The distance of the point (7, -2, 11)$(7, -2, 11)$ from the line fracx-61=fracy-40=fracz-83$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line fracx-52=fracy-1-3=fracz-56$\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is:
A.12$12$
B.14$14$
C.18$18$
D.21$21$
Solution & Explanation
### Related Formula
fracx-x_1a = fracy-y_1b = fracz-z_1c = lambda$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \lambda$
### Core Logic
Let point A = (7, -2, 11)$A = (7, -2, 11)$. We want the distance from A$A$ to the line L_1: fracx-61=fracy-40=fracz-83$L_1: \frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ measured *along* a line parallel to L_2: fracx-52=fracy-1-3=fracz-56$L_2: \frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$.
The line passing through A$A$ parallel to L_2$L_2$ will have the equation:
fracx-72 = fracy+2-3 = fracz-116 = lambda$\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6} = \lambda$
Any general point B$B$ on this new line can be represented as:
B equiv (2lambda + 7, -3lambda - 2, 6lambda + 11)$B \equiv (2\lambda + 7, -3\lambda - 2, 6\lambda + 11)$
### Step 1: Finding Intersection Point B
Since B$B$ lies on the given line L_1$L_1$, its coordinates must satisfy the equation of L_1$L_1$:
frac(2lambda + 7) - 61 = frac(-3lambda - 2) - 40 = frac(6lambda + 11) - 83$\frac{(2\lambda + 7) - 6}{1} = \frac{(-3\lambda - 2) - 4}{0} = \frac{(6\lambda + 11) - 8}{3}$
Focus on the middle term (since denominator is 0, numerator must equal 0 for intersection):
-3lambda - 6 = 0 Rightarrow lambda = -2$-3\lambda - 6 = 0 \Rightarrow \lambda = -2$
### Step 2: Coordinates of B and Distance Calculation
Substitute lambda = -2$\lambda = -2$ back into the coordinates of point B$B$:
B = (2(-2)+7, -3(-2)-2, 6(-2)+11) = (3, 4, -1)$B = (2(-2)+7, -3(-2)-2, 6(-2)+11) = (3, 4, -1)$
Now, calculate the distance AB$AB$ using the 3D distance formula:
AB = sqrt(7-3)^2 + (-2-4)^2 + (11 - (-1))^2$AB = \sqrt{(7-3)^2 + (-2-4)^2 + (11 - (-1))^2}$AB = sqrt4^2 + (-6)^2 + (12)^2$AB = \sqrt{4^2 + (-6)^2 + (12)^2}$AB = sqrt16 + 36 + 144$AB = \sqrt{16 + 36 + 144}$AB = sqrt196 = 14$AB = \sqrt{196} = 14$
### Pattern Recognition
Distance of a point from a line *along* another direction means finding the intersection of the given line and a new line passing through the point parallel to the direction vector. The zero in the direction ratio is a massive shortcut—just equate the corresponding numerator to zero.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Three Dimensional Geometry
Keywords:#distance of the point from the line#JEE Main 2024 Morning Q3#Three Dimensional Geometry JEE Main 2024#Distance of a Point from a Line JEE Main 2024
More Three Dimensional Geometry Previous-Year Questions — Page 10
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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