Let veca=-5hati+hatj-3hatk$\vec{a}=-5\hat{i}+\hat{j}-3\hat{k}$, vecb=hati+2hatj-4hatk$\vec{b}=\hat{i}+2\hat{j}-4\hat{k}$ and vecc=(((vecatimesvecb)timeshati)timeshati)timeshati.$\vec{c}=(((\vec{a}\times\vec{b})\times\hat{i})\times\hat{i})\times\hat{i}.$ Then vecccdot(-hati+hatj+hatk)$\vec{c}\cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to:

### Core Logic Let the vertices of the triangle be A$A$, B$B$, and C$C$. The vectors representing the side paths are: overlineAB = 2hati - hatj + hatk$$\overline{AB} = 2\hat{i} - \hat{j} + \hat{k}$$ overlineCA = hati - 3hatj - 5hatk$$\overline{CA} = \hat{i} - 3\hat{j} - 5\hat{k}$$ overlineCB = 3hati - 4hatj - 4hatk$$\overline{CB} = 3\hat{i} - 4\hat{j} - 4\hat{k}$$ Notice that overlineAB + overlineCA = (2+1)hati + (-1-3)hatj + (1-5)hatk = 3hati - 4hatj - 4hatk = overlineCB$\overline{AB} + \overline{CA} = (2+1)\hat{i} + (-1-3)\hat{j} + (1-5)\hat{k} = 3\hat{i} - 4\hat{j} - 4\hat{k} = \overline{CB}$. This structurally validates vector addition rules.

### Step 1: Finding Position Vectors relative to A Let's set vertex A$A$ as the origin origin point (Position vector vecA = vec0$\vec{A} = \vec{0}$): - Position vector of B$B$: vecB = 2hati - hatj + hatk$\vec{B} = 2\hat{i} - \hat{j} + \hat{k}$ - Position vector of C$C$: Since overlineCA = vecA - vecC = -vecC implies vecC = -hati + 3hatj + 5hatk$\overline{CA} = \vec{A} - \vec{C} = -\vec{C} \implies \vec{C} = -\hat{i} + 3\hat{j} + 5hat{k}$ Now, calculate the position vector of the centroid G$G$: vecG = fracvecA + vecB + vecC3 = fracvec0 + (2hati - hatj + hatk) + (-hati + 3hatj + 5hatk)3 = frac13left(hati + 2hatj + 6hatkright)$$\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} = \frac{\vec{0} + (2\hat{i} - \hat{j} + \hat{k}) + (-\hat{i} + 3\hat{j} + 5\hat{k})}{3} = \frac{1}{3}\left(\hat{i} + 2\hat{j} + 6\hat{k}\right)$$ ### Step 2: Calculating Squared Lengths to the Centroid Let's find each individual vector distance block: - overlineAG = vecG - vecA = frac13(hati + 2hatj + 6hatk) implies |overlineAG|^2 = frac19(1^2 + 2^2 + 6^2) = frac419$\overline{AG} = \vec{G} - \vec{A} = \frac{1}{3}(\hat{i} + 2\hat{j} + 6\hat{k}) \implies |\overline{AG}|^2 = \frac{1}{9}(1^2 + 2^2 + 6^2) = \frac{41}{9}$ - overlineBG = vecG - vecB = left(frac13-2right)hati + left(frac23+1right)hatj + left(2-1right)hatk = -frac53hati + frac53hatj + 1hatk$\overline{BG} = \vec{G} - \vec{B} = \left(\frac{1}{3}-2\right)\hat{i} + \left(\frac{2}{3}+1\right)\hat{j} + \left(2-1\right)\hat{k} = -\frac{5}{3}\hat{i} + \frac{5}{3}\hat{j} + 1\hat{k}$ |overlineBG|^2 = left(-frac53right)^2 + left(frac53right)^2 + 1^2 = frac259 + frac259 + 1 = frac599$$|\overline{BG}|^2 = \left(-\frac{5}{3}\right)^2 + \left(\frac{5}{3}\right)^2 + 1^2 = \frac{25}{9} + \frac{25}{9} + 1 = \frac{59}{9}$$ - overlineCG = vecG - vecC = left(frac13+1right)hati + left(frac23-3right)hatj + left(2-5right)hatk = frac43hati - frac73hatj - 3hatk$\overline{CG} = \vec{G} - \vec{C} = \left(\frac{1}{3}+1\right)\hat{i} + \left(\frac{2}{3}-3\right)\hat{j} + \left(2-5\right)\hat{k} = \frac{4}{3}\hat{i} - \frac{7}{3}\hat{j} - 3\hat{k}$ |overlineCG|^2 = left(frac43right)^2 + left(-frac73right)^2 + (-3)^2 = frac169 + frac499 + 9 = frac1469$$|\overline{CG}|^2 = \left(\frac{4}{3}\right)^2 + \left(-\frac{7}{3}\right)^2 + (-3)^2 = \frac{16}{9} + \frac{49}{9} + 9 = \frac{146}{9}$$ ### Step 3: Final Targeted Evaluation Summing the squared values and multiplying by 6$6$: textValue = 6 left[ |overlineAG|^2 + |overlineBG|^2 + |overlineCG|^2 right] = 6 left[ frac419 + frac599 + frac1469 right]$$\text{Value} = 6 \left[ |\overline{AG}|^2 + |\overline{BG}|^2 + |\overline{CG}|^2 \right] = 6 \left[ \frac{41}{9} + \frac{59}{9} + \frac{146}{9} \right]$$ textValue = 6 times frac2469 = 2 times frac2463 = 2 times 82 = 164$$\text{Value} = 6 \times \frac{246}{9} = 2 \times \frac{246}{3} = 2 \times 82 = 164$$ ### Pattern Recognition Setting one vector node as the origin point (vecA = vec0$\vec{A} = \vec{0}$) heavily dampens intermediate coordinate math steps, avoiding dealing with an absolute baseline origin orientation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula By orthogonal vector decomposition (Pythagorean property): |vecv|^2 = |vecv_1|^2 + |vecv_2|^2 quad textwhen vecv_1 cdot vecv_2 = 0$$|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2 \quad \text{when } \vec{v}_1 \cdot \vec{v}_2 = 0$$ ### Core Logic Compute lambda$\lambda$ using dot product formula: costheta = fracvecu cdot vecv|vecu||vecv| implies fracsqrt52sqrt7 = frac3(2) + (-1)(1)sqrt3^2 + (-1)^2 sqrt2^2 + 1^2 + (-lambda)^2$$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \implies \frac{\sqrt{5}}{2\sqrt{7}} = \frac{3(2) + (-1)(1)}{\sqrt{3^2 + (-1)^2} \sqrt{2^2 + 1^2 + (-\lambda)^2}}$$ fracsqrt52sqrt7 = frac5sqrt10sqrt5 + lambda^2 implies frac12sqrt7 = fracsqrt5sqrt10sqrt5 + lambda^2 = frac1sqrt2sqrt5 + lambda^2$$\frac{\sqrt{5}}{2\sqrt{7}} = \frac{5}{\sqrt{10}\sqrt{5 + \lambda^2}} \implies \frac{1}{2\sqrt{7}} = \frac{\sqrt{5}}{\sqrt{10}\sqrt{5 + \lambda^2}} = \frac{1}{\sqrt{2}\sqrt{5 + \lambda^2}}$$ ### Step 1: Solve for lambda Square both sides of equation: frac128 = frac12(5 + lambda^2) implies 2(5 + lambda^2) = 28 implies 5 + lambda^2 = 14 implies lambda^2 = 9 implies lambda = 3$$\frac{1}{28} = \frac{1}{2(5 + \lambda^2)} \implies 2(5 + \lambda^2) = 28 \implies 5 + \lambda^2 = 14 \implies \lambda^2 = 9 \implies \lambda = 3$$ Since vecv = 2hati + hatj - 3hatk$\vec{v} = 2\hat{i} + \hat{j} - 3\hat{k}$. ### Step 2: Apply Identity Since components are orthogonal, direct magnitude squared holds: |vecv_1|^2 + |vecv_2|^2 = |vecv|^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14$$|\vec{v}_1|^2 + |\vec{v}_2|^2 = |\vec{v}|^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14$$ ### Pattern Recognition Do not waste time explicitly projecting components vecv_1$\vec{v}_1$ and vecv_2$\vec{v}_2$ if only the sum of their squared magnitudes is requested. The scalar length matches the total vector length invariant under any orthogonal basis change. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula Componendo and Dividendo rule states that if fracxy = fracpq$\frac{x}{y} = \frac{p}{q}$, then: fracx+yx-y = fracp+qp-q$$\frac{x+y}{x-y} = \frac{p+q}{p-q}$$ ### Core Logic Given expression: frac|veca + vecb| + |veca - vecb||veca + vecb| - |veca - vecb| = fracsqrt2 + 11$$\frac{|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|}{|\vec{a} + \vec{b}| - |\vec{a} - \vec{b}|} = \frac{\sqrt{2} + 1}{1}$$ Applying Componendo and Dividendo: frac2|veca + vecb|2|veca - vecb| = frac(sqrt2 + 1) + 1(sqrt2 + 1) - 1 = fracsqrt2 + 2sqrt2 = 1 + sqrt2$$\frac{2|\vec{a} + \vec{b}|}{2|\vec{a} - \vec{b}|} = \frac{(\sqrt{2} + 1) + 1}{(\sqrt{2} + 1) - 1} = \frac{\sqrt{2} + 2}{\sqrt{2}} = 1 + \sqrt{2}$$ Squaring both sides: |veca + vecb|^2 = (1 + sqrt2)^2 |veca - vecb|^2$$|\vec{a} + \vec{b}|^2 = (1 + \sqrt{2})^2 |\vec{a} - \vec{b}|^2$$ |veca + vecb|^2 = (3 + 2sqrt2) |veca - vecb|^2$$|\vec{a} + \vec{b}|^2 = (3 + 2\sqrt{2}) |\vec{a} - \vec{b}|^2$$ ### Step 1: Vector Expansion Expanding using dot products, keeping in mind that |veca| = |vecb|$|\vec{a}| = |\vec{b}|$: |veca|^2 + |vecb|^2 + 2vecacdotvecb = (3 + 2sqrt2)(|veca|^2 + |vecb|^2 - 2vecacdotvecb)$$|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b} = (3 + 2\sqrt{2})(|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b})$$ 2|veca|^2 + 2vecacdotvecb = (3 + 2sqrt2)(2|veca|^2 - 2vecacdotvecb)$$2|\vec{a}|^2 + 2\vec{a}\cdot\vec{b} = (3 + 2\sqrt{2})(2|\vec{a}|^2 - 2\vec{a}\cdot\vec{b})$$ 2|veca|^2 (1 - (3 + 2sqrt2)) = -2vecacdotvecb (1 + 3 + 2sqrt2)$$2|\vec{a}|^2 (1 - (3 + 2\sqrt{2})) = -2\vec{a}\cdot\vec{b} (1 + 3 + 2\sqrt{2})$$ Simplifying directly leads to: fracvecacdotvecb|veca|^2 = frac2 + 2sqrt24 + 2sqrt2 = frac1sqrt2$$\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2} = \frac{2 + 2\sqrt{2}}{4 + 2\sqrt{2}} = \frac{1}{\sqrt{2}}$$ ### Step 2: Final Calculation We need to find frac|veca + vecb|^2|veca|^2$\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2}$: frac|veca + vecb|^2|veca|^2 = frac|veca|^2 + |vecb|^2 + 2vecacdotvecb|veca|^2 = 1 + 1 + frac2vecacdotvecb|veca|^2$$\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2} = \frac{|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}}{|\vec{a}|^2} = 1 + 1 + \frac{2\vec{a}\cdot\vec{b}}{|\vec{a}|^2}$$ = 2 + 2left(frac1sqrt2right) = 2 + sqrt2$$= 2 + 2\left(\frac{1}{\sqrt{2}}\right) = 2 + \sqrt{2}$$ ### Pattern Recognition Whenever symmetric sums and differences like |vecx|+|vecy|$|\vec{x}|+|\vec{y}|$ and |vecx|-|vecy|$|\vec{x}|-|\vec{y}|$ occur in ratios, Componendo-Dividendo should be applied immediately to isolate the ratio of the individual magnitudes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Vector Algebra Class 12 Mathematics: Vector Algebra

### 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