### Related Formula Any vector veca$\vec{a}$ can be written as the sum of its component parallel to vecb$\vec{b}$ (along vecb$\vec{b}$) and its component perpendicular to vecb$\vec{b}$: veca = veca_parallel + veca_perp$\vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp}$ Magnitude squared: |veca|^2 = alpha^2 + beta^2 + gamma^2$|\vec{a}|^2 = \alpha^2 + \beta^2 + \gamma^2$ ### Core Logic Given: veca_parallel = frac1611(3hati+hatj-hatk)$\vec{a}_{\parallel} = \frac{16}{11}(3\hat{i}+\hat{j}-\hat{k})$ veca_perp = frac111(-4hati-5hatj-17hatk)$\vec{a}_{\perp} = \frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})$ ### Step 1: Reconstruct Vector a Add both components to find veca$\vec{a}$: veca = frac1611(3hati+hatj-hatk) + frac111(-4hati-5hatj-17hatk)$\vec{a} = \frac{16}{11}(3\hat{i}+\hat{j}-\hat{k}) + \frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})$ veca = frac111 left[ (48 - 4)hati + (16 - 5)hatj + (-16 - 17)hatk right]$\vec{a} = \frac{1}{11} \left[ (48 - 4)\hat{i} + (16 - 5)\hat{j} + (-16 - 17)\hat{k} \right]$ veca = frac111 left[ 44hati + 11hatj - 33hatk right] = 4hati + hatj - 3hatk$\vec{a} = \frac{1}{11} \left[ 44\hat{i} + 11\hat{j} - 33\hat{k} \right] = 4\hat{i} + \hat{j} - 3\hat{k}$ Therefore, alpha = 4$\alpha = 4$, beta = 1$\beta = 1$, gamma = -3$\gamma = -3$. ### Step 2: Calculate Sum of Squares alpha^2 + beta^2 + gamma^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26$\alpha^2 + \beta^2 + \gamma^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26$ ### Pattern Recognition Since parallel and perpendicular components are orthogonal vectors, you can also use directly |veca|^2 = |veca_parallel|^2 + |veca_perp|^2$|\vec{a}|^2 = |\vec{a}_{\parallel}|^2 + |\vec{a}_{\perp}|^2$ to save algebra step: |veca_parallel|^2 = frac16^211^2(9+1+1) = frac25611$|\vec{a}_{\parallel}|^2 = \frac{16^2}{11^2}(9+1+1) = \frac{256}{11}$, |veca_perp|^2 = frac111^2(16+25+289) = frac330121 = frac3011$|\vec{a}_{\perp}|^2 = \frac{1}{11^2}(16+25+289) = \frac{330}{121} = \frac{30}{11}$. Total = frac28611 = 26$\frac{286}{11} = 26$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula vecu times vecv = -vecv times vecu$\vec{u} \times \vec{v} = -\vec{v} \times \vec{u}$ textIf vecu times vecv = 0 implies vecu parallel vecv implies vecu = lambda vecv$\text{If } \vec{u} \times \vec{v} = 0 \implies \vec{u} \parallel \vec{v} \implies \vec{u} = \lambda \vec{v}$ ### Core Logic Given veca times vecc = vecc times vecb implies veca times vecc + vecb times vecc = 0$\vec{a} \times \vec{c} = \vec{c} \times \vec{b} \implies \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = 0$ (veca + vecb) times vecc = 0 implies vecc = lambda(veca + vecb)$(\vec{a} + \vec{b}) \times \vec{c} = 0 \implies \vec{c} = \lambda(\vec{a} + \vec{b})$ ### Step 1: Compute a + b veca + vecb = (2+3)hati + (-1-5)hatj + (3+1)hatk = 5hati - 6hatj + 4hatk$\vec{a} + \vec{b} = (2+3)\hat{i} + (-1-5)\hat{j} + (3+1)\hat{k} = 5\hat{i} - 6\hat{j} + 4\hat{k}$ vecc = lambda(5hati - 6hatj + 4hatk)$\vec{c} = \lambda(5\hat{i} - 6\hat{j} + 4\hat{k})$ |vecc|^2 = lambda^2(25 + 36 + 16) = 77lambda^2$|\vec{c}|^2 = \lambda^2(25 + 36 + 16) = 77\lambda^2$ ### Step 2: Expand the Dot Product Condition (veca + vecc) cdot (vecb + vecc) = 168$(\vec{a} + \vec{c}) \cdot (\vec{b} + \vec{c}) = 168$ veca cdot vecb + vecc cdot (veca + vecb) + |vecc|^2 = 168$\vec{a} \cdot \vec{b} + \vec{c} \cdot (\vec{a} + \vec{b}) + |\vec{c}|^2 = 168$ Evaluate veca cdot vecb = (2)(3) + (-1)(-5) + (3)(1) = 6 + 5 + 3 = 14$\vec{a} \cdot \vec{b} = (2)(3) + (-1)(-5) + (3)(1) = 6 + 5 + 3 = 14$. Substitute vecc cdot (veca + vecb) = lambda |veca + vecb|^2 = 77lambda$\vec{c} \cdot (\vec{a} + \vec{b}) = \lambda |\vec{a} + \vec{b}|^2 = 77\lambda$: 14 + 77lambda + 77lambda^2 = 168 implies 77lambda^2 + 77lambda - 154 = 0$14 + 77\lambda + 77\lambda^2 = 168 \implies 77\lambda^2 + 77\lambda - 154 = 0$ lambda^2 + lambda - 2 = 0 implies lambda = 1 text or lambda = -2$\lambda^2 + \lambda - 2 = 0 \implies \lambda = 1 \text{ or } \lambda = -2$ ### Step 3: Maximize |vecc|^2$|\vec{c}|^2$ Maximum value occurs when lambda = -2$\lambda = -2$: |vecc|^2 = 77(-2)^2 = 77 times 4 = 308$|\vec{c}|^2 = 77(-2)^2 = 77 \times 4 = 308$ ### Pattern Recognition Recognize the cross-product rule inversion immediately: vecx times vecy = vecy times vecz implies (vecx+vecz) parallel vecy$\vec{x} \times \vec{y} = \vec{y} \times \vec{z} \implies (\vec{x}+\vec{z}) \parallel \vec{y}$. This linear reduction circumvents solving complex linear systems. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula Vector Triple Product Identity: (vecu times vecv) times vecw = (vecu cdot vecw)vecv - (vecv cdot vecw)veca$(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{a}$ ### Core Logic Given vectors veca$\vec{a}$ and vecb$\vec{b}$: First, let's find the inner vector triple product (veca times vecb) times hati$(\vec{a} \times \vec{b}) \times \hat{i}$: (veca times vecb) times hati = (veca cdot hati)vecb - (vecb cdot hati)veca $(\vec{a} \times \vec{b}) \times \hat{i} = (\vec{a} \cdot \hat{i})\vec{b} - (\vec{b} \cdot \hat{i})\vec{a} $ From the given components: - veca cdot hati = -5$\vec{a} \cdot \hat{i} = -5$ - vecb cdot hati = 1$\vec{b} \cdot \hat{i} = 1$ (veca times vecb) times hati = -5vecb - veca $(\vec{a} \times \vec{b}) \times \hat{i} = -5\vec{b} - \vec{a} $ ### Step 1: Substitute Vector Coordinates Substitute coordinates of veca$\vec{a}$ and vecb$\vec{b}$ into the expression: -5vecb - veca = -5(hati + 2hatj - 4hatk) - (-5hati + hatj - 3hatk)$-5\vec{b} - \vec{a} = -5(\hat{i} + 2\hat{j} - 4\hat{k}) - (-5\hat{i} + \hat{j} - 3\hat{k})$ = -5hati - 10hatj + 20hatk + 5hati - hatj + 3hatk = -11hatj + 23hatk $= -5\hat{i} - 10\hat{j} + 20\hat{k} + 5\hat{i} - \hat{j} + 3\hat{k} = -11\hat{j} + 23\hat{k} $ Now, perform successive cross products with hati$\hat{i}$: textNext Step = (-11hatj + 23hatk) times hati = -11(hatj times hati) + 23(hatk times hati) = 11hatk + 23hatj $\text{Next Step} = (-11\hat{j} + 23\hat{k}) \times \hat{i} = -11(\hat{j} \times \hat{i}) + 23(\hat{k} \times \hat{i}) = 11\hat{k} + 23\hat{j} $ textFinal Vector vecc = (23hatj + 11hatk) times hati = 23(hatj times hati) + 11(hatk times hati) = -23hatk + 11hatj $\text{Final Vector } \vec{c} = (23\hat{j} + 11\hat{k}) \times \hat{i} = 23(\hat{j} \times \hat{i}) + 11(\hat{k} \times \hat{i}) = -23\hat{k} + 11\hat{j} $ ### Step 2: Calculate Dot Product Now evaluate vecc cdot (-hati + hatj + hatk)$\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$: vecc = 11hatj - 23hatk $\vec{c} = 11\hat{j} - 23\hat{k} $ vecc cdot (-hati + hatj + hatk) = (0)( -1) + (11)(1) + (-23)(1) = 11 - 23 = -12 $\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k}) = (0)( -1) + (11)(1) + (-23)(1) = 11 - 23 = -12 $ ### Pattern Recognition Sees: Repeated vector cross products with standard unit vectors. Shortcut: Observe that crossing a vector in the y-z$y-z$ plane with hati$\hat{i}$ rotates its components by 90^circ$90^{\circ}$ in that plane. Doing it twice returns it to the original plane with flipped coefficients and signs (vecv times hati times hati = -vecv$\vec{v} \times \hat{i} \times \hat{i} = -\vec{v}$). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula vecA cdot vecB = |vecA| |vecB| cos phi$\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \phi$ ### Core Logic Let the given baseline vectors be vecp_1, vecp_2, vecp_3$\vec{p}_1, \vec{p}_2, \vec{p}_3$. Note that |vecp_1| = |vecp_2| = |vecp_3| = 1$|\vec{p}_1| = |\vec{p}_2| = |\vec{p}_3| = 1$. 1. Angle with vecp_1$\vec{p}_1$ is fracpi2$\frac{\pi}{2}$: hatu cdot vecp_1 = 0 implies fracxsqrt2 + fraczsqrt2 = 0 implies x + z = 0 quad dots (i)$\hat{u} \cdot \vec{p}_1 = 0 \implies \frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0 \implies x + z = 0 \quad \dots (i)$ 2. Angle with vecp_2$\vec{p}_2$ is fracpi3$\frac{\pi}{3}$: hatu cdot vecp_2 = cosfracpi3 implies fracysqrt2 + fraczsqrt2 = frac12 implies y + z = frac1sqrt2 quad dots (ii)$\hat{u} \cdot \vec{p}_2 = \cos\frac{\pi}{3} \implies \frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2} \implies y + z = \frac{1}{\sqrt{2}} \quad \dots (ii)$ 3. Angle with vecp_3$\vec{p}_3$ is frac2pi3$\frac{2\pi}{3}$: hatu cdot vecp_3 = cosfrac2pi3 implies fracxsqrt2 + fracysqrt2 = -frac12 implies x + y = -frac1sqrt2 quad dots (iii)$\hat{u} \cdot \vec{p}_3 = \cos\frac{2\pi}{3} \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2} \implies x + y = -\frac{1}{\sqrt{2}} \quad \dots (iii)$ ### Step 1: Finding Vector Coordinates Subtracting (ii) from (iii): (x + y) - (y + z) = -frac1sqrt2 - frac1sqrt2 implies x - z = -sqrt2 quad dots (iv)$(x + y) - (y + z) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \implies x - z = -\sqrt{2} \quad \dots (iv)$ Solving (i) and (iv): 2x = -sqrt2 implies x = -frac1sqrt2$2x = -\sqrt{2} \implies x = -\frac{1}{\sqrt{2}}$ z = frac1sqrt2$z = \frac{1}{\sqrt{2}}$ From (ii): y = frac1sqrt2 - frac1sqrt2 = 0$y = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$. So, hatu = -frac1sqrt2hati + 0hatj + frac1sqrt2hatk$\hat{u} = -\frac{1}{\sqrt{2}}\hat{i} + 0\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$. ### Step 2: Evaluating the Norm Difference Given vecv = frac1sqrt2hati + frac1sqrt2hatj + frac1sqrt2hatk$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$: hatu - vecv = left(-frac1sqrt2 - frac1sqrt2right)hati + left(0 - frac1sqrt2right)hatj + left(frac1sqrt2 - frac1sqrt2right)hatk$\hat{u} - \vec{v} = \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{i} + \left(0 - \frac{1}{\sqrt{2}}\right)\hat{j} + \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{k}$ hatu - vecv = -sqrt2hati - frac1sqrt2hatj + 0hatk$\hat{u} - \vec{v} = -\sqrt{2}\hat{i} - \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k}$ Evaluating the squared magnitude: |hatu - vecv|^2 = (-sqrt2)^2 + left(-frac1sqrt2right)^2 = 2 + frac12 = frac52$|\hat{u} - \vec{v}|^2 = (-\sqrt{2})^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 = 2 + \frac{1}{2} = \frac{5}{2}$ ### Pattern Recognition Setting up dot products systematically transforms descriptive geometric angles into solvable sets of linear equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra