### Related Formula Cross product for vector perpendicular direction alignment: vecu = vecb times vecc$\vec{u} = \vec{b} \times \vec{c}$ Angle projection formula: costheta = fracveca cdot vecv|veca||vecv|$\cos\theta = \frac{\vec{a} \cdot \vec{v}}{|\vec{a}||\vec{v}|}$ ### Core Logic Compute cross product of vecb$\vec{b}$ and vecc$\vec{c}$: vecb times vecc = beginvmatrix hati & hatj & hatk \\ 1 & -2 & 3 \\ 2 & 3 & -1 endvmatrix = -7hati + 7hatj + 7hatk = -7(hati - hatj - hatk)$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix} = -7\hat{i} + 7\hat{j} + 7\hat{k} = -7(\hat{i} - \hat{j} - \hat{k})$ Hence, unit vector hata$\hat{a}$ matches form: hata = pm frachati - hatj - hatksqrt3$\hat{a} = \pm \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$ ### Step 1: Isolate Core Angle Direction Check conditions against vector vecv = hati + hatj + hatk$\vec{v} = \hat{i} + \hat{j} + \hat{k}$: Using hata = frachati - hatj - hatksqrt3$\hat{a} = \frac{\hat{i} - \hat{j} - \hat{k}}{\sqrt{3}}$: costheta = frac1 - 1 - 1sqrt3sqrt3 = -frac13$\cos\theta = \frac{1 - 1 - 1}{\sqrt{3}\sqrt{3}} = -\frac{1}{3}$ This confirms the direction for hata$\hat{a}$. ### Step 2: Solve for Unknown Scalar Variable Now compute angle with vector hati + alphahatj + hatk$\hat{i} + \alpha\hat{j} + \hat{k}$ for theta = fracpi3$\theta = \frac{\pi}{3}$: cosfracpi3 = frac1sqrt3 cdot frac1 - alpha - 1sqrt2 + alpha^2$\cos\frac{\pi}{3} = \frac{1}{\sqrt{3}} \cdot \frac{1 - \alpha - 1}{\sqrt{2 + \alpha^2}}$ frac12 = frac-alphasqrt3sqrtalpha^2 + 2$\frac{1}{2} = \frac{-\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}}$ Since left hand side is positive, alpha$\alpha$ must be strictly negative. Squaring both sides: frac14 = fracalpha^23(alpha^2 + 2) implies 3alpha^2 + 6 = 4alpha^2 implies alpha^2 = 6$\frac{1}{4} = \frac{\alpha^2}{3(\alpha^2 + 2)} \implies 3\alpha^2 + 6 = 4\alpha^2 \implies \alpha^2 = 6$ Since alpha < 0$\alpha < 0$, alpha = -sqrt6$\alpha = -\sqrt{6}$. ### Pattern Recognition Keep strict track of signs when dealing with algebra containing square roots. Checking value constraints early on allows you to drop phantom positive/negative branches seamlessly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula Vector cross distribution property: (veca - vecb) times hatd = 0 implies hatd parallel (veca - vecb)$(\vec{a} - \vec{b}) \times \hat{d} = 0 \implies \hat{d} \parallel (\vec{a} - \vec{b})$ ### Core Logic Gather cross products to solve for collineation lines [cite: 1472, 1473]: (veca - vecb) times hatd = 0 implies hatd = t(veca - vecb)$(\vec{a} - \vec{b}) \times \hat{d} = 0 \implies \hat{d} = t(\vec{a} - \vec{b})$ [cite: 1473, 1474] Compute the baseline difference vector [cite: 1475]: veca - vecb = (1-3)hati + (1-2)hatj + (1 - (-1))hatk = -2hati - hatj + 2hatk$\vec{a} - \vec{b} = (1-3)\hat{i} + (1-2)\hat{j} + (1 - (-1))\hat{k} = -2\hat{i} - \hat{j} + 2\hat{k}$ [cite: 1475] hatd = t(-2hati - hatj + 2hatk)$\hat{d} = t(-2\hat{i} - \hat{j} + 2\hat{k})$ [cite: 1475] Since hatd$\hat{d}$ is a unit vector[cite: 1476]: |t| cdot sqrt(-2)^2 + (-1)^2 + 2^2 = 1 implies 3|t| = 1 implies |t| = frac13$|t| \cdot \sqrt{(-2)^2 + (-1)^2 + 2^2} = 1 \implies 3|t| = 1 \implies |t| = \frac{1}{3}$ [cite: 1476, 1479] ### Step 1: Applying orthogonality constraints Using orthogonal information given for vectors vecc$\vec{c}$ and veca$\vec{a}$ [cite: 1482]: vecc cdot veca = 0 implies (0)(1) + lambda(1) + mu(1) = 0 implies mu = -lambda$\vec{c} \cdot \vec{a} = 0 \implies (0)(1) + \lambda(1) + \mu(1) = 0 \implies \mu = -\lambda$ [cite: 1483, 1484] vecc = lambda(hatj - hatk) implies |vecc|^2 = 2lambda^2$\vec{c} = \lambda(\hat{j} - \hat{k}) \implies |\vec{c}|^2 = 2\lambda^2$ [cite: 1485] Use product condition vecc cdot hatd = 1$\vec{c} \cdot \hat{d} = 1$ to isolate scalar values [cite: 1486]: t(-2hati - hatj + 2hatk) cdot lambda(hatj - hatk) = 1$t(-2\hat{i} - \hat{j} + 2\hat{k}) \cdot \lambda(\hat{j} - \hat{k}) = 1$ [cite: 1487] tlambda(-1 - 2) = 1 implies -3tlambda = 1 implies tlambda = -frac13$t\lambda(-1 - 2) = 1 \implies -3t\lambda = 1 \implies t\lambda = -\frac{1}{3}$ [cite: 1488] Since |t|^2 = frac19$|t|^2 = \frac{1}{9}$, squaring components yields [cite: 1488]: lambda^2 = 1$\lambda^2 = 1$ [cite: 1488] ### Step 2: Vector magnitude resolution Expand target expression using standard inner dot product expansions [cite: 1488]: |3lambda hatd + mu vecc|^2 = 9lambda^2|hatd|^2 + mu^2|vecc|^2 + 6lambdamu(hatd cdot vecc)$|3\lambda \hat{d} + \mu \vec{c}|^2 = 9\lambda^2|\hat{d}|^2 + \mu^2|\vec{c}|^2 + 6\lambda\mu(\hat{d} \cdot \vec{c})$ [cite: 1488] Substitute values evaluated throughout sections [cite: 1488, 1489]: = 9(1)(1) + (lambda)^2(2lambda^2) + 6lambda(-lambda)(1)$= 9(1)(1) + (\lambda)^2(2\lambda^2) + 6\lambda(-\lambda)(1)$ = 9 + 2lambda^4 - 6lambda^2 = 9 + 2(1) - 6(1) = 5$= 9 + 2\lambda^4 - 6\lambda^2 = 9 + 2(1) - 6(1) = 5$ [cite: 1489, 1490] ### Pattern Recognition Translating vector cross equalities directly into linear scale parameter multipliers prevents manual determinant expansions, leaving clean system variables behind. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### 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