Let veca = 2hati - 3hatj + hatk$\vec{a} = 2\hat{i} - 3\hat{j} + \hat{k}$, vecb = 3hati + 2hatj + 5hatk$\vec{b} = 3\hat{i} + 2\hat{j} + 5\hat{k}$ and a vector vecc$\vec{c}$ be such that (veca - vecc) times vecb = -18hati - 3hatj + 12hatk$(\vec{a} - \vec{c}) \times \vec{b} = -18\hat{i} - 3\hat{j} + 12\hat{k}$ and veca cdot vecc = 3$\vec{a} \cdot \vec{c} = 3$. If vecb times vecc = vecd$\vec{b} \times \vec{c} = \vec{d}$, then |veca cdot vecd|$|\vec{a} \cdot \vec{d}|$ is equal to:

### Related Formula A vector vecmathbfc$\vec{\mathbf{c}}$ coplanar with vecmathbfa$\vec{\mathbf{a}}$ and vecmathbfb$\vec{\mathbf{b}}$ and perpendicular to vecmathbfb$\vec{\mathbf{b}}$ can be expressed using the vector triple product command template: vecmathbfc = lambda (vecmathbfb times (vecmathbfa times vecmathbfb)) = lambda [(vecmathbfb cdot vecmathbfb)vecmathbfa - (vecmathbfa cdot vecmathbfb)vecmathbfb]$$\vec{\mathbf{c}} = \lambda (\vec{\mathbf{b}} \times (\vec{\mathbf{a}} \times \vec{\mathbf{b}})) = \lambda [(\vec{\mathbf{b}} \cdot \vec{\mathbf{b}})\vec{\mathbf{a}} - (\vec{\mathbf{a}} \cdot \vec{\mathbf{b}})\vec{\mathbf{b}}]$$ ### Core Logic First, find the dot products of the given vectors: vecmathbfb cdot vecmathbfb = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11$$\vec{\mathbf{b}} \cdot \vec{\mathbf{b}} = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11$$ vecmathbfa cdot vecmathbfb = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2$$\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2$$ Substituting these values into the expression for vecmathbfc$\vec{\mathbf{c}}$: vecmathbfc = lambda [11vecmathbfa - 2vecmathbfb]$$\vec{\mathbf{c}} = \lambda [11\vec{\mathbf{a}} - 2\vec{\mathbf{b}}]$$ vecmathbfc = lambda [11(hatmathbfi + 2hatmathbfj + 3hatmathbfk) - 2(3hatmathbfi + hatmathbfj - hatmathbfk)]$$\vec{\mathbf{c}} = \lambda [11(\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}}) - 2(3\hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}})]$$ vecmathbfc = lambda (5hatmathbfi + 20hatmathbfj + 35hatmathbfk) = 5lambda (hatmathbfi + 4hatmathbfj + 7hatmathbfk)$$\vec{\mathbf{c}} = \lambda (5\hat{\mathbf{i}} + 20\hat{\mathbf{j}} + 35\hat{\mathbf{k}}) = 5\lambda (\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})$$ ### Step 1: Determine Lambda Using the given condition vecmathbfa cdot vecmathbfc = 5$\vec{\mathbf{a}} \cdot \vec{\mathbf{c}} = 5$: vecmathbfa cdot [5lambda (hatmathbfi + 4hatmathbfj + 7hatmathbfk)] = 5$$\vec{\mathbf{a}} \cdot [5\lambda (\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})] = 5$$ 5lambda (1 cdot 1 + 2 cdot 4 + 3 cdot 7) = 5$$5\lambda (1 \cdot 1 + 2 \cdot 4 + 3 \cdot 7) = 5$$ 5lambda (1 + 8 + 21) = 5 implies 30lambda = 1 implies lambda = frac130$$5\lambda (1 + 8 + 21) = 5 \implies 30\lambda = 1 \implies \lambda = \frac{1}{30}$$ Thus, the vector vecmathbfc$\vec{\mathbf{c}}$ is: vecmathbfc = frac530(hatmathbfi + 4hatmathbfj + 7hatmathbfk) = frac16(hatmathbfi + 4hatmathbfj + 7hatmathbfk)$$\vec{\mathbf{c}} = \frac{5}{30}(\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}}) = \frac{1}{6}(\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})$$ ### Step 2: Calculate Magnitude The magnitude of vecmathbfc$\vec{\mathbf{c}}$ is evaluated as: \|vecmathbfc\| = fracsqrt1^2 + 4^2 + 7^26 = fracsqrt1 + 16 + 496 = fracsqrt666 = sqrtfrac6636 = sqrtfrac116$$\|\vec{\mathbf{c}}\| = \frac{\sqrt{1^2 + 4^2 + 7^2}}{6} = \frac{\sqrt{1 + 16 + 49}}{6} = \frac{\sqrt{66}}{6} = \sqrt{\frac{66}{36}} = \sqrt{\frac{11}{6}}$$ ### Pattern Recognition Whenever a vector is specified to be coplanar with vecmathbfa, vecmathbfb$\vec{\mathbf{a}}, \vec{\mathbf{b}}$ and perpendicular to vecmathbfb$\vec{\mathbf{b}}$, direct setup with the cross-product template vecmathbfb times (vecmathbfa times vecmathbfb)$\vec{\mathbf{b}} \times (\vec{\mathbf{a}} \times \vec{\mathbf{b}})$ circumvents solving cumbersome linear systems of scalar variables. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Vector Algebra

### Related Formula The line bisecting the angle between two symmetric vectors passing through the origin in the first quadrant is given by y = x$y = x$ or x - y = 0$x - y = 0$. Distance from point (x_1, y_1)$(x_1, y_1)$ to line Ax + By + C = 0$Ax + By + C = 0$ is: d = frac|Ax_1 + By_1 + C|sqrtA^2 + B^2$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$ ### Core Logic Vectors overlineOA = sqrt3hati+hatj$\overline{OA} = \sqrt{3}\hat{i}+\hat{j}$ and overlineOB = hati+sqrt3hatj$\overline{OB} = \hat{i}+\sqrt{3}\hat{j}$ are symmetric about the line y = x$y = x$. Therefore, the angle bisector of overlineOA$\overline{OA}$ and overlineOB$\overline{OB}$ is the line x - y = 0$x - y = 0$. Point C has coordinates (a, 1 - a)$(a, 1 - a)$. ### Step 1: Calculate Distance and Solve for a The perpendicular distance from C(a, 1 - a)$C(a, 1 - a)$ to x - y = 0$x - y = 0$ is: d = frac|a - (1 - a)|sqrt1^2 + (-1)^2 = frac|2a - 1|sqrt2$$d = \frac{|a - (1 - a)|}{\sqrt{1^2 + (-1)^2}} = \frac{|2a - 1|}{\sqrt{2}}$$ Given that this distance is frac9sqrt2$\frac{9}{\sqrt{2}}$: frac|2a - 1|sqrt2 = frac9sqrt2 implies |2a - 1| = 9$$\frac{|2a - 1|}{\sqrt{2}} = \frac{9}{\sqrt{2}} \implies |2a - 1| = 9$$ This gives two solutions: 1) 2a - 1 = 9 implies 2a = 10 implies a = 5$2a - 1 = 9 \implies 2a = 10 \implies a = 5$ 2) 2a - 1 = -9 implies 2a = -8 implies a = -4$2a - 1 = -9 \implies 2a = -8 \implies a = -4$ ### Step 2: Find the Sum of Values Sum of all possible values of a$a$: textSum = 5 + (-4) = 1$$\text{Sum} = 5 + (-4) = 1$$ ### Pattern Recognition Notice that overlineOA$\overline{OA}$ and overlineOB$\overline{OB}$ have swapped coordinates, meaning they are symmetric with respect to y=x$y=x$. Thus, the angle bisector equation is immediate (x-y=0$x-y=0$), simplifying the problem to a standard point-to-line distance calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Straight Lines Class 12 Mathematics: Vector Algebra

### 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