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 veca times vecc = beginvmatrix hati & hatj & hatk \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 endvmatrix$$\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$$ ### Core Logic Let the unknown vector be vecc = c_1hati + c_2hatj + c_3hatk$\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$. Given veca = -hati + 2hatj + 2hatk$\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$ and vecb = 8hati + 7hatj - 3hatk$\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$. Compute the cross product veca times vecc$\vec{a} \times \vec{c}$: veca times vecc = (2c_3 - 2c_2)hati + (-1 cdot c_3 - 2c_1)(-1)hatj + (-1 cdot c_2 - 2c_1)hatk$$\vec{a} \times \vec{c} = (2c_3 - 2c_2)\hat{i} + (-1 \cdot c_3 - 2c_1)(-1)\hat{j} + (-1 \cdot c_2 - 2c_1)\hat{k}$$ = (2c_3 - 2c_2)hati + (c_3 + 2c_1)hatj - (c_2 + 2c_1)hatk$$= (2c_3 - 2c_2)\hat{i} + (c_3 + 2c_1)\hat{j} - (c_2 + 2c_1)\hat{k}$$ ### Step 1: Equate components to find c Equating this to vecb$\vec{b}$: 2c_3 - 2c_2 = 8 Rightarrow c_3 - c_2 = 4$2c_3 - 2c_2 = 8 \Rightarrow c_3 - c_2 = 4$ c_3 + 2c_1 = 7$c_3 + 2c_1 = 7$ -(c_2 + 2c_1) = -3 Rightarrow c_2 + 2c_1 = 3$-(c_2 + 2c_1) = -3 \Rightarrow c_2 + 2c_1 = 3$ We are also given vecc cdot (hati + hatj + hatk) = 4$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$: c_1 + c_2 + c_3 = 4$$c_1 + c_2 + c_3 = 4$$

### Step 2: Solve the linear system From the dot product equation: c_1 = 4 - c_2 - c_3$c_1 = 4 - c_2 - c_3$. Substitute into c_2 + 2c_1 = 3$c_2 + 2c_1 = 3$: c_2 + 2(4 - c_2 - c_3) = 3 Rightarrow 8 - c_2 - 2c_3 = 3 Rightarrow c_2 + 2c_3 = 5$$c_2 + 2(4 - c_2 - c_3) = 3 \Rightarrow 8 - c_2 - 2c_3 = 3 \Rightarrow c_2 + 2c_3 = 5$$ We have the system: c_3 - c_2 = 4 Rightarrow c_2 = c_3 - 4$c_3 - c_2 = 4 \Rightarrow c_2 = c_3 - 4$ Substitute into above: (c_3 - 4) + 2c_3 = 5 Rightarrow 3c_3 = 9 Rightarrow c_3 = 3$(c_3 - 4) + 2c_3 = 5 \Rightarrow 3c_3 = 9 \Rightarrow c_3 = 3$ Then c_2 = 3 - 4 = -1$c_2 = 3 - 4 = -1$. Then c_1 = 4 - (-1) - 3 = 2$c_1 = 4 - (-1) - 3 = 2$. So, vecc = 2hati - hatj + 3hatk$\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$. ### Step 3: Evaluate the final expression veca + vecc = (-hati + 2hatj + 2hatk) + (2hati - hatj + 3hatk) = hati + hatj + 5hatk$$\vec{a} + \vec{c} = (-\hat{i} + 2\hat{j} + 2\hat{k}) + (2\hat{i} - \hat{j} + 3\hat{k}) = \hat{i} + \hat{j} + 5\hat{k}$$ |veca + vecc|^2 = 1^2 + 1^2 + 5^2 = 1 + 1 + 25 = 27$$|\vec{a} + \vec{c}|^2 = 1^2 + 1^2 + 5^2 = 1 + 1 + 25 = 27$$ ### Pattern Recognition Whenever veca times vecc = vecb$\vec{a} \times \vec{c} = \vec{b}$ and a dot product constraint is given, simply expand vecc$\vec{c}$ algebraically as (c_1, c_2, c_3)$(c_1, c_2, c_3)$. The cross product and dot product create an easily solvable 3x3 linear system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra

### Related Formula vecu times vecv = -(vecv times vecu)$$\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$$ General equation of a circle requires coefficient of x^2$x^2$ = coefficient of y^2$y^2$ and coefficient of xy$xy$ = 0$0$. ### Core Logic Given vecc times (2hati + 3hatj + 4hatk) = (2hati + 3hatj + 4hatk) times vecd$\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \vec{d}$ Rearranging: vecc times (2hati + 3hatj + 4hatk) + vecd times (2hati + 3hatj + 4hatk) = vec0$$\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) + \vec{d} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \vec{0}$$ (vecc + vecd) times (2hati + 3hatj + 4hatk) = vec0$$(\vec{c} + \vec{d}) \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \vec{0}$$ This implies vecc + vecd$\vec{c} + \vec{d}$ is parallel to 2hati + 3hatj + 4hatk$2\hat{i} + 3\hat{j} + 4\hat{k}$. Let vecc + vecd = lambda (2hati + 3hatj + 4hatk)$\vec{c} + \vec{d} = \lambda (2\hat{i} + 3\hat{j} + 4\hat{k})$. ### Step 1: Find the scaling factor Given |vecc + vecd| = sqrt29$|\vec{c} + \vec{d}| = \sqrt{29}$ |lambda| sqrt2^2 + 3^2 + 4^2 = sqrt29$$|\lambda| \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$$ |lambda| sqrt4 + 9 + 16 = sqrt29$$|\lambda| \sqrt{4 + 9 + 16} = \sqrt{29}$$ |lambda| sqrt29 = sqrt29 Rightarrow lambda = pm 1$$|\lambda| \sqrt{29} = \sqrt{29} \Rightarrow \lambda = \pm 1$$ Therefore, vecc + vecd = pm(2hati + 3hatj + 4hatk)$\vec{c} + \vec{d} = \pm(2\hat{i} + 3\hat{j} + 4\hat{k})$. ### Step 2: Evaluate the dot product limits We need values of (vecc + vecd) cdot (-7hati + 2hatj + 3hatk)$(\vec{c} + \vec{d}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$. Dot product = lambda(2(-7) + 3(2) + 4(3)) = lambda(-14 + 6 + 12) = 4lambda$= \lambda(2(-7) + 3(2) + 4(3)) = \lambda(-14 + 6 + 12) = 4\lambda$. Since lambda = pm 1$\lambda = \pm 1$, the possible values are 4$4$ and -4$-4$. Given lambda_1 > lambda_2$\lambda_1 > \lambda_2$, we have lambda_1 = 4$\lambda_1 = 4$ and lambda_2 = -4$\lambda_2 = -4$. ### Step 3: Apply circle constraints Substitute lambda_1$\lambda_1$ and lambda_2$\lambda_2$ into the conic equation: K^2 x^2 + (K^2 - 5K + 4)xy + left(3K - 2right)y^2 - 8x + 12y - 4 = 0$$K^2 x^2 + (K^2 - 5K + 4)xy + \left(3K - 2\right)y^2 - 8x + 12y - 4 = 0$$ For this to represent a circle: 1) Coefficient of xy$xy$ must be 0$0$: K^2 - 5K + 4 = 0 Rightarrow K = 1, 4$$K^2 - 5K + 4 = 0 \Rightarrow K = 1, 4$$ 2) Coefficient of x^2$x^2$ must equal coefficient of y^2$y^2$: K^2 = 3K - 2 Rightarrow K^2 - 3K + 2 = 0 Rightarrow K = 1, 2$$K^2 = 3K - 2 \Rightarrow K^2 - 3K + 2 = 0 \Rightarrow K = 1, 2$$ The common value satisfying both conditions is K = 1$K = 1$. ### Pattern Recognition Cross-product equations mapping to X times A = -Y times A$X \times A = -Y \times A$ perfectly factor out to (X+Y) times A = 0$(X+Y) \times A = 0$, guaranteeing (X+Y)$(X+Y)$ is a scalar multiple of A$A$. This immediately unlocks vector magnitudes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra Class 11 Maths: Conic Sections

### Related Formula textVector Cross product distributes over subtraction: (veca - vecc) times vecb = veca times vecb - vecc times vecb$$\text{Vector Cross product distributes over subtraction: } (\vec{a} - \vec{c}) \times \vec{b} = \vec{a} \times \vec{b} - \vec{c} \times \vec{b}$$ textScalar Triple Product cyclic identity: veca cdot (vecb times vecc) = (veca times vecb) cdot vecc$$\text{Scalar Triple Product cyclic identity: } \vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}$$ textAntisymmetry: vecc times vecb = - vecb times vecc$$\text{Antisymmetry: } \vec{c} \times \vec{b} = - \vec{b} \times \vec{c}$$ ### Core Logic Instead of solving for the individual coordinates of vector vecc$\vec{c}$, we apply vector algebraic identities to compute the target scalar triple product directly. ### Step 1: Expand and rewrite the cross product Given (veca - vecc) times vecb = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk$(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}$: veca times vecb - vecc times vecb = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk$$\vec{a} \times \vec{b} - \vec{c} \times \vec{b} = -18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}$$ veca times vecb + vecb times vecc = -18hatmathrmi - 3hatmathrmj + 12hatmathrmk$$\vec{a} \times \vec{b} + \vec{b} \times \vec{c} = -18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}$$ vecb times vecc = (-18hatmathrmi - 3hatmathrmj + 12hatmathrmk) - (veca times vecb) quad text--- (1)$$\vec{b} \times \vec{c} = (-18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}) - (\vec{a} \times \vec{b}) \quad \text{--- (1)}$$ ### Step 2: Calculate a x b Evaluate the cross product: veca times vecb = beginvmatrix hatmathrmi & hatmathrmj & hatmathrmk \\ 2 & -3 & 1 \\ 3 & 2 & 5 endvmatrix$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -3 & 1 \\ 3 & 2 & 5 \end{vmatrix}$$ veca times vecb = hatmathrmi(-15 - 2) - hatmathrmj(10 - 3) + hatmathrmk(4 - (-9)) = -17hatmathrmi - 7hatmathrmj + 13hatmathrmk$$\vec{a} \times \vec{b} = \hat{\mathrm{i}}(-15 - 2) - \hat{\mathrm{j}}(10 - 3) + \hat{\mathrm{k}}(4 - (-9)) = -17\hat{\mathrm{i}} - 7\hat{\mathrm{j}} + 13\hat{\mathrm{k}}$$ ### Step 3: Solve for the vector d Substitute veca times vecb$\vec{a} \times \vec{b}$ back into equation (1): vecd = vecb times vecc = (-18hatmathrmi - 3hatmathrmj + 12hatmathrmk) - (-17hatmathrmi - 7hatmathrmj + 13hatmathrmk)$$\vec{d} = \vec{b} \times \vec{c} = (-18\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + 12\hat{\mathrm{k}}) - (-17\hat{\mathrm{i}} - 7\hat{\mathrm{j}} + 13\hat{\mathrm{k}})$$ vecd = -hatmathrmi + 4hatmathrmj - hatmathrmk$$\vec{d} = -\hat{\mathrm{i}} + 4\hat{\mathrm{j}} - \hat{\mathrm{k}}$$ ### Step 4: Compute the final dot product Now compute the requested dot product: veca cdot vecd = (2hatmathrmi - 3hatmathrmj + hatmathrmk) cdot (-hatmathrmi + 4hatmathrmj - hatmathrmk)$$\vec{a} \cdot \vec{d} = (2\hat{\mathrm{i}} - 3\hat{\mathrm{j}} + \hat{\mathrm{k}}) \cdot (-\hat{\mathrm{i}} + 4\hat{\mathrm{j}} - \hat{\mathrm{k}})$$ veca cdot vecd = 2(-1) + (-3)(4) + 1(-1) = -2 - 12 - 1 = -15$$\vec{a} \cdot \vec{d} = 2(-1) + (-3)(4) + 1(-1) = -2 - 12 - 1 = -15$$ left| veca cdot vecd right| = 15$$\left| \vec{a} \cdot \vec{d} \right| = 15$$ ### Pattern Recognition Scalar triple product shortcut: Recognizing that veca cdot vecd = veca cdot (vecb times vecc) = [ veca \, vecb \, vecc ]$\vec{a} \cdot \vec{d} = \vec{a} \cdot (\vec{b} \times \vec{c}) = [ \vec{a} \, \vec{b} \, \vec{c} ]$ allows you to find the scalar value through simple determinants and linear equations instead of solving for the vector components. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula The projection of vector veca$\vec{a}$ onto vector vecb$\vec{b}$ is: textProj_vecbveca = fracveca cdot vecb|vecb|$$\text{Proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$$ ### Core Logic Let veca = a_1hatmathbfi + a_2hatmathbfj + a_3hatmathbfk$\vec{a} = a_1\hat{\mathbf{i}} + a_2\hat{\mathbf{j}} + a_3\hat{\mathbf{k}}$. Set up equations equating the three projections to determine the ratios of components. ### Step 1: Set up Projection Ratios Let vecb = 2hatmathbfi-hatmathbfj+2hatmathbfk implies |vecb| = 3$\vec{b} = 2\hat{\mathbf{i}}-\hat{\mathbf{j}}+2\hat{\mathbf{k}} \implies |\vec{b}| = 3$ Let vecc = hatmathbfi+2hatmathbfj-2hatmathbfk implies |vecc| = 3$\vec{c} = \hat{\mathbf{i}}+2\hat{\mathbf{j}}-2\hat{\mathbf{k}} \implies |\vec{c}| = 3$ Let vecd = hatmathbfk implies |vecd| = 1$\vec{d} = \hat{\mathbf{k}} \implies |\vec{d}| = 1$ Equating projections: frac2a_1 - a_2 + 2a_33 = fraca_1 + 2a_2 - 2a_33 = fraca_31$$\frac{2a_1 - a_2 + 2a_3}{3} = \frac{a_1 + 2a_2 - 2a_3}{3} = \frac{a_3}{1}$$ ### Step 2: Solve System of Linear Equations From the last equality: 2a_1 - a_2 + 2a_3 = 3a_3 implies 2a_1 - a_2 = a_3 quad dots (1)$$2a_1 - a_2 + 2a_3 = 3a_3 \implies 2a_1 - a_2 = a_3 \quad \dots (1)$$ a_1 + 2a_2 - 2a_3 = 3a_3 implies a_1 + 2a_2 = 5a_3 quad dots (2)$$a_1 + 2a_2 - 2a_3 = 3a_3 \implies a_1 + 2a_2 = 5a_3 \quad \dots (2)$$ Multiply (1) by 2$2$ and add to (2): 4a_1 - 2a_2 + a_1 + 2a_2 = 2a_3 + 5a_3 implies 5a_1 = 7a_3 implies a_1 = frac75a_3$$4a_1 - 2a_2 + a_1 + 2a_2 = 2a_3 + 5a_3 \implies 5a_1 = 7a_3 \implies a_1 = \frac{7}{5}a_3$$ Substitute back to find a_2$a_2$: a_2 = 2a_1 - a_3 = 2left(frac75a_3right) - a_3 = frac95a_3$$a_2 = 2a_1 - a_3 = 2\left(\frac{7}{5}a_3\right) - a_3 = \frac{9}{5}a_3$$ ### Step 3: Normalize to Unit Vector The vector components are in ratio a_1 : a_2 : a_3 = frac75 : frac95 : 1 = 7 : 9 : 5$a_1 : a_2 : a_3 = \frac{7}{5} : \frac{9}{5} : 1 = 7 : 9 : 5$. textMagnitude factor = sqrt7^2 + 9^2 + 5^2 = sqrt49 + 81 + 25 = sqrt155$$\text{Magnitude factor} = \sqrt{7^2 + 9^2 + 5^2} = \sqrt{49 + 81 + 25} = \sqrt{155}$$ Thus, the unit vector is: hata = frac1sqrt155left(7hatmathbfi + 9hatmathbfj + 5hatmathbfkright)$$\hat{a} = \frac{1}{\sqrt{155}}\left(7\hat{\mathbf{i}} + 9\hat{\mathbf{j}} + 5\hat{\mathbf{k}}\right)$$ ### Pattern Recognition Looking at the options, only option (3) provides components with positive signs for all three unit basis elements matching our proportional derivation of 7:9:5$7:9:5$ directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Vector Algebra

### Related Formula Lagrange's Identity: |vecu times vecv|^2 = |vecu|^2 |vecv|^2 - (vecu cdot vecv)^2$$|\vec{u} \times \vec{v}|^2 = |\vec{u}|^2 |\vec{v}|^2 - (\vec{u} \cdot \vec{v})^2$$ Also, if vecu times vecw = vecv times vecw implies (vecu - vecv) times vecw = 0 implies vecw$\vec{u} \times \vec{w} = \vec{v} \times \vec{w} \implies (\vec{u} - \vec{v}) \times \vec{w} = 0 \implies \vec{w}$ is parallel/collinear to vecu - vecv$\vec{u} - \vec{v}$. ### Core Logic Rearranging the cross product: vecb times vecd - vecc times vecd = vec0 implies (vecb - vecc) times vecd = vec0$$\vec{b} \times \vec{d} - \vec{c} \times \vec{d} = \vec{0} \implies (\vec{b} - \vec{c}) \times \vec{d} = \vec{0}$$ Thus, vecd = lambda (vecb - vecc)$\vec{d} = \lambda (\vec{b} - \vec{c})$. ### Step 1: Finding vector vecd$\vec{d}$ Calculate (vecb - vecc)$(\vec{b} - \vec{c})$: vecb - vecc = (3hati - 3hatj + 3hatk) - (2hati - hatj + 2hatk) = hati - 2hatj + hatk$$\vec{b} - \vec{c} = (3\hat{i} - 3\hat{j} + 3\hat{k}) - (2\hat{i} - \hat{j} + 2\hat{k}) = \hat{i} - 2\hat{j} + \hat{k}$$ Therefore: vecd = lambda(hati - 2hatj + hatk)$$\vec{d} = \lambda(\hat{i} - 2\hat{j} + \hat{k})$$ Substitute into veca cdot vecd = 4$\vec{a} \cdot \vec{d} = 4$: lambda(hati + 2hatj + hatk) cdot (hati - 2hatj + hatk) = 4$$\lambda(\hat{i} + 2\hat{j} + \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = 4$$ lambda(1 - 4 + 1) = 4 implies -2lambda = 4 implies lambda = -2$$\lambda(1 - 4 + 1) = 4 \implies -2\lambda = 4 \implies \lambda = -2$$ Thus: vecd = -2(hati - 2hatj + hatk) = -2hati + 4hatj - 2hatk$$\vec{d} = -2(\hat{i} - 2\hat{j} + \hat{k}) = -2\hat{i} + 4\hat{j} - 2\hat{k}$$ |vecd|^2 = (-2)^2 + 4^2 + (-2)^2 = 4 + 16 + 4 = 24$$|\vec{d}|^2 = (-2)^2 + 4^2 + (-2)^2 = 4 + 16 + 4 = 24$$ ### Step 2: Calculating |veca times vecd|^2$|\vec{a} \times \vec{d}|^2$ using Lagrange's Identity Given |veca|^2 = 1^2 + 2^2 + 1^2 = 6$|\vec{a}|^2 = 1^2 + 2^2 + 1^2 = 6$: |veca times vecd|^2 = |veca|^2 |vecd|^2 - (veca cdot vecd)^2$$|\vec{a} \times \vec{d}|^2 = |\vec{a}|^2 |\vec{d}|^2 - (\vec{a} \cdot \vec{d})^2$$ |veca times vecd|^2 = 6 times 24 - 4^2 = 144 - 16 = 128$$|\vec{a} \times \vec{d}|^2 = 6 \times 24 - 4^2 = 144 - 16 = 128$$ ### Pattern Recognition Applying Lagrange's Identity avoids computing the actual cross product determinants vector-by-vector. This is extremely efficient and reduces chances of sign errors. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra