Let the three sides of a triangle ABC$ABC$ be given by the vectors 2hati - hatj + hatk$2\hat{i} - \hat{j} + \hat{k}$, hati - 3hatj - 5hatk$\hat{i} - 3\hat{j} - 5\hat{k}$, and 3hati - 4hatj - 4hatk$3\hat{i} - 4\hat{j} - 4\hat{k}$. Let G$G$ be the centroid of the triangle ABC$ABC$. Then 6left(|overlineAG|^2 + |overlineBG|^2 + |overlineCG|^2right)$6\left(|\overline{AG}|^2 + |\overline{BG}|^2 + |\overline{CG}|^2\right)$ is equal to

### Related Formula Properties of dot and cross vector setups: hata cdot hata = |hata|^2 = 1$$\hat{a} \cdot \hat{a} = |\hat{a}|^2 = 1$$ hata cdot hatb = |hata||hatb|costheta = costheta$$\hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}|\cos\theta = \cos\theta$$ hata cdot (hata times hatb) = 0, quad hatb cdot (hata times hatb) = 0$$\hat{a} \cdot (\hat{a} \times \hat{b}) = 0, \quad \hat{b} \cdot (\hat{a} \times \hat{b}) = 0$$ ### Core Logic Given sintheta = fracsqrt659$\sin\theta = \frac{\sqrt{65}}{9}$. Since theta$\theta$ lies in the first quadrant: costheta = sqrt1 - sin^2theta = sqrt1 - frac6581 = sqrtfrac1681 = frac49$$\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{65}{81}} = \sqrt{\frac{16}{81}} = \frac{4}{9}$$ Now, let's take individual dot product equations using vecc = 3hata + 6hatb + 9(hata times hatb)$\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$: 1. Find vecc cdot hata$\vec{c} \cdot \hat{a}$: vecc cdot hata = 3(hata cdot hata) + 6(hatb cdot hata) + 9((hata times hatb) cdot hata)$$\vec{c} \cdot \hat{a} = 3(\hat{a} \cdot \hat{a}) + 6(\hat{b} \cdot \hat{a}) + 9((\hat{a} \times \hat{b}) \cdot \hat{a})$$ vecc cdot hata = 3(1) + 6costheta + 0 = 3 + 6left(frac49right) = 3 + frac249 = frac519$$\vec{c} \cdot \hat{a} = 3(1) + 6\cos\theta + 0 = 3 + 6\left(\frac{4}{9}\right) = 3 + \frac{24}{9} = \frac{51}{9}$$ ### Step 1: Compute Second Dot Product Term 2. Find vecc cdot hatb$\vec{c} \cdot \hat{b}$: vecc cdot hatb = 3(hata cdot hatb) + 6(hatb cdot hatb) + 9((hata times hatb) cdot hatb)$$\vec{c} \cdot \hat{b} = 3(\hat{a} \cdot \hat{b}) + 6(\hat{b} \cdot \hat{b}) + 9((\hat{a} \times \hat{b}) \cdot \hat{b})$$ vecc cdot hatb = 3costheta + 6(1) + 0 = 3left(frac49right) + 6 = frac129 + 6 = frac12 + 549 = frac669 = frac223$$\vec{c} \cdot \hat{b} = 3\cos\theta + 6(1) + 0 = 3\left(\frac{4}{9}\right) + 6 = \frac{12}{9} + 6 = \frac{12 + 54}{9} = \frac{66}{9} = \frac{22}{3}$$ ### Step 2: Substitute and Finalize Result Evaluate the target expression layout: textValue = 9(vecc cdot hata) - 3(vecc cdot hatb)$$\text{Value} = 9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$$ textValue = 9left(frac519right) - 3left(frac223right) = 51 - 22 = 29$$\text{Value} = 9\left(\frac{51}{9}\right) - 3\left(\frac{22}{3}\right) = 51 - 22 = 29$$ ### Pattern Recognition Remember that a cross product vector (hata times hatb)$(\hat{a} \times \hat{b})$ is orthogonal to both constituent vectors. Thus, taking their dot product evaluates to 0$0$ immediately, allowing you to ignore that entire component during computation steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula vecp = K(veca + lambdavecb)$$\vec{p} = K(\vec{a} + \lambda\vec{b})$$ vecp cdot veca = 0$$\vec{p} \cdot \vec{a} = 0$$ ### Core Logic Formulate a coplanar parameterization vector, apply the zero dot-product geometric orthogonality constraint to pin down the linear parameter, and then normalize. ### Step 1: Define Coplanar Structural Form Let the targeting vector path be: vecp = K(veca + lambdavecb) = Kleft( (1+2lambda)hati + (2+lambda)hatj + (1-lambda)hatk right)$$\vec{p} = K(\vec{a} + \lambda\vec{b}) = K\left( (1+2\lambda)\hat{i} + (2+\lambda)\hat{j} + (1-\lambda)\hat{k} \right)$$ ### Step 2: Force Orthogonality Constraint Impose vecp cdot veca = 0$\vec{p} \cdot \vec{a} = 0$: 1(1+2lambda) + 2(2+lambda) + 1(1-lambda) = 0$$1(1+2\lambda) + 2(2+\lambda) + 1(1-\lambda) = 0$$ 1 + 2lambda + 4 + 2lambda + 1 - lambda = 0 implies 6 + 3lambda = 0 implies lambda = -2$$1 + 2\lambda + 4 + 2\lambda + 1 - \lambda = 0 \implies 6 + 3\lambda = 0 \implies \lambda = -2$$ ### Step 3: Substitute and Normalize Substitute lambda = -2$\lambda = -2$ back into the base formulation: vecp = K(-3hati + 3hatk)$$\vec{p} = K(-3\hat{i} + 3\hat{k})$$ Normalizing to turn this vector into a proper unit scale form: hatc = pm frac-hati + hatksqrt2$$\hat{c} = \pm \frac{-\hat{i} + \hat{k}}{\sqrt{2}}$$ ### Pattern Recognition Finding coplanar vectors orthogonal to one base component matches taking cross expansions like (veca times vecb) times veca$(\vec{a} \times \vec{b}) \times \vec{a}$ up to scalar metrics. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### 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 magnitude expansion identity: |vecu - vecv|^2 = |vecu|^2 + |vecv|^2 - 2(vecu cdot vecv)$$|\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2(\vec{u} \cdot \vec{v})$$ ### Core Logic Calculate the reference vector values: veca = hati + hatj + hatk implies |veca|^2 = 3$\vec{a} = \hat{i} + \hat{j} + \hat{k} \implies |\vec{a}|^2 = 3$ vecb = 2hati + 2hatj + hatk$\vec{b} = 2\hat{i} + 2\hat{j} + \hat{k}$ vecd = veca times vecb = -hati + hatj implies |vecd|^2 = 2$\vec{d} = \vec{a} \times \vec{b} = -\hat{i} + \hat{j} \implies |\vec{d}|^2 = 2$ Expanding |vecc - 2veca|^2 = 8$|\vec{c} - 2\vec{a}|^2 = 8$: |vecc|^2 + 4|veca|^2 - 4(veca cdot vecc) = 8$$|\vec{c}|^2 + 4|\vec{a}|^2 - 4(\vec{a} \cdot \vec{c}) = 8$$ Substituting |veca|^2 = 3$|\vec{a}|^2 = 3$ and veca cdot vecc = |vecc|$\vec{a} \cdot \vec{c} = |\vec{c}|$: |vecc|^2 - 4|vecc| + 4 = 0 implies (|vecc| - 2)^2 = 0 implies |vecc| = 2$$|\vec{c}|^2 - 4|\vec{c}| + 4 = 0 \implies (|\vec{c}| - 2)^2 = 0 \implies |\vec{c}| = 2$$ ### Step 1: Evaluating the Cross Product Vector Component Using the given angle fracpi4$\frac{\pi}{4}$ between vecd$\vec{d}$ and vecc$\vec{c}$: |vecd times vecc|^2 = (|vecd||vecc|sinfracpi4)^2 = left(sqrt2 cdot 2 cdot frac1sqrt2right)^2 = 4$$|\vec{d} \times \vec{c}|^2 = (|\vec{d}||\vec{c}|\sin\frac{\pi}{4})^2 = \left(\sqrt{2} \cdot 2 \cdot \frac{1}{\sqrt{2}}\right)^2 = 4$$ ### Step 2: Solving for the Vector Dot Product Expanding using standard vector identities yields vecb cdot vecc = frac83$\vec{b} \cdot \vec{c} = \frac{8}{3}$. Substituting this back into the target expression: textValue = left|10 - 3left(frac83 ight)right| + 4 = |10 - 8| + 4 = 6$$\text{Value} = \left|10 - 3\left(\frac{8}{3} ight)\right| + 4 = |10 - 8| + 4 = 6$$ ### Pattern Recognition Recognizing a perfect square trinomial (x^2 - 4x + 4 = 0$x^2 - 4x + 4 = 0$) instantly isolates the unknown vector magnitude. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vectors

### 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