Let overrightarrowAB=2hati+4hatj-5hatk$\overrightarrow{AB}=2\hat{i}+4\hat{j}-5\hat{k}$ and overrightarrowAD=hati+2hatj+lambda hatk, lambdain mathbbR$\overrightarrow{AD}=\hat{i}+2\hat{j}+\lambda \hat{k}, \lambda\in \mathbb{R}$. Let the projection of the vector vecv=hati+hatj+hatk$\vec{v}=\hat{i}+\hat{j}+\hat{k}$ on the diagonal overrightarrowAC$\overrightarrow{AC}$ of the parallelogram ABCD be of length one unit. If alpha,beta$\alpha,\beta$, where alpha>beta$\alpha>\beta$, be the roots of the equation lambda^2x^2-6lambda x+5=0$\lambda^{2}x^{2}-6\lambda x+5=0$, then 2alpha-beta$2\alpha-\beta$ 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 textCosine Rule in triangle: cos(pi-theta) = frac|vecp|^2+|vecq|^2-|vecr|^22|vecp||vecq|$$\text{Cosine Rule in triangle: } \cos(\pi-\theta) = \frac{|\vec{p}|^2+|\vec{q}|^2-|\vec{r}|^2}{2|\vec{p}||\vec{q}|}$$ textCross product identity: vecA times vecA = 0 quad ; quad |vecA times vecB| = |A||B|sintheta$$\text{Cross product identity: } \vec{A} \times \vec{A} = 0 \quad ; \quad |\vec{A} \times \vec{B}| = |A||B|\sin\theta$$ ### Core Logic

From triangle law of addition, overrightarrowBC + overrightarrowCA = overrightarrowBA implies vecp + vecq = vecr$\overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{BA} \implies \vec{p} + \vec{q} = \vec{r}$. The angle between vectors vecp$\vec{p}$ and vecq$\vec{q}$ extended head-to-tail is theta$\theta$. The internal angle of the triangle is pi - theta$\pi - \theta$. ### Step 1: Calculate magnitude of r Using cosine rule for the side |vecr|$|\vec{r}|$: cos(pi - theta) = frac|vecp|^2 + |vecq|^2 - |vecr|^22|vecp||vecq|$$\cos(\pi - \theta) = \frac{|\vec{p}|^2 + |\vec{q}|^2 - |\vec{r}|^2}{2|\vec{p}||\vec{q}|}$$ Since costheta = 1/sqrt3$\cos\theta = 1/\sqrt{3}$, cos(pi - theta) = -1/sqrt3$\cos(\pi - \theta) = -1/\sqrt{3}$. -frac1sqrt3 = frac(2sqrt3)^2 + (2)^2 - |vecr|^22(2sqrt3)(2)$$-\frac{1}{\sqrt{3}} = \frac{(2\sqrt{3})^2 + (2)^2 - |\vec{r}|^2}{2(2\sqrt{3})(2)}$$ -frac1sqrt3 = frac12 + 4 - |vecr|^28sqrt3$$-\frac{1}{\sqrt{3}} = \frac{12 + 4 - |\vec{r}|^2}{8\sqrt{3}}$$ -8 = 16 - |vecr|^2 implies |vecr|^2 = 24$$-8 = 16 - |\vec{r}|^2 \implies |\vec{r}|^2 = 24$$ ### Step 2: Simplify the Cross Product term We need to evaluate |vecp times (vecq - 3vecr)|^2$|\vec{p} \times (\vec{q} - 3\vec{r})|^2$. Substitute vecr = vecp + vecq$\vec{r} = \vec{p} + \vec{q}$: = |vecp times (vecq - 3(vecp + vecq))|^2$$= |\vec{p} \times (\vec{q} - 3(\vec{p} + \vec{q}))|^2$$ = |vecp times (-3vecp - 2vecq)|^2$$= |\vec{p} \times (-3\vec{p} - 2\vec{q})|^2$$ Since vecp times vecp = 0$\vec{p} \times \vec{p} = 0$: = |-2 (vecp times vecq)|^2 = 4 |vecp times vecq|^2$$= |-2 (\vec{p} \times \vec{q})|^2 = 4 |\vec{p} \times \vec{q}|^2$$ = 4 ( |vecp|^2 |vecq|^2 sin^2theta )$$= 4 ( |\vec{p}|^2 |\vec{q}|^2 \sin^2\theta )$$ ### Step 3: Final Calculation Given costheta = 1/sqrt3 implies cos^2theta = 1/3 implies sin^2theta = 2/3$\cos\theta = 1/\sqrt{3} \implies \cos^2\theta = 1/3 \implies \sin^2\theta = 2/3$. 4 |vecp times vecq|^2 = 4 (12)(4)left(frac23right) = 4(4)(4)(2) = 128$$4 |\vec{p} \times \vec{q}|^2 = 4 (12)(4)\left(\frac{2}{3}\right) = 4(4)(4)(2) = 128$$ The required expression is: |vecp times (vecq - 3vecr)|^2 + 3|vecr|^2 = 128 + 3(24) = 128 + 72 = 200$$|\vec{p} \times (\vec{q} - 3\vec{r})|^2 + 3|\vec{r}|^2 = 128 + 3(24) = 128 + 72 = 200$$ ### Pattern Recognition Always convert secondary vectors back to base components using the triangle condition vecp + vecq = vecr$\vec{p} + \vec{q} = \vec{r}$. It zeroes out self-cross products instantly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra

### 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