### Related Formula |vecx - vecy|^2 = |vecx|^2 + |vecy|^2 - 2(vecx cdot vecy)$|\vec{x} - \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2(\vec{x} \cdot \vec{y})$ (vecb times veca) cdot vecb = 0 quad text(Scalar triple product with repeated vectors is zero)$(\vec{b} \times \vec{a}) \cdot \vec{b} = 0 \quad \text{(Scalar triple product with repeated vectors is zero)}$ ### Core Logic Given |vecb| = 1$|\vec{b}| = 1$ and |vecb times veca| = 2$|\vec{b} \times \vec{a}| = 2$. Expand the requested expression: |(vecb times veca) - vecb|^2 = |vecb times veca|^2 + |vecb|^2 - 2((vecb times veca) cdot vecb)$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2 - 2((\vec{b} \times \vec{a}) \cdot \vec{b})$ ### Step 1: Simplify using Vector Properties The cross product (vecb times veca)$(\vec{b} \times \vec{a})$ produces a vector orthogonal to both vecb$\vec{b}$ and veca$\vec{a}$. Therefore, (vecb times veca) cdot vecb = 0$(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$. Substituting this back: |(vecb times veca) - vecb|^2 = |vecb times veca|^2 + |vecb|^2 - 0$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2 - 0$ |(vecb times veca) - vecb|^2 = (2)^2 + (1)^2$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = (2)^2 + (1)^2$ |(vecb times veca) - vecb|^2 = 4 + 1 = 5$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = 4 + 1 = 5$ ### Pattern Recognition A cross product vecu times vecv$\vec{u} \times \vec{v}$ is inherently perpendicular to vecu$\vec{u}$. Any length squared involving (vecu times vecv) pm vecu$(\vec{u} \times \vec{v}) \pm \vec{u}$ resolves simply via Pythagorean sum. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra

### Related Formula |veca times vecb|^2 + (veca cdot vecb)^2 = |veca|^2 |vecb|^2$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$ cos theta = fracvecb cdot vecc|vecb| |vecc|$\cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|}$ ### Core Logic Given |veca| = 1, |vecb| = 4, veca cdot vecb = 2$|\vec{a}| = 1, |\vec{b}| = 4, \vec{a} \cdot \vec{b} = 2$. Let's evaluate |veca times vecb|^2$|\vec{a} \times \vec{b}|^2$ using Lagrange's identity: |veca times vecb|^2 = |veca|^2 |vecb|^2 - (veca cdot vecb)^2$|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$ |veca times vecb|^2 = (1)(16) - (2)^2 = 16 - 4 = 12$|\vec{a} \times \vec{b}|^2 = (1)(16) - (2)^2 = 16 - 4 = 12$ ### Step 1: Dot products with c We are given: vecc = 2(veca times vecb) - 3vecb$\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$ To find the angle between vecb$\vec{b}$ and vecc$\vec{c}$, we need vecb cdot vecc$\vec{b} \cdot \vec{c}$ and |vecc|$|\vec{c}|$. Taking the dot product with vecb$\vec{b}$ on both sides: vecb cdot vecc = 2(vecb cdot (veca times vecb)) - 3(vecb cdot vecb)$\vec{b} \cdot \vec{c} = 2(\vec{b} \cdot (\vec{a} \times \vec{b})) - 3(\vec{b} \cdot \vec{b})$ Since vecb cdot (veca times vecb) = 0$\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ (scalar triple product with repeated vector): vecb cdot vecc = 0 - 3|vecb|^2 = -3(16) = -48 quad dots (1)$\vec{b} \cdot \vec{c} = 0 - 3|\vec{b}|^2 = -3(16) = -48 \quad \dots (1)$ ### Step 2: Finding magnitude of c Now, let's find |vecc|^2$|\vec{c}|^2$: |vecc|^2 = vecc cdot vecc = (2(veca times vecb) - 3vecb) cdot (2(veca times vecb) - 3vecb)$|\vec{c}|^2 = \vec{c} \cdot \vec{c} = (2(\vec{a} \times \vec{b}) - 3\vec{b}) \cdot (2(\vec{a} \times \vec{b}) - 3\vec{b})$ Since (veca times vecb) cdot vecb = 0$(\vec{a} \times \vec{b}) \cdot \vec{b} = 0$, the cross terms vanish: |vecc|^2 = 4|veca times vecb|^2 + 9|vecb|^2$|\vec{c}|^2 = 4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2$ |vecc|^2 = 4(12) + 9(16) = 48 + 144 = 192$|\vec{c}|^2 = 4(12) + 9(16) = 48 + 144 = 192$ |vecc| = sqrt192 = 8sqrt3$|\vec{c}| = \sqrt{192} = 8\sqrt{3}$ ### Step 3: Calculating angle Now apply the angle formula: cos theta = fracvecb cdot vecc|vecb| |vecc| = frac-48(4)(8sqrt3)$\cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|} = \frac{-48}{(4)(8\sqrt{3})}$ cos theta = frac-4832sqrt3 = frac-32sqrt3 = -fracsqrt32$\cos \theta = \frac{-48}{32\sqrt{3}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$ theta = cos^-1left(-fracsqrt32right)$\theta = \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ ### Pattern Recognition Always remember that the cross product vector (veca times vecb)$(\vec{a} \times \vec{b})$ is orthogonal to both veca$\vec{a}$ and vecb$\vec{b}$. This immediately eliminates cross terms when finding the magnitude of linear combinations like x(veca times vecb) + yvecb$x(\vec{a} \times \vec{b}) + y\vec{b}$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra

### Related Formula textArea of parallelogram = frac12 |overrightarrowd_1 times overrightarrowd_2|$\text{Area of parallelogram} = \frac{1}{2} |\overrightarrow{d_1} \times \overrightarrow{d_2}|$ ### Core Logic The diagonals of the parallelogram are overrightarrowAC$\overrightarrow{AC}$ and overrightarrowBD$\overrightarrow{BD}$. First, calculate the diagonal vector overrightarrowAC$\overrightarrow{AC}$: overrightarrowAC = textPosition vector of C - textPosition vector of A$\overrightarrow{AC} = \text{Position vector of } C - \text{Position vector of } A$ overrightarrowAC = (-3 - 2)hati + (4 - 3)hatj + (-2 - 5)hatk = -5hati + 1hatj - 7hatk$\overrightarrow{AC} = (-3 - 2)\hat{i} + (4 - 3)\hat{j} + (-2 - 5)\hat{k} = -5\hat{i} + 1\hat{j} - 7\hat{k}$ Alternatively, taking overrightarrowCA = 5hati - hatj + 7hatk$\overrightarrow{CA} = 5\hat{i} - \hat{j} + 7\hat{k}$. Let's use overrightarrowCA$\overrightarrow{CA}$ or overrightarrowAC$\overrightarrow{AC}$, the magnitude of the cross product will be the same. The second diagonal is given: overrightarrowBD = hati + 2hatj + 3hatk$\overrightarrow{BD} = \hat{i} + 2\hat{j} + 3\hat{k}$ ### Step 1: Finding Cross Product overrightarrowArea = frac12 |overrightarrowAC times overrightarrowBD|$\overrightarrow{Area} = \frac{1}{2} |\overrightarrow{AC} \times \overrightarrow{BD}|$ overrightarrowAC times overrightarrowBD = beginvmatrix hati & hatj & hatk \\ -5 & 1 & -7 \\ 1 & 2 & 3 endvmatrix$\overrightarrow{AC} \times \overrightarrow{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -7 \\ 1 & 2 & 3 \end{vmatrix}$ = hati(3 - (-14)) - hatj(-15 - (-7)) + hatk(-10 - 1)$= \hat{i}(3 - (-14)) - \hat{j}(-15 - (-7)) + \hat{k}(-10 - 1)$ = hati(17) - hatj(-8) + hatk(-11)$= \hat{i}(17) - \hat{j}(-8) + \hat{k}(-11)$ = 17hati + 8hatj - 11hatk$= 17\hat{i} + 8\hat{j} - 11\hat{k}$ ### Step 2: Calculating Magnitude textArea = frac12 |17hati + 8hatj - 11hatk|$\text{Area} = \frac{1}{2} |17\hat{i} + 8\hat{j} - 11\hat{k}|$ = frac12 sqrt17^2 + 8^2 + (-11)^2$= \frac{1}{2} \sqrt{17^2 + 8^2 + (-11)^2}$ = frac12 sqrt289 + 64 + 121$= \frac{1}{2} \sqrt{289 + 64 + 121}$ = frac12 sqrt474$= \frac{1}{2} \sqrt{474}$ ### Pattern Recognition When opposite vertices and one full diagonal vector are provided, immediately calculate the second diagonal vector via displacement and evaluate half the magnitude of their cross product. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra

### Core Logic Evaluate base vectors: veca + vecb = (5, 1, 4)$\vec{a} + \vec{b} = (5, 1, 4)$ veca times vecb = beginvmatrix hati & hatj & hatk \\ 3 & 2 & 1 \\ 2 & -1 & 3 endvmatrix = (7, -7, -7)$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 2 & -1 & 3 \end{vmatrix} = (7, -7, -7)$ Substitute into given cross product equation, letting vecc = (x, y, z)$\vec{c} = (x, y, z)$: (5hati + hatj + 4hatk) times (xhati + yhatj + zhatk) = 2(7hati - 7hatj - 7hatk) + 24hatj - 6hatk$(5\hat{i} + \hat{j} + 4\hat{k}) \times (x\hat{i} + y\hat{j} + z\hat{k}) = 2(7\hat{i} - 7\hat{j} - 7\hat{k}) + 24\hat{j} - 6\hat{k}$ beginvmatrix hati & hatj & hatk \\ 5 & 1 & 4 \\ x & y & z endvmatrix = (14, -14, -14) + (0, 24, -6)$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z \end{vmatrix} = (14, -14, -14) + (0, 24, -6)$ (z-4y)hati - (5z-4x)hatj + (5y-x)hatk = (14, 10, -20)$(z-4y)\hat{i} - (5z-4x)\hat{j} + (5y-x)\hat{k} = (14, 10, -20)$ Equating components: z - 4y = 14 implies z = 4y + 14$z - 4y = 14 \implies z = 4y + 14$ 4x - 5z = 10$4x - 5z = 10$ 5y - x = -20 implies x = 5y + 20$5y - x = -20 \implies x = 5y + 20$ Now use the dot product constraint: veca - vecb + hati = (3-2+1)hati + (2+1)hatj + (1-3)hatk = (2, 3, -2)$\vec{a} - \vec{b} + \hat{i} = (3-2+1)\hat{i} + (2+1)\hat{j} + (1-3)\hat{k} = (2, 3, -2)$. (veca - vecb + hati) cdot vecc = -3$(\vec{a} - \vec{b} + \hat{i}) \cdot \vec{c} = -3$ 2x + 3y - 2z = -3$2x + 3y - 2z = -3$ Substitute x = 5y + 20$x = 5y + 20$ and z = 4y + 14$z = 4y + 14$: 2(5y + 20) + 3y - 2(4y + 14) = -3$2(5y + 20) + 3y - 2(4y + 14) = -3$ 10y + 40 + 3y - 8y - 28 = -3$10y + 40 + 3y - 8y - 28 = -3$ 5y + 12 = -3 implies 5y = -15 implies y = -3$5y + 12 = -3 \implies 5y = -15 \implies y = -3$ Calculate x$x$ and z$z$: x = 5(-3) + 20 = 5$x = 5(-3) + 20 = 5$ z = 4(-3) + 14 = 2$z = 4(-3) + 14 = 2$ So, vecc = (5, -3, 2)$\vec{c} = (5, -3, 2)$. |vecc|^2 = 5^2 + (-3)^2 + 2^2 = 25 + 9 + 4 = 38$|\vec{c}|^2 = 5^2 + (-3)^2 + 2^2 = 25 + 9 + 4 = 38$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra

### Core Logic Given vecp times vecb = vecc times vecb$\vec{p} \times \vec{b} = \vec{c} \times \vec{b}$. (vecp - vecc) times vecb = vec0$(\vec{p} - \vec{c}) \times \vec{b} = \vec{0}$ Thus, vecp - vecc = lambda vecb implies vecp = vecc + lambda vecb$\vec{p} - \vec{c} = \lambda \vec{b} \implies \vec{p} = \vec{c} + \lambda \vec{b}$. ### Step 1: Utilize Dot Product Condition Given vecp cdot veca = 0$\vec{p} \cdot \vec{a} = 0$. (vecc + lambda vecb) cdot veca = 0$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0$ vecc cdot veca + lambda (vecb cdot veca) = 0$\vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$ Calculate vecc cdot veca$\vec{c} \cdot \vec{a}$: (1)(3) + (-3)(1) + (4)(-2) = 3 - 3 - 8 = -8$(1)(3) + (-3)(1) + (4)(-2) = 3 - 3 - 8 = -8$. Calculate vecb cdot veca$\vec{b} \cdot \vec{a}$: (4)(3) + (1)(1) + (7)(-2) = 12 + 1 - 14 = -1$(4)(3) + (1)(1) + (7)(-2) = 12 + 1 - 14 = -1$. -8 + lambda(-1) = 0 implies lambda = -8$-8 + \lambda(-1) = 0 \implies \lambda = -8$ ### Step 2: Substitute and Solve Substitute lambda$\lambda$ back to find vecp$\vec{p}$: vecp = vecc - 8vecb = (hati - 3hatj + 4hatk) - 8(4hati + hatj + 7hatk)$\vec{p} = \vec{c} - 8\vec{b} = (\hat{i} - 3\hat{j} + 4\hat{k}) - 8(4\hat{i} + \hat{j} + 7\hat{k})$ vecp = -31hati - 11hatj - 52hatk$\vec{p} = -31\hat{i} - 11\hat{j} - 52\hat{k}$ Now compute vecp cdot (hati - hatj - hatk)$\vec{p} \cdot (\hat{i} - \hat{j} - \hat{k})$: = (-31)(1) + (-11)(-1) + (-52)(-1) = -31 + 11 + 52 = 32$= (-31)(1) + (-11)(-1) + (-52)(-1) = -31 + 11 + 52 = 32$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra