### Related Formula For any line making angles alpha, beta, gamma$\alpha, \beta, \gamma$ with the coordinate axes, the direction cosines satisfy: cos^2 alpha + cos^2 beta + cos^2 gamma = 1$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$ ### Core Logic Given that: beta = fracalpha2 quad textand quad gamma = fracalpha2$\beta = \frac{\alpha}{2} \quad \text{and} \quad \gamma = \frac{\alpha}{2}$ Substituting these values into the identity: cos^2 alpha + 2cos^2left(fracalpha2right) = 1$\cos^2 \alpha + 2\cos^2\left(\frac{\alpha}{2}\right) = 1$ ### Step 1: Solving the Trigonometric Equation Using the half-angle identity 2cos^2left(fracalpha2right) = 1 + cos alpha$2\cos^2\left(\frac{\alpha}{2}\right) = 1 + \cos \alpha$: cos^2 alpha + 1 + cos alpha = 1$\cos^2 \alpha + 1 + \cos \alpha = 1$ cos^2 alpha + cos alpha = 0$\cos^2 \alpha + \cos \alpha = 0$ cos alpha (cos alpha + 1) = 0$\cos \alpha (\cos \alpha + 1) = 0$ This yields two possible cases: 1. cos alpha = 0 implies alpha = fracpi2$\cos \alpha = 0 \implies \alpha = \frac{\pi}{2}$ 2. cos alpha = -1 implies alpha = pi$\cos \alpha = -1 \implies \alpha = \pi$ ### Step 2: Finding Values of beta$\beta$ and their Sum Now we find corresponding values for beta = fracalpha2$\beta = \frac{\alpha}{2}$: - If alpha = fracpi2 implies beta_1 = fracpi4$\alpha = \frac{\pi}{2} \implies \beta_1 = \frac{\pi}{4}$ - If alpha = pi implies beta_2 = fracpi2$\alpha = \pi \implies \beta_2 = \frac{\pi}{2}$ Sum of all possible values: beta_1 + beta_2 = fracpi4 + fracpi2 = frac3pi4$\beta_1 + \beta_2 = \frac{\pi}{4} + \frac{\pi}{2} = \frac{3\pi}{4}$ ### Pattern Recognition Direction cosines are bounded between [-1, 1]$[-1, 1]$. Always utilize the identities relating double angles or half angles (2cos^2theta = 1 + cos 2theta$2\cos^2\theta = 1 + \cos 2\theta$) to simplify quadratic forms involving different multiples of the coordinate angles. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Trigonometric Functions

### Related Formula Distance 'along a line' means the direction vector of the line segment joining target point P$P$ to intersecting point Q$Q$ must be parallel to the given direction ratios: vecPQ = k cdot vecd$\vec{PQ} = k \cdot \vec{d}$ ### Core Logic Let point P = (7, 10, 11)$P = (7, 10, 11)$. Any general point Q$Q$ on the first line fracx-41 = fracy-40 = fracz-23 = lambda$\frac{x-4}{1} = \frac{y-4}{0} = \frac{z-2}{3} = \lambda$ is: Q = (lambda + 4, 4, 3lambda + 2)$Q = (\lambda + 4, 4, 3\lambda + 2)$ Direction ratios of PQ$PQ$: vecPQ = (lambda + 4 - 7, 4 - 10, 3lambda + 2 - 11) = (lambda - 3, -6, 3lambda - 9)$\vec{PQ} = (\lambda + 4 - 7, 4 - 10, 3\lambda + 2 - 11) = (\lambda - 3, -6, 3\lambda - 9)$ ### Step 1: Equating direction vectors Since distance is measured parallel to the line with direction vector (2, 3, 6)$(2, 3, 6)$, vecPQ$\vec{PQ}$ must be parallel to (2, 3, 6)$(2, 3, 6)$: fraclambda - 32 = frac-63 = frac3lambda - 96$\frac{\lambda - 3}{2} = \frac{-6}{3} = \frac{3\lambda - 9}{6}$ From the middle term: fraclambda - 32 = -2 implies lambda - 3 = -4 implies lambda = -1$\frac{\lambda - 3}{2} = -2 \implies \lambda - 3 = -4 \implies \lambda = -1$ Let's check consistency with the third term: frac3(-1) - 96 = -2 quad (textConsistent!)$\frac{3(-1) - 9}{6} = -2 \quad (\text{Consistent!})$

### Step 2: Distance calculation For lambda = -1$\lambda = -1$, the intersection point Q$Q$ is: Q = (3, 4, -1)$Q = (3, 4, -1)$ Distance PQ$PQ$: PQ = sqrt(7 - 3)^2 + (10 - 4)^2 + (11 - (-1))^2$PQ = \sqrt{(7 - 3)^2 + (10 - 4)^2 + (11 - (-1))^2}$ PQ = sqrt4^2 + 6^2 + 12^2 = sqrt16 + 36 + 144 = sqrt196 = 14$PQ = \sqrt{4^2 + 6^2 + 12^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$ ### Pattern Recognition When solving distance parallel to a line in 3D, always write down the parametric coordinates of the general point first. Match the ratio of direction cosines directly to avoid setting up complicated systems of coordinate planes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Vector Algebra

### Related Formula Shortest distance between two skewed lines with vector equations vecr = veca_1 + lambda vecb_1$\vec{r} = \vec{a}_1 + \lambda \vec{b}_1$ and \vec{r} = \vec{a}_2 + \mu \vec{b}_2 is: S.D. = left| frac(veca_2 - veca_1) cdot (vecb_1 times vecb_2)|vecb_1 times vecb_2| right|$S.D. = \left| \frac{(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)}{|\vec{b}_1 \times \vec{b}_2|} \right|$ ### Core Logic From the given lines: Line 1 passes through A(1, 2, 3)$A(1, 2, 3)$ with direction vector vecb_1 = 2hati + 3hatj + 4hatk$\vec{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$. Line 2 passes through B(0, 0, 5)$B(0, 0, 5)$ with direction vector vecb_2 = hati + alphahatj + hatk$\vec{b}_2 = \hat{i} + \alpha\hat{j} + \hat{k}$. The vector connecting the two fixed points is: vecBA = (1-0)hati + (2-0)hatj + (3-5)hatk = hati + 2hatj - 2hatk$\vec{BA} = (1-0)\hat{i} + (2-0)\hat{j} + (3-5)\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k}$ ### Step 1: Compute Cross Product of Direction Vectors vecn = vecb_1 times vecb_2 = left| beginmatrix hati & hatj & hatk \\ 2 & 3 & 4 \\ 1 & alpha & 1 endmatrix right| = hati(3 - 4alpha) - hatj(2 - 4) + hatk(2alpha - 3)$\vec{n} = \vec{b}_1 \times \vec{b}_2 = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & \alpha & 1 \end{matrix} \right| = \hat{i}(3 - 4\alpha) - \hat{j}(2 - 4) + \hat{k}(2\alpha - 3)$ vecn = (3 - 4alpha)hati + 2hatj + (2alpha - 3)hatk$\vec{n} = (3 - 4\alpha)\hat{i} + 2\hat{j} + (2\alpha - 3)\hat{k}$ ### Step 2: Apply Shortest Distance Formula

S.D. = left| frac(hati + 2hatj - 2hatk) cdot vecn|vecn| right| = frac5sqrt6$S.D. = \left| \frac{(\hat{i} + 2\hat{j} - 2\hat{k}) \cdot \vec{n}}{|\vec{n}|} \right| = \frac{5}{\sqrt{6}}$ Taking dot product in numerator: (hati + 2hatj - 2hatk) cdot vecn = 1(3-4alpha) + 2(2) - 2(2alpha-3) = 3 - 4alpha + 4 - 4alpha + 6 = 13 - 8alpha$(\hat{i} + 2\hat{j} - 2\hat{k}) \cdot \vec{n} = 1(3-4\alpha) + 2(2) - 2(2\alpha-3) = 3 - 4\alpha + 4 - 4\alpha + 6 = 13 - 8\alpha$ Squaring both sides: frac(13 - 8alpha)^2(3 - 4alpha)^2 + 4 + (2alpha - 3)^2 = frac256$\frac{(13 - 8\alpha)^2}{(3 - 4\alpha)^2 + 4 + (2\alpha - 3)^2} = \frac{25}{6}$ 6(64alpha^2 - 208alpha + 169) = 25(16alpha^2 - 24alpha + 9 + 4 + 4alpha^2 - 12alpha + 9)$6(64\alpha^2 - 208\alpha + 169) = 25(16\alpha^2 - 24\alpha + 9 + 4 + 4\alpha^2 - 12\alpha + 9)$ 6(64alpha^2 - 208alpha + 169) = 25(20alpha^2 - 36alpha + 22)$6(64\alpha^2 - 208\alpha + 169) = 25(20\alpha^2 - 36\alpha + 22)$ 384alpha^2 - 1248alpha + 1014 = 500alpha^2 - 900alpha + 550$384\alpha^2 - 1248\alpha + 1014 = 500\alpha^2 - 900\alpha + 550$ 116alpha^2 + 348alpha - 464 = 0$116\alpha^2 + 348\alpha - 464 = 0$ alpha^2 + 3alpha - 4 = 0$\alpha^2 + 3\alpha - 4 = 0$ ### Step 3: Calculate the Sum of Roots The sum of all possible values of alpha$\alpha$ is given by the relation: alpha_1 + alpha_2 = -frac31 = -3$\alpha_1 + \alpha_2 = -\frac{3}{1} = -3$ ### Pattern Recognition Whenever asked for the sum of all possible parameter values resulting from a vector condition, look to directly read the linear term coefficient via Vieta's relations rather than explicitly factoring or computing the roots individually. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 12 Mathematics: Vector Algebra

### Related Formula General coordinates of any variable point on a 3D line given symmetric form equations: P_1 = (2lambda + 1, \, 3lambda - 1, \, 4lambda + 1)$P_1 = (2\lambda + 1, \, 3\lambda - 1, \, 4\lambda + 1)$ P_2 = (mu + 3, \, 2mu + 4, \, mu)$P_2 = (\mu + 3, \, 2\mu + 4, \, \mu)$ ### Core Logic Let line L intersect Line 1 at point A(2lambda + 1, 3lambda - 1, 4lambda + 1)$A(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$ and Line 2 at point B(mu + 3, 2mu + 4, mu)$B(\mu + 3, 2\mu + 4, \mu)$. Since line L passes through C(1, 1, 1)$C(1, 1, 1)$, the direction ratios computed from vector segment AC$AC$ must be proportional to the direction ratios computed from vector segment BC$BC$. ### Step 1: Determine Direction Ratio Parameters

Direction ratios of AC$AC$ segment: vecAC = (2lambda + 1 - 1, \, 3lambda - 1 - 1, \, 4lambda + 1 - 1) = (2lambda, \, 3lambda - 2, \, 4lambda)$\vec{AC} = (2\lambda + 1 - 1, \, 3\lambda - 1 - 1, \, 4\lambda + 1 - 1) = (2\lambda, \, 3\lambda - 2, \, 4\lambda)$ Direction ratios of BC$BC$ segment: vecBC = (mu + 3 - 1, \, 2mu + 4 - 1, \, mu - 1) = (mu + 2, \, 2mu + 3, \, mu - 1)$\vec{BC} = (\mu + 3 - 1, \, 2\mu + 4 - 1, \, \mu - 1) = (\mu + 2, \, 2\mu + 3, \, \mu - 1)$ Equating directional proportionality ratios: fracmu + 22lambda = frac2mu + 33lambda - 2 = fracmu - 14lambda$\frac{\mu + 2}{2\lambda} = \frac{2\mu + 3}{3\lambda - 2} = \frac{\mu - 1}{4\lambda}$ ### Step 2: Solve for Parameter Intersection values From the first and third fractional groups: fracmu + 22lambda = fracmu - 14lambda implies 2(mu + 2) = mu - 1$\frac{\mu + 2}{2\lambda} = \frac{\mu - 1}{4\lambda} \implies 2(\mu + 2) = \mu - 1$ 2mu + 4 = mu - 1 implies mu = -5$2\mu + 4 = \mu - 1 \implies \mu = -5$ Substitute mu = -5$\mu = -5$ back into the second parameter group linkage to evaluate the target structural direction indicators, which gives the simplified direction ratio vector for BC$BC$ as: textD.R.s = (-5 + 2, \, 2(-5) + 3, \, -5 - 1) = (-3, \, -7, \, -6) equiv (3, \, 7, \, 6)$\text{D.R.s} = (-5 + 2, \, 2(-5) + 3, \, -5 - 1) = (-3, \, -7, \, -6) \equiv (3, \, 7, \, 6)$ ### Step 3: Construct Line Equation and Verify Choice Equation of line L passing through C(1, 1, 1)$C(1, 1, 1)$ with direction vector (3, 7, 6)$(3, 7, 6)$: fracx - 13 = fracy - 17 = fracz - 16$\frac{x - 1}{3} = \frac{y - 1}{7} = \frac{z - 1}{6}$ Let's check option (7, 15, 13)$(7, 15, 13)$: frac7 - 13 = frac63 = 2$\frac{7 - 1}{3} = \frac{6}{3} = 2$ frac15 - 17 = frac147 = 2$\frac{15 - 1}{7} = \frac{14}{7} = 2$ frac13 - 16 = frac126 = 2$\frac{13 - 1}{6} = \frac{12}{6} = 2$ Since all values match perfectly, (7, 15, 13)$(7, 15, 13)$ lies on the line L. ### Pattern Recognition By comparing the first and third fractional terms containing lambda$\lambda$ in the denominator, you can solve for mu$\mu$ independently without tracking complex cross-multiplied quadratic lambdamu$\lambda\mu$ variations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry

### Related Formula textShortest Distance = left| fracvecAB cdot (vecp times vecq)|vecp times vecq| right|$\text{Shortest Distance} = \left| \frac{\vec{AB} \cdot (\vec{p} \times \vec{q})}{|\vec{p} \times \vec{q}|} \right|$ ### Core Logic Identify points A(1, 2, 3)$A(1, 2, 3)$ and B(lambda, 4, 5)$B(\lambda, 4, 5)$ on the lines with directions vecp = 2hati + 3hatj + 4hatk$\vec{p} = 2\hat{i} + 3\hat{j} + 4\hat{k}$ and vecq = 3hati + 4hatj + 5hatk$\vec{q} = 3\hat{i} + 4\hat{j} + 5\hat{k}$ respectively. Use the shortest distance formula to determine lambda_1$\lambda_1$ and lambda_2$\lambda_2$. ### Step 1: Calculate Cross Product and Direction Vector vecp times vecq = beginvmatrix hati & hatj & hatk \\ 2 & 3 & 4 \\ 3 & 4 & 5 endvmatrix = -hati + 2hatj - hatk$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = -\hat{i} + 2\hat{j} - \hat{k}$ |vecp times vecq| = sqrt(-1)^2 + 2^2 + (-1)^2 = sqrt6$|\vec{p} \times \vec{q}| = \sqrt{(-1)^2 + 2^2 + (-1)^2} = \sqrt{6}$ vecAB = (lambda - 1)hati + 2hatj + 2hatk$\vec{AB} = (\lambda - 1)\hat{i} + 2\hat{j} + 2\hat{k}$ ### Step 2: Solve for Lambda frac1sqrt6 = left| frac((lambda - 1)hati + 2hatj + 2hatk) cdot (-hati + 2hatj - hatk)sqrt6 right|$\frac{1}{\sqrt{6}} = \left| \frac{((\lambda - 1)\hat{i} + 2\hat{j} + 2\hat{k}) \cdot (-\hat{i} + 2\hat{j} - \hat{k})}{\sqrt{6}} \right|$ implies |-lambda + 1 + 4 - 2| = 1 implies |lambda - 3| = 1$\implies |-\lambda + 1 + 4 - 2| = 1 \implies |\lambda - 3| = 1$ implies lambda = 4 text or 2$\implies \lambda = 4 \text{ or } 2$ ### Step 3: Radius of the Passing Circle The circle passes through (0,0)$(0,0)$, (4,2)$(4,2)$ and (2,4)$(2,4)$. Using the circumradius formula R = fracabc4Delta$R = \frac{abc}{4\Delta}$: a = sqrt20, quad b = sqrt20, quad c = sqrt8$a = \sqrt{20}, \quad b = \sqrt{20}, \quad c = \sqrt{8}$ Delta = frac12 beginvmatrix 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 endvmatrix = 6$\Delta = \frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4 \end{vmatrix} = 6$ R = fracsqrt20 times sqrt20 times sqrt84 times 6 = frac40sqrt224 = frac5sqrt23$R = \frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times 6} = \frac{40\sqrt{2}}{24} = \frac{5\sqrt{2}}{3}$ ### Pattern Recognition Shortest distance values create symmetric configurations. When finding a circle passing through (0,0)$(0,0)$, (x,y)$(x,y)$, and (y,x)$(y,x)$, the symmetry about y=x$y=x$ simplifies radius calculations immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry Class 11 Mathematics: Circles