Keywords:#image of a point in a 3D line#JEE Main 2025 Morning Q62#Three Dimensional Geometry JEE Main 2025#Image of a Point in a Line JEE Main 2025
More Three Dimensional Geometry Previous-Year Questions — Page 10
Q28jee_main_2024_31_jan_eveningDistance of a point on a line
A line passes through A(4, -6, -2)$A(4, -6, -2)$ and B(16, -2, 4)$B(16, -2, 4)$. The point P(a, b, c)$P(a, b, c)$ where a, b, c$a, b, c$ are non-negative integers, on the line AB$AB$ lies at a distance of 21$21$ units, from the point A$A$. The distance between the points P(a, b, c)$P(a, b, c)$ and Q(4, -12, 3)$Q(4, -12, 3)$ is equal to
Numerical Answer.Answer: 22 to 22
Solution
### 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
Q14jee_main_2024_31_jan_morningDistance of a Point from a Line
The distance of the point Q(0, 2, -2)$Q(0, 2, -2)$ form the line passing through the point P(5, -4, 3)$P(5, -4, 3)$ and perpendicular to the lines vecr = (-3hati + 2hatk) + lambda(2hati + 3hatj + 5hatk), lambda in mathbbR$\vec{r} = (-3\hat{i} + 2\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 5\hat{k}), \lambda \in \mathbb{R}$ and vecr = (hati - 2hatj + hatk) + mu(-hati + 3hatj + 2hatk), mu in mathbbR$\vec{r} = (\hat{i} - 2\hat{j} + \hat{k}) + \mu(-\hat{i} + 3\hat{j} + 2\hat{k}), \mu \in \mathbb{R}$
A.sqrt86$\sqrt{86}$
B.sqrt20$\sqrt{20}$
C.sqrt54$\sqrt{54}$
D.sqrt74$\sqrt{74}$
Solution
### Core Logic
A vector in the direction of the required line is perpendicular to both given lines. We obtain it via cross product of their direction vectors:
vecn = beginvmatrix hati & hatj & hatk \\ 2 & 3 & 5 \\ -1 & 3 & 2 endvmatrix = -9hati - 9hatj + 9hatk$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix} = -9\hat{i} - 9\hat{j} + 9\hat{k}$$
Taking the direction vector as hati + hatj - hatk$\hat{i} + \hat{j} - \hat{k}$.
### Step 1: Required Line Equation
The line passes through P(5, -4, 3)$P(5, -4, 3)$ with direction hati + hatj - hatk$\hat{i} + \hat{j} - \hat{k}$.
Equation: vecr = (5hati - 4hatj + 3hatk) + alpha(hati + hatj - hatk)$\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \alpha(\hat{i} + \hat{j} - \hat{k})$.
### Step 2: Projection & Distance
Any point on the line is M(5+alpha, -4+alpha, 3-alpha)$M(5+\alpha, -4+\alpha, 3-\alpha)$.
We need distance from Q(0, 2, -2)$Q(0, 2, -2)$.
Vector vecQM = (5+alpha)hati + (alpha-6)hatj + (5-alpha)hatk$\vec{QM} = (5+\alpha)\hat{i} + (\alpha-6)\hat{j} + (5-\alpha)\hat{k}$.
Since vecQM$\vec{QM}$ is perpendicular to the line direction (hati + hatj - hatk)$(\hat{i} + \hat{j} - \hat{k})$:
(5+alpha)(1) + (alpha-6)(1) + (5-alpha)(-1) = 0$$(5+\alpha)(1) + (\alpha-6)(1) + (5-\alpha)(-1) = 0$$5 + alpha + alpha - 6 - 5 + alpha = 0 implies 3alpha = 6 implies alpha = 2.$$5 + \alpha + \alpha - 6 - 5 + \alpha = 0 \implies 3\alpha = 6 \implies \alpha = 2.$$Distance of a Point from a Line diagram for Q14 - JEE Main 2024 Morning
### Step 3: Distance calculation
Substitute alpha = 2$\alpha = 2$ in vecQM$\vec{QM}$:
vecQM = 7hati - 4hatj + 3hatk$\vec{QM} = 7\hat{i} - 4\hat{j} + 3\hat{k}$.
Distance |vecQM| = sqrt7^2 + (-4)^2 + 3^2 = sqrt49 + 16 + 9 = sqrt74$|\vec{QM}| = \sqrt{7^2 + (-4)^2 + 3^2} = \sqrt{49 + 16 + 9} = \sqrt{74}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Three Dimensional Geometry
Class 12 Maths: Vector Algebra
Q24jee_main_2024_31_jan_morningFoot of Perpendicular and Angle
Let Q$Q$ and R$R$ be the feet of perpendiculars from the point P(a, a, a)$P(a, a, a)$ on the lines x = y, z = 1$x = y, z = 1$ and x = -y, z = -1$x = -y, z = -1$ respectively. If angle QPR$\angle QPR$ is a right angle, then 12a^2$12a^2$ is equal to
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