### Related Formula 1. Centroid (G$G$) of a \triangle with vertices A, B, C$A, B, C$ is given by: vecG = fracvecA + vecB + vecC3$\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}$ 2. Euler\'s line property: The orthocenter (O$O$), centroid (G$G$), and circumcenter (C$C$) are collinear, and G$G$ divides the segment OC$OC$ internally in the ratio 2:1$2:1$. ### Step 1: Compute the Centroid Vector Sum the vectors of the three given vertices [cite: 3266, 3268]: vecA = 4vecp+vecq-3vecr$\vec{A} = 4\vec{p}+\vec{q}-3\vec{r}$ vecB = -5vecp+vecq+2vecr$\vec{B} = -5\vec{p}+\vec{q}+2\vec{r}$ vecC = 2vecp-vecq+2vecr$\vec{C} = 2\vec{p}-\vec{q}+2\vec{r}$ vecG = frac(4 - 5 + 2)vecp + (1 + 1 - 1)vecq + (-3 + 2 + 2)vecr3 = fracvecp + vecq + vecr3$\vec{G} = \frac{(4 - 5 + 2)\vec{p} + (1 + 1 - 1)\vec{q} + (-3 + 2 + 2)\vec{r}}{3} = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$ ### Step 2: Apply Euler Line Section Ratio Using the section formula ratio O-G-C$O-G-C$ as 2:1$2:1$ [cite: 3931, 3932]:

vecG = frac2vecC + vecO3 Rightarrow 3vecG = 2vecC + vecO$\vec{G} = \frac{2\vec{C} + \vec{O}}{3} \Rightarrow 3\vec{G} = 2\vec{C} + \vec{O}$ 2vecC = 3vecG - vecO = 3left(fracvecp + vecq + vecr3right) - fracvecp + vecq + vecr4$2\vec{C} = 3\vec{G} - \vec{O} = 3\left(\frac{\vec{p} + \vec{q} + \vec{r}}{3}\right) - \frac{\vec{p} + \vec{q} + \vec{r}}{4}$ 2vecC = (vecp + vecq + vecr) - frac14(vecp + vecq + vecr) = frac34(vecp + vecq + vecr)$2\vec{C} = (\vec{p} + \vec{q} + \vec{r}) - \frac{1}{4}(\vec{p} + \vec{q} + \vec{r}) = \frac{3}{4}(\vec{p} + \vec{q} + \vec{r})$ vecC = frac38vecp + frac38vecq + frac38vecr$\vec{C} = \frac{3}{8}\vec{p} + \frac{3}{8}\vec{q} + \frac{3}{8}\vec{r}$ ### Step 3: Coefficient Matching Compare with the given circumcenter format alphavecp + betavecq + gammavecr$\alpha\vec{p} + \beta\vec{q} + \gamma\vec{r}$ [cite: 3270, 3939]: alpha = frac38, quad beta = frac38, quad gamma = frac38$\alpha = \frac{3}{8}, \quad \beta = \frac{3}{8}, \quad \gamma = \frac{3}{8}$ Calculate alpha + 2beta + 5gamma$\alpha + 2\beta + 5\gamma$ [cite: 3272, 3949]: frac38 + 2left(frac38right) + 5left(frac38right) = frac3 + 6 + 158 = frac248 = 3$\frac{3}{8} + 2\left(\frac{3}{8}\right) + 5\left(\frac{3}{8}\right) = \frac{3 + 6 + 15}{8} = \frac{24}{8} = 3$ ### Pattern Recognition Euler line configuration is universally O-G-C$O-G-C$ in 2:1$2:1$. Remember the mnemonic 'Oil-Gas-Company' or simply 3G = 2C + O$3G = 2C + O$ to prevent swapping structural coefficients under exam stress. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra Class 11 Mathematics: Properties of Triangles

### Related Formula The scalar projection of vector vecv$\vec{v}$ onto vector vecw$\vec{w}$ is calculated as: textProjection = fracvecv cdot vecw|vecw|$\text{Projection} = \frac{\vec{v} \cdot \vec{w}}{|\vec{w}|}$ ### Step 1: Calculate vecb$\vec{b}$ Compute the cross product using standard matrix expansion [cite: 4013, 4015]: vecb = veca times (hati - 2hatk) = beginvmatrix hati & hatj & hatk \\ 3 & -1 & 2 \\ 1 & 0 & -2 endvmatrix$\vec{b} = \vec{a} \times (\hat{i} - 2\hat{k}) = \begin{\vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{\vmatrix}$ vecb = hati(2 - 0) - hatj(-6 - 2) + hatk(0 - (-1)) = 2hati + 8hatj + hatk$\vec{b} = \hat{i}(2 - 0) - \hat{j}(-6 - 2) + \hat{k}(0 - (-1)) = 2\hat{i} + 8\hat{j} + \hat{k}$ ### Step 2: Calculate vecc$\vec{c}$ and vecc - 2hatj$\vec{c} - 2\hat{j}$ Perform the second cross product with unit vector hatk$\hat{k}$ [cite: 3359, 4016]: vecc = vecb times hatk = (2hati + 8hatj + hatk) times hatk = 2(hati times hatk) + 8(hatj times hatk) + vec0$\vec{c} = \vec{b} \times \hat{k} = (2\hat{i} + 8\hat{j} + \hat{k}) \times \hat{k} = 2(\hat{i} \times \hat{k}) + 8(\hat{j} \times \hat{k}) + \vec{0}$ vecc = 2(-hatj) + 8(hati) = 8hati - 2hatj$\vec{c} = 2(-\hat{j}) + 8(\hat{i}) = 8\hat{i} - 2\hat{j}$ Subtract 2hatj$2\hat{j}$ [cite: 3364, 4016]: vecc - 2hatj = (8hati - 2hatj) - 2hatj = 8hati - 4hatj$\vec{c} - 2\hat{j} = (8\hat{i} - 2\hat{j}) - 2\hat{j} = 8\hat{i} - 4\hat{j}$ ### Step 3: Compute the projection onto veca$\vec{a}$ Using the \dot product formula : textProjection = frac(vecc - 2hatj) cdot veca|veca| = fraclangle 8, -4, 0 rangle cdot langle 3, -1, 2 ranglesqrt3^2 + (-1)^2 + 2^2$\text{Projection} = \frac{(\vec{c} - 2\hat{j}) \cdot \vec{a}}{|\vec{a}|} = \frac{\langle 8, -4, 0 \rangle \cdot \langle 3, -1, 2 \rangle}{\sqrt{3^2 + (-1)^2 + 2^2}}$ [cite: 3358, 3995] textProjection = frac24 + 4 + 0sqrt9 + 1 + 4 = frac28sqrt14 = 2sqrt14$\text{Projection} = \frac{24 + 4 + 0}{\sqrt{9 + 1 + 4}} = \frac{28}{\sqrt{14}} = 2\sqrt{14}$ ### Pattern Recognition Keep cyclic unit cross products clear: hati times hatk = -hatj$\hat{i} \times \hat{k} = -\hat{j}$ and hatj times hatk = hati$\hat{j} \times \hat{k} = \hat{i}$. Missing a negative sign during basic cross multiplications ruins multi-step vector projections easily. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula A vector vecmathbfc$\vec{\mathbf{c}}$ coplanar with vecmathbfa$\vec{\mathbf{a}}$ and vecmathbfb$\vec{\mathbf{b}}$ and perpendicular to vecmathbfb$\vec{\mathbf{b}}$ can be expressed using the vector triple product command template: vecmathbfc = lambda (vecmathbfb times (vecmathbfa times vecmathbfb)) = lambda [(vecmathbfb cdot vecmathbfb)vecmathbfa - (vecmathbfa cdot vecmathbfb)vecmathbfb]$\vec{\mathbf{c}} = \lambda (\vec{\mathbf{b}} \times (\vec{\mathbf{a}} \times \vec{\mathbf{b}})) = \lambda [(\vec{\mathbf{b}} \cdot \vec{\mathbf{b}})\vec{\mathbf{a}} - (\vec{\mathbf{a}} \cdot \vec{\mathbf{b}})\vec{\mathbf{b}}]$ ### Core Logic First, find the dot products of the given vectors: vecmathbfb cdot vecmathbfb = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11$\vec{\mathbf{b}} \cdot \vec{\mathbf{b}} = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11$ vecmathbfa cdot vecmathbfb = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2$\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2$ Substituting these values into the expression for vecmathbfc$\vec{\mathbf{c}}$: vecmathbfc = lambda [11vecmathbfa - 2vecmathbfb]$\vec{\mathbf{c}} = \lambda [11\vec{\mathbf{a}} - 2\vec{\mathbf{b}}]$ vecmathbfc = lambda [11(hatmathbfi + 2hatmathbfj + 3hatmathbfk) - 2(3hatmathbfi + hatmathbfj - hatmathbfk)]$\vec{\mathbf{c}} = \lambda [11(\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}}) - 2(3\hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}})]$ vecmathbfc = lambda (5hatmathbfi + 20hatmathbfj + 35hatmathbfk) = 5lambda (hatmathbfi + 4hatmathbfj + 7hatmathbfk)$\vec{\mathbf{c}} = \lambda (5\hat{\mathbf{i}} + 20\hat{\mathbf{j}} + 35\hat{\mathbf{k}}) = 5\lambda (\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})$ ### Step 1: Determine Lambda Using the given condition vecmathbfa cdot vecmathbfc = 5$\vec{\mathbf{a}} \cdot \vec{\mathbf{c}} = 5$: vecmathbfa cdot [5lambda (hatmathbfi + 4hatmathbfj + 7hatmathbfk)] = 5$\vec{\mathbf{a}} \cdot [5\lambda (\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})] = 5$ 5lambda (1 cdot 1 + 2 cdot 4 + 3 cdot 7) = 5$5\lambda (1 \cdot 1 + 2 \cdot 4 + 3 \cdot 7) = 5$ 5lambda (1 + 8 + 21) = 5 implies 30lambda = 1 implies lambda = frac130$5\lambda (1 + 8 + 21) = 5 \implies 30\lambda = 1 \implies \lambda = \frac{1}{30}$ Thus, the vector vecmathbfc$\vec{\mathbf{c}}$ is: vecmathbfc = frac530(hatmathbfi + 4hatmathbfj + 7hatmathbfk) = frac16(hatmathbfi + 4hatmathbfj + 7hatmathbfk)$\vec{\mathbf{c}} = \frac{5}{30}(\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}}) = \frac{1}{6}(\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})$ ### Step 2: Calculate Magnitude The magnitude of vecmathbfc$\vec{\mathbf{c}}$ is evaluated as: \|vecmathbfc\| = fracsqrt1^2 + 4^2 + 7^26 = fracsqrt1 + 16 + 496 = fracsqrt666 = sqrtfrac6636 = sqrtfrac116$\|\vec{\mathbf{c}}\| = \frac{\sqrt{1^2 + 4^2 + 7^2}}{6} = \frac{\sqrt{1 + 16 + 49}}{6} = \frac{\sqrt{66}}{6} = \sqrt{\frac{66}{36}} = \sqrt{\frac{11}{6}}$ ### Pattern Recognition Whenever a vector is specified to be coplanar with vecmathbfa, vecmathbfb$\vec{\mathbf{a}}, \vec{\mathbf{b}}$ and perpendicular to vecmathbfb$\vec{\mathbf{b}}$, direct setup with the cross-product template vecmathbfb times (vecmathbfa times vecmathbfb)$\vec{\mathbf{b}} \times (\vec{\mathbf{a}} \times \vec{\mathbf{b}})$ circumvents solving cumbersome linear systems of scalar variables. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula The line bisecting the angle between two symmetric vectors passing through the origin in the first quadrant is given by y = x$y = x$ or x - y = 0$x - y = 0$. Distance from point (x_1, y_1)$(x_1, y_1)$ to line Ax + By + C = 0$Ax + By + C = 0$ is: d = frac|Ax_1 + By_1 + C|sqrtA^2 + B^2$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$ ### Core Logic Vectors overlineOA = sqrt3hati+hatj$\overline{OA} = \sqrt{3}\hat{i}+\hat{j}$ and overlineOB = hati+sqrt3hatj$\overline{OB} = \hat{i}+\sqrt{3}\hat{j}$ are symmetric about the line y = x$y = x$. Therefore, the angle bisector of overlineOA$\overline{OA}$ and overlineOB$\overline{OB}$ is the line x - y = 0$x - y = 0$. Point C has coordinates (a, 1 - a)$(a, 1 - a)$. ### Step 1: Calculate Distance and Solve for a The perpendicular distance from C(a, 1 - a)$C(a, 1 - a)$ to x - y = 0$x - y = 0$ is: d = frac|a - (1 - a)|sqrt1^2 + (-1)^2 = frac|2a - 1|sqrt2$d = \frac{|a - (1 - a)|}{\sqrt{1^2 + (-1)^2}} = \frac{|2a - 1|}{\sqrt{2}}$ Given that this distance is frac9sqrt2$\frac{9}{\sqrt{2}}$: frac|2a - 1|sqrt2 = frac9sqrt2 implies |2a - 1| = 9$\frac{|2a - 1|}{\sqrt{2}} = \frac{9}{\sqrt{2}} \implies |2a - 1| = 9$ This gives two solutions: 1) 2a - 1 = 9 implies 2a = 10 implies a = 5$2a - 1 = 9 \implies 2a = 10 \implies a = 5$ 2) 2a - 1 = -9 implies 2a = -8 implies a = -4$2a - 1 = -9 \implies 2a = -8 \implies a = -4$ ### Step 2: Find the Sum of Values Sum of all possible values of a$a$: textSum = 5 + (-4) = 1$\text{Sum} = 5 + (-4) = 1$ ### Pattern Recognition Notice that overlineOA$\overline{OA}$ and overlineOB$\overline{OB}$ have swapped coordinates, meaning they are symmetric with respect to y=x$y=x$. Thus, the angle bisector equation is immediate (x-y=0$x-y=0$), simplifying the problem to a standard point-to-line distance calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines Class 12 Mathematics: Vector Algebra

### Related Formula Any vector veca$\vec{a}$ can be written as the sum of its component parallel to vecb$\vec{b}$ (along vecb$\vec{b}$) and its component perpendicular to vecb$\vec{b}$: veca = veca_parallel + veca_perp$\vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp}$ Magnitude squared: |veca|^2 = alpha^2 + beta^2 + gamma^2$|\vec{a}|^2 = \alpha^2 + \beta^2 + \gamma^2$ ### Core Logic Given: veca_parallel = frac1611(3hati+hatj-hatk)$\vec{a}_{\parallel} = \frac{16}{11}(3\hat{i}+\hat{j}-\hat{k})$ veca_perp = frac111(-4hati-5hatj-17hatk)$\vec{a}_{\perp} = \frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})$ ### Step 1: Reconstruct Vector a Add both components to find veca$\vec{a}$: veca = frac1611(3hati+hatj-hatk) + frac111(-4hati-5hatj-17hatk)$\vec{a} = \frac{16}{11}(3\hat{i}+\hat{j}-\hat{k}) + \frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})$ veca = frac111 left[ (48 - 4)hati + (16 - 5)hatj + (-16 - 17)hatk right]$\vec{a} = \frac{1}{11} \left[ (48 - 4)\hat{i} + (16 - 5)\hat{j} + (-16 - 17)\hat{k} \right]$ veca = frac111 left[ 44hati + 11hatj - 33hatk right] = 4hati + hatj - 3hatk$\vec{a} = \frac{1}{11} \left[ 44\hat{i} + 11\hat{j} - 33\hat{k} \right] = 4\hat{i} + \hat{j} - 3\hat{k}$ Therefore, alpha = 4$\alpha = 4$, beta = 1$\beta = 1$, gamma = -3$\gamma = -3$. ### Step 2: Calculate Sum of Squares alpha^2 + beta^2 + gamma^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26$\alpha^2 + \beta^2 + \gamma^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26$ ### Pattern Recognition Since parallel and perpendicular components are orthogonal vectors, you can also use directly |veca|^2 = |veca_parallel|^2 + |veca_perp|^2$|\vec{a}|^2 = |\vec{a}_{\parallel}|^2 + |\vec{a}_{\perp}|^2$ to save algebra step: |veca_parallel|^2 = frac16^211^2(9+1+1) = frac25611$|\vec{a}_{\parallel}|^2 = \frac{16^2}{11^2}(9+1+1) = \frac{256}{11}$, |veca_perp|^2 = frac111^2(16+25+289) = frac330121 = frac3011$|\vec{a}_{\perp}|^2 = \frac{1}{11^2}(16+25+289) = \frac{330}{121} = \frac{30}{11}$. Total = frac28611 = 26$\frac{286}{11} = 26$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra