### Related Formula vecx cdot (vecy times vecz) = [vecx \ vecy \ vecz] = (vecx times vecy) cdot vecz$\vec{x} \cdot (\vec{y} \times \vec{z}) = [\vec{x} \ \vec{y} \ \vec{z}] = (\vec{x} \times \vec{y}) \cdot \vec{z}$ ### Core Logic The expression to evaluate is: E = veca cdot ((vecc times vecb) - vecb - vecc)$E = \vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} - \vec{c})$ Distributing the dot product over the terms gives: E = veca cdot (vecc times vecb) - veca cdot vecb - veca cdot vecc$E = \vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$ ### Step 1: Evaluating the Scalar Triple Product The first term is a scalar triple product: veca cdot (vecc times vecb) = (veca times vecc) cdot vecb$\vec{a} \cdot (\vec{c} \times \vec{b}) = (\vec{a} \times \vec{c}) \cdot \vec{b}$ We are given that veca times vecc = vecb$\vec{a} \times \vec{c} = \vec{b}$. Substitute this in: (vecb) cdot vecb = |vecb|^2$(\vec{b}) \cdot \vec{b} = |\vec{b}|^2$ Given vecb = 3hati - 3hatj + 3hatk$\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$, its magnitude squared is: |vecb|^2 = 3^2 + (-3)^2 + 3^2 = 9 + 9 + 9 = 27$|\vec{b}|^2 = 3^2 + (-3)^2 + 3^2 = 9 + 9 + 9 = 27$ ### Step 2: Evaluating the remaining Dot Products For the second term, calculate veca cdot vecb$\vec{a} \cdot \vec{b}$: veca = 1hati + 2hatj + 1hatk$\vec{a} = 1\hat{i} + 2\hat{j} + 1\hat{k}$ vecb = 3hati - 3hatj + 3hatk$\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$ veca cdot vecb = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0$ For the third term, we are explicitly given: veca cdot vecc = 3$\vec{a} \cdot \vec{c} = 3$ ### Step 3: Final Output Calculation Substitute all individual values back into the expanded expression: E = 27 - 0 - 3 = 24$E = 27 - 0 - 3 = 24$ ### Pattern Recognition When asked to evaluate complex vector expressions containing veca cdot (dots times dots)$\vec{a} \cdot (\dots \times \dots)$, immediately distribute and convert them into Scalar Triple Products [veca \ vecb \ vecc]$[\vec{a} \ \vec{b} \ \vec{c}]$. Cyclic permutations and given cross-product relationships will rapidly collapse the expression into simple magnitudes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra

### Related Formula cos theta = fracveca cdot vecb|veca||vecb|$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$ For an acute angle, cos theta > 0$\cos \theta > 0$, which strictly means veca cdot vecb > 0$\vec{a} \cdot \vec{b} > 0$. ### Core Logic Given vectors vecA = alphahati - 2hatj + 2hatk$\vec{A} = \alpha\hat{i} - 2\hat{j} + 2\hat{k}$ and vecB = alphahati + 2alphahatj - 2hatk$\vec{B} = \alpha\hat{i} + 2\alpha\hat{j} - 2\hat{k}$. For the angle to be acute, their dot product must be strictly positive: vecA cdot vecB > 0$\vec{A} \cdot \vec{B} > 0$ (alpha)(alpha) + (-2)(2alpha) + (2)(-2) > 0$(\alpha)(\alpha) + (-2)(2\alpha) + (2)(-2) > 0$ alpha^2 - 4alpha - 4 > 0$\alpha^2 - 4\alpha - 4 > 0$ ### Step 1: Solving the Inequality Complete the square to find critical points: alpha^2 - 4alpha + 4 > 8$\alpha^2 - 4\alpha + 4 > 8$ (alpha - 2)^2 > 8$(\alpha - 2)^2 > 8$ Extracting roots gives: alpha - 2 > 2sqrt2 quad textor quad alpha - 2 < -2sqrt2$\alpha - 2 > 2\sqrt{2} \quad \text{or} \quad \alpha - 2 < -2\sqrt{2}$ alpha > 2 + 2sqrt2 quad textor quad alpha < 2 - 2sqrt2$\alpha > 2 + 2\sqrt{2} \quad \text{or} \quad \alpha < 2 - 2\sqrt{2}$ ### Step 2: Finding Least Positive Integer Approximate the boundary values. Since sqrt2 approx 1.414$\sqrt{2} \approx 1.414$: 2 + 2(1.414) = 4.828$2 + 2(1.414) = 4.828$ The ranges are alpha in (-infty, -0.828) cup (4.828, infty)$\alpha \in (-\infty, -0.828) \cup (4.828, \infty)$. We need the *least positive integral value* of alpha$\alpha$. Looking at the interval (4.828, infty)$(4.828, \infty)$, the smallest integer present is 5$5$. ### Pattern Recognition Acute angle translates directly to a positive dot product. Set up the quadratic inequality, compute numerical bounds of irrational roots, and select the immediate next integer. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra Class 11 Maths: Linear Inequalities

### Related Formula textIf vecX text and vecY text are collinear, then vecX = lambda vecY text for some scalar lambda.$\text{If } \vec{X} \text{ and } \vec{Y} \text{ are collinear, then } \vec{X} = \lambda \vec{Y} \text{ for some scalar } \lambda.$ Linear Independence: If vecu$\vec{u}$ and vecv$\vec{v}$ are non-collinear, then xvecu + yvecv = 0 implies x = 0 text and y = 0$x\vec{u} + y\vec{v} = 0 \implies x = 0 \text{ and } y = 0$. ### Core Logic Based on the problem statement: 1) veca + 5vecb$\vec{a} + 5\vec{b}$ is collinear with vecc$\vec{c}$: * quad veca + 5vecb = lambdavecc quad dots(1)$* \quad \vec{a} + 5\vec{b} = \lambda\vec{c} \quad \dots(1)$ 2) vecb + 6vecc$\vec{b} + 6\vec{c}$ is collinear with veca$\vec{a}$: * quad vecb + 6vecc = muveca quad dots(2)$* \quad \vec{b} + 6\vec{c} = \mu\vec{a} \quad \dots(2)$ ### Step 1: Eliminate Vector a From equation (1), isolate veca$\vec{a}$: veca = lambdavecc - 5vecb$\vec{a} = \lambda\vec{c} - 5\vec{b}$ Substitute this into equation (2): vecb + 6vecc = mu(lambdavecc - 5vecb)$\vec{b} + 6\vec{c} = \mu(\lambda\vec{c} - 5\vec{b})$ vecb + 6vecc = mulambdavecc - 5muvecb$\vec{b} + 6\vec{c} = \mu\lambda\vec{c} - 5\mu\vec{b}$ Rearrange to group coefficients for vecb$\vec{b}$ and vecc$\vec{c}$: (1 + 5mu)vecb + (6 - mulambda)vecc = 0$(1 + 5\mu)\vec{b} + (6 - \mu\lambda)\vec{c} = 0$ ### Step 2: Determine Scalars Since vectors vecb$\vec{b}$ and vecc$\vec{c}$ are given as non-collinear, their linear combination resolving to zero dictates that their scalar coefficients must both individually be zero. 1 + 5mu = 0 Rightarrow mu = -frac15$1 + 5\mu = 0 \Rightarrow \mu = -\frac{1}{5}$ 6 - mulambda = 0 Rightarrow 6 - left(-frac15right)lambda = 0 Rightarrow 6 + fraclambda5 = 0$6 - \mu\lambda = 0 \Rightarrow 6 - \left(-\frac{1}{5}\right)\lambda = 0 \Rightarrow 6 + \frac{\lambda}{5} = 0$ fraclambda5 = -6 Rightarrow lambda = -30$\frac{\lambda}{5} = -6 \Rightarrow \lambda = -30$ ### Step 3: Match the Final Equation Substitute lambda = -30$\lambda = -30$ back into equation (1): veca + 5vecb = -30vecc$\vec{a} + 5\vec{b} = -30\vec{c}$ veca + 5vecb + 30vecc = 0$\vec{a} + 5\vec{b} + 30\vec{c} = 0$ The problem provides the structure veca + alphavecb + betavecc = 0$\vec{a} + \alpha\vec{b} + \beta\vec{c} = 0$. Comparing the two yields: alpha = 5, quad beta = 30$\alpha = 5, \quad \beta = 30$ Thus: alpha + beta = 5 + 30 = 35$\alpha + \beta = 5 + 30 = 35$ ### Pattern Recognition Double-collinearity equations should always be mapped out by eliminating the "third" vector (veca$\vec{a}$ in this case) and funneling everything into the two known non-collinear vectors. The resulting zero-equation perfectly exposes the scalar unknowns. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula textInternal Angle Bisector Theorem: fracACCB = frac|vecOA||vecOB|$\text{Internal Angle Bisector Theorem: } \frac{AC}{CB} = \frac{|\vec{OA}|}{|\vec{OB}|}$ textSection Formula: vecOC = fracmvecOB + nvecOAm+n$\text{Section Formula: } \vec{OC} = \frac{m\vec{OB} + n\vec{OA}}{m+n}$ ### Core Logic Find the magnitudes of the position vectors vecOA$\vec{OA}$ and vecOB$\vec{OB}$: |vecOA| = sqrt2^2 + 2^2 + 1^2 = sqrt4+4+1 = sqrt9 = 3$|\vec{OA}| = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3$ |vecOB| = sqrt2^2 + 4^2 + 4^2 = sqrt4+16+16 = sqrt36 = 6$|\vec{OB}| = \sqrt{2^2 + 4^2 + 4^2} = \sqrt{4+16+16} = \sqrt{36} = 6$ According to the internal angle bisector theorem in Delta AOB$\Delta AOB$, the point C divides the segment AB in the ratio of the adjacent sides: fracACCB = frac|vecOA||vecOB| = frac36 = frac12$\frac{AC}{CB} = \frac{|\vec{OA}|}{|\vec{OB}|} = \frac{3}{6} = \frac{1}{2}$

### Step 1: Apply Section Formula Using the section formula to find the position vector of C, dividing AB internally in ratio m:n = 1:2$m:n = 1:2$: vecOC = frac1(vecOB) + 2(vecOA)1 + 2$\vec{OC} = \frac{1(\vec{OB}) + 2(\vec{OA})}{1 + 2}$ vecOC = frac1(2hati+4hatj+4hatk) + 2(2hati+2hatj+hatk)3$\vec{OC} = \frac{1(2\hat{i}+4\hat{j}+4\hat{k}) + 2(2\hat{i}+2\hat{j}+\hat{k})}{3}$ vecOC = frac(2+4)hati + (4+4)hatj + (4+2)hatk3$\vec{OC} = \frac{(2+4)\hat{i} + (4+4)\hat{j} + (4+2)\hat{k}}{3}$ vecOC = frac6hati + 8hatj + 6hatk3 = 2hati + frac83hatj + 2hatk$\vec{OC} = \frac{6\hat{i} + 8\hat{j} + 6\hat{k}}{3} = 2\hat{i} + \frac{8}{3}\hat{j} + 2\hat{k}$ ### Step 2: Compute Length of OC Now, find the magnitude (length) of the vector vecOC$\vec{OC}$: |vecOC| = sqrt2^2 + left(frac83right)^2 + 2^2$|\vec{OC}| = \sqrt{2^2 + \left(\frac{8}{3}\right)^2 + 2^2}$ = sqrt4 + frac649 + 4 = sqrt8 + frac649$= \sqrt{4 + \frac{64}{9} + 4} = \sqrt{8 + \frac{64}{9}}$ = sqrtfrac72 + 649 = sqrtfrac1369$= \sqrt{\frac{72 + 64}{9}} = \sqrt{\frac{136}{9}}$ = fracsqrt4 times 343 = frac2sqrt343$= \frac{\sqrt{4 \times 34}}{3} = \frac{2\sqrt{34}}{3}$ ### Pattern Recognition Vector angle bisector questions invariably test the geometric property that the bisector divides the opposite side in the ratio of the side lengths. Combine this directly with the 3D coordinate section formula. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula vecacdotvecb = |veca||vecb| cos theta$\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}| \cos \theta$ |veca times vecb|^2 = |veca|^2 |vecb|^2 sin^2 theta$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$ |veca|^2 = a_x^2 + a_y^2 + a_z^2$|\vec{a}|^2 = a_x^2 + a_y^2 + a_z^2$ ### Core Logic Given |vecb|^2 = 6$|\vec{b}|^2 = 6$, angle theta = fracpi4$\theta = \frac{\pi}{4}$, and vecacdotvecb = 3sqrt2$\vec{a}\cdot\vec{b} = 3\sqrt{2}$. |veca||vecb| cos theta = 3sqrt2$|\vec{a}||\vec{b}| \cos \theta = 3\sqrt{2}$ Squaring both sides: |veca|^2 |vecb|^2 cos^2 theta = 18$|\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta = 18$ |veca|^2 (6) left(frac1sqrt2right)^2 = 18$|\vec{a}|^2 (6) \left(\frac{1}{\sqrt{2}}\right)^2 = 18$ |veca|^2 (6) left(frac12right) = 18 Rightarrow 3|veca|^2 = 18 Rightarrow |veca|^2 = 6$|\vec{a}|^2 (6) \left(\frac{1}{2}\right) = 18 \Rightarrow 3|\vec{a}|^2 = 18 \Rightarrow |\vec{a}|^2 = 6$ ### Step 1: Finding alpha and beta relation We have veca = hati + alpha hatj + beta hatk$\vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k}$. |veca|^2 = 1^2 + alpha^2 + beta^2$|\vec{a}|^2 = 1^2 + \alpha^2 + \beta^2$ 1 + alpha^2 + beta^2 = 6 Rightarrow alpha^2 + beta^2 = 5$1 + \alpha^2 + \beta^2 = 6 \Rightarrow \alpha^2 + \beta^2 = 5$ ### Step 2: Evaluating the Target Expression We need to find the value of (alpha^2 + beta^2)|veca times vecb|^2$(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2$. |veca times vecb|^2 = |veca|^2 |vecb|^2 sin^2 theta$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$ |veca times vecb|^2 = (6) (6) sin^2left(fracpi4right) = 36 left(frac12right) = 18$|\vec{a} \times \vec{b}|^2 = (6) (6) \sin^2\left(\frac{\pi}{4}\right) = 36 \left(\frac{1}{2}\right) = 18$ Thus, the required value is: (alpha^2 + beta^2)|veca times vecb|^2 = (5)(18) = 90$(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2 = (5)(18) = 90$ ### Pattern Recognition Vector magnitude and dot product give direct length values. |vecatimesvecb|^2 + (vecacdotvecb)^2 = |veca|^2|vecb|^2$|\vec{a}\times\vec{b}|^2 + (\vec{a}\cdot\vec{b})^2 = |\vec{a}|^2|\vec{b}|^2$ (Lagrange's Identity) skips sin^2theta$\sin^2\theta$ calculations completely. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra