2}{2} = \lambda$$

### Step 2: Find Perpendicular Foot F Let F$F$ be the foot of the perpendicular from point A(1, 2, 12)$A(1, 2, 12)$ to the line PQ$PQ$. Any general point on line PQ$PQ$ is F(2lambda+2, lambda+2, 2lambda+2)$F(2\lambda+2, \lambda+2, 2\lambda+2)$. The direction ratios of vector vecAF$\vec{AF}$ are: (2lambda+2-1, lambda+2-2, 2lambda+2-12) = (2lambda+1, lambda, 2lambda-10)$$(2\lambda+2-1, \lambda+2-2, 2\lambda+2-12) = (2\lambda+1, \lambda, 2\lambda-10)$$ Since vecAF$\vec{AF}$ is perpendicular to line PQ$PQ$ (which has direction ratios 2, 1, 2$2, 1, 2$), their dot product must be zero: 2(2lambda+1) + 1(lambda) + 2(2lambda-10) = 0$$2(2\lambda+1) + 1(\lambda) + 2(2\lambda-10) = 0$$ 4lambda + 2 + lambda + 4lambda - 20 = 0$$4\lambda + 2 + \lambda + 4\lambda - 20 = 0$$ 9lambda = 18 Rightarrow lambda = 2$$9\lambda = 18 \Rightarrow \lambda = 2$$ Substitute lambda=2$\lambda=2$ to find the exact coordinates of foot F$F$: F = (2(2)+2, 2+2, 2(2)+2) = (6, 4, 6)$$F = (2(2)+2, 2+2, 2(2)+2) = (6, 4, 6)$$ ### Step 3: Compute Final Distance Squared Calculate the squared distance l^2$l^2$ between A(1, 2, 12)$A(1, 2, 12)$ and F(6, 4, 6)$F(6, 4, 6)$: l^2 = (6-1)^2 + (4-2)^2 + (6-12)^2$$l^2 = (6-1)^2 + (4-2)^2 + (6-12)^2$$ l^2 = 5^2 + 2^2 + (-6)^2$$l^2 = 5^2 + 2^2 + (-6)^2$$ l^2 = 25 + 4 + 36 = 65$$l^2 = 25 + 4 + 36 = 65$$ ### Pattern Recognition Whenever you must link two skew lines with a third intersecting line given constant direction ratios, immediately construct generic parametric points on each skew line. Subtraction yields a vector that is directly proportional to the given constants, instantly solving the system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula textTwo lines are perpendicular if their direction vectors' dot product is 0: vecd_1 cdot vecd_2 = 0$$\text{Two lines are perpendicular if their direction vectors' dot product is 0: } \vec{d}_1 \cdot \vec{d}_2 = 0$$ textDirection of a line perpendicular to two lines is their cross product: vecd_3 = vecd_1 times vecd_2$$\text{Direction of a line perpendicular to two lines is their cross product: } \vec{d}_3 = \vec{d}_1 \times \vec{d}_2$$ ### Core Logic Direction vectors of L_1$L_1$ and L_2$L_2$: vecd_1 = hati - hatj + 2hatk$\vec{d}_1 = \hat{i} - \hat{j} + 2\hat{k}$ vecd_2 = 3hati + hatj + phatk$\vec{d}_2 = 3\hat{i} + \hat{j} + p\hat{k}$ Since L_1 perp L_2$L_1 \perp L_2$: vecd_1 cdot vecd_2 = (1)(3) + (-1)(1) + (2)(p) = 0$\vec{d}_1 \cdot \vec{d}_2 = (1)(3) + (-1)(1) + (2)(p) = 0$ 3 - 1 + 2p = 0 Rightarrow 2p = -2 Rightarrow p = -1$3 - 1 + 2p = 0 \Rightarrow 2p = -2 \Rightarrow p = -1$ So, vecd_2 = 3hati + hatj - hatk$\vec{d}_2 = 3\hat{i} + \hat{j} - \hat{k}$. ### Step 1: Finding Direction of L3 Line L_3$L_3$ is perpendicular to both L_1$L_1$ and L_2$L_2$, so its direction vector vecd_3$\vec{d}_3$ is parallel to vecd_1 times vecd_2$\vec{d}_1 \times \vec{d}_2$: vecd_3 = beginvmatrix hati & hatj & hatk \\ 1 & -1 & 2 \\ 3 & 1 & -1 endvmatrix$$\vec{d}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & -1 \end{vmatrix}$$ = hati(1 - 2) - hatj(-1 - 6) + hatk(1 + 3)$$= \hat{i}(1 - 2) - \hat{j}(-1 - 6) + \hat{k}(1 + 3)$$ = -hati + 7hatj + 4hatk$$= -\hat{i} + 7\hat{j} + 4\hat{k}$$ The given equation of L_3$L_3$ is vecr = delta (ell hati + m hatj + n hatk)$\vec{r} = \delta (\ell \hat{i} + m \hat{j} + n \hat{k})$, which means it passes through the origin with direction proportional to vecd_3$\vec{d}_3$. Therefore, points on L_3$L_3$ are of the form (-delta, 7delta, 4delta)$(-\delta, 7\delta, 4\delta)$. ### Step 2: Checking Options Substitute delta = 1$\delta = 1$ into our generalized point (-delta, 7delta, 4delta)$(-\delta, 7\delta, 4\delta)$: Point = (-1, 7, 4)$= (-1, 7, 4)$ This perfectly matches option (1). ### Pattern Recognition A line orthogonal to two known lines always carries a direction vector equivalent to the cross product of the two known direction vectors. Here, the cross product directly generated the required scalar multiple form. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry Class 12 Maths: Vector Algebra