### Related Formula a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 quad text(Condition for perpendicular vectors)$a_1 a_2 + b_1 b_2 + c_1 c_2 = 0 \quad \text{(Condition for perpendicular vectors)}$ ### Core Logic

Let the given point be A(1, 2, 3)$A(1, 2, 3)$. Let the foot of the perpendicular on the line be P(alpha, beta, gamma)$P(\alpha, \beta, \gamma)$. The equation of the line is fracx + 35 = fracy - 12 = fracz + 43 = k$\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} = k$. So, any point on the line can be written as (5k - 3, 2k + 1, 3k - 4)$(5k - 3, 2k + 1, 3k - 4)$. Let this point be P$P$. Thus, P(alpha, beta, gamma) equiv (5k - 3, 2k + 1, 3k - 4)$P(\alpha, \beta, \gamma) \equiv (5k - 3, 2k + 1, 3k - 4)$. ### Step 1: Finding Direction Ratios Direction ratios (DR's) of the line segment AP$AP$ are: ((5k - 3) - 1, (2k + 1) - 2, (3k - 4) - 3) equiv (5k - 4, 2k - 1, 3k - 7)$((5k - 3) - 1, (2k + 1) - 2, (3k - 4) - 3) \equiv (5k - 4, 2k - 1, 3k - 7)$ Direction ratios of the given line are (5, 2, 3)$(5, 2, 3)$. ### Step 2: Using Perpendicularity Condition Since AP$AP$ is perpendicular to the given line, the dot product of their direction ratios is zero: 5(5k - 4) + 2(2k - 1) + 3(3k - 7) = 0$5(5k - 4) + 2(2k - 1) + 3(3k - 7) = 0$ (25k - 20) + (4k - 2) + (9k - 21) = 0$(25k - 20) + (4k - 2) + (9k - 21) = 0$ 38k - 43 = 0$38k - 43 = 0$ k = frac4338$k = \frac{43}{38}$ ### Step 3: Calculating target expression We need to find 19(alpha + beta + gamma)$19(\alpha + \beta + \gamma)$. alpha + beta + gamma = (5k - 3) + (2k + 1) + (3k - 4)$\alpha + \beta + \gamma = (5k - 3) + (2k + 1) + (3k - 4)$ alpha + beta + gamma = 10k - 6$\alpha + \beta + \gamma = 10k - 6$ Substituting k = frac4338$k = \frac{43}{38}$: 19(alpha + beta + gamma) = 19 left(10left(frac4338right) - 6right)$19(\alpha + \beta + \gamma) = 19 \left(10\left(\frac{43}{38}\right) - 6\right)$ = 19 left( frac43038 - 6 right) = frac4302 - 114 = 215 - 114 = 101$= 19 \left( \frac{430}{38} - 6 \right) = \frac{430}{2} - 114 = 215 - 114 = 101$ ### Pattern Recognition To find coordinates of a foot of perpendicular, define a general point on the line using parameter k$k$, create the direction vector, and dot it with the line's direction vector set to zero. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry

### Related Formula textDot product constraint: vecPR cdot vecd = 0 text where vecd text is direction of the line.$\text{Dot product constraint: } \vec{PR} \cdot \vec{d} = 0 \text{ where } \vec{d} \text{ is direction of the line.}$ ### Core Logic

Let P = (2, 3, 5)$P = (2, 3, 5)$ and its mirror image be R(alpha, beta, gamma)$R(\alpha, \beta, \gamma)$. The vector connecting the point and its mirror image, vecPR$\vec{PR}$, is strictly perpendicular to the given line. The direction ratios of the line are vecd = (2, 3, 4)$\vec{d} = (2, 3, 4)$. Therefore, the dot product must be zero: vecPR cdot vecd = 0$\vec{PR} \cdot \vec{d} = 0$ (alpha - 2, beta - 3, gamma - 5) cdot (2, 3, 4) = 0$(\alpha - 2, \beta - 3, \gamma - 5) \cdot (2, 3, 4) = 0$ 2(alpha - 2) + 3(beta - 3) + 4(gamma - 5) = 0$2(\alpha - 2) + 3(\beta - 3) + 4(\gamma - 5) = 0$ 2alpha - 4 + 3beta - 9 + 4gamma - 20 = 0$2\alpha - 4 + 3\beta - 9 + 4\gamma - 20 = 0$ 2alpha + 3beta + 4gamma = 33$2\alpha + 3\beta + 4\gamma = 33$ ### Pattern Recognition When asked for a linear combination of image coordinates matching the direction ratios of the line, skip finding the exact foot of perpendicular. Just use the orthogonality condition directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry

### Related Formula textShortest distance = frac|(veca_2 - veca_1) cdot (vecn_1 times vecn_2)||vecn_1 times vecn_2|$\text{Shortest distance } = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{n}_1 \times \vec{n}_2)|}{|\vec{n}_1 \times \vec{n}_2|}$ ### Core Logic For L_1$L_1$: point a_1(1, -1, -4)$a_1(1, -1, -4)$ and direction vecn_1 = (2, -3, 2)$\vec{n}_1 = (2, -3, 2)$. For L_2$L_2$: passes through A(-4, 4, 3)$A(-4, 4, 3)$ and B(-1, 6, 3)$B(-1, 6, 3)$. Direction vecn_2 = vecAB = (-1 - (-4), 6 - 4, 3 - 3) = (3, 2, 0)$\vec{n}_2 = \vec{AB} = (-1 - (-4), 6 - 4, 3 - 3) = (3, 2, 0)$. The condition "perpendicular to..." is extra confirming information, since (3,2,0) cdot (-2,3,1) = 0$(3,2,0) \cdot (-2,3,1) = 0$. Cross product of directions: vecn_1 times vecn_2 = beginvmatrix hati & hatj & hatk \\ 2 & -3 & 2 \\ 3 & 2 & 0 endvmatrix = (-4, 6, 13)$\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 2 \\ 3 & 2 & 0 \end{vmatrix} = (-4, 6, 13)$ Magnitude: |vecn_1 times vecn_2| = sqrt(-4)^2 + 6^2 + 13^2 = sqrt16 + 36 + 169 = sqrt221$|\vec{n}_1 \times \vec{n}_2| = \sqrt{(-4)^2 + 6^2 + 13^2} = \sqrt{16 + 36 + 169} = \sqrt{221}$. Vector connecting lines: veca_2 - veca_1 = (-4 - 1, 4 - (-1), 3 - (-4)) = (-5, 5, 7)$\vec{a}_2 - \vec{a}_1 = (-4 - 1, 4 - (-1), 3 - (-4)) = (-5, 5, 7)$ Numerator of distance formula: |(veca_2 - veca_1) cdot (vecn_1 times vecn_2)| = |-5(-4) + 5(6) + 7(13)|$|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{n}_1 \times \vec{n}_2)| = |-5(-4) + 5(6) + 7(13)|$ = |20 + 30 + 91| = 141$= |20 + 30 + 91| = 141$ Shortest Distance = frac141sqrt221$= \frac{141}{\sqrt{221}}$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry

### Related Formula textDistance of point P text on line from A(x_1,y_1,z_1): P = (x_1 pm rd_x, y_1 pm rd_y, z_1 pm rd_z)$\text{Distance of point } P \text{ on line from } A(x_1,y_1,z_1): P = (x_1 \pm rd_x, y_1 \pm rd_y, z_1 \pm rd_z)$ textwhere (d_x,d_y,d_z) text are direction cosines and r text is distance.$\text{where } (d_x,d_y,d_z) \text{ are direction cosines and } r \text{ is distance.}$ ### Core Logic Direction ratios of AB = (16-4, -2 - (-6), 4 - (-2)) = (12, 4, 6)$AB = (16-4, -2 - (-6), 4 - (-2)) = (12, 4, 6)$. Magnitude of this vector = sqrt144 + 16 + 36 = sqrt196 = 14$= \sqrt{144 + 16 + 36} = \sqrt{196} = 14$. Direction cosines are left(frac1214, frac414, frac614right) = left(frac67, frac27, frac37right)$\left(\frac{12}{14}, \frac{4}{14}, \frac{6}{14}\right) = \left(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\right)$. Point P$P$ is at a distance of 21 units from A(4, -6, -2)$A(4, -6, -2)$: P = left(4 pm 21left(frac67right), -6 pm 21left(frac27right), -2 pm 21left(frac37right)right)$P = \left(4 \pm 21\left(\frac{6}{7}\right), -6 \pm 21\left(\frac{2}{7}\right), -2 \pm 21\left(\frac{3}{7}\right)\right)$ P = (4 pm 18, -6 pm 6, -2 pm 9)$P = (4 \pm 18, -6 \pm 6, -2 \pm 9)$ Since coordinates a,b,c$a,b,c$ of P$P$ are non-negative integers, we take the '+' sign: P = (4+18, -6+6, -2+9) = (22, 0, 7)$P = (4+18, -6+6, -2+9) = (22, 0, 7)$ Calculate distance from Q(4, -12, 3)$Q(4, -12, 3)$: PQ = sqrt(22 - 4)^2 + (0 - (-12))^2 + (7 - 3)^2$PQ = \sqrt{(22 - 4)^2 + (0 - (-12))^2 + (7 - 3)^2}$ PQ = sqrt18^2 + 12^2 + 4^2 = sqrt324 + 144 + 16 = sqrt484 = 22$PQ = \sqrt{18^2 + 12^2 + 4^2} = \sqrt{324 + 144 + 16} = \sqrt{484} = 22$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry