### 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

### 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