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JEE Main · Mathematics · 3D Geometry
MCQ · Mathematics · Vector Lines
Q. Let the line $L_1$ be parallel to the vector $-3\hat{i} + 2\hat{j} + 4\hat{k}$ and pass through the point $(2, 6, 7)$, and the line $L_2$ be parallel to the vector $2\hat{i} + \hat{j} + 3\hat{k}$ and pass through the point $(4, 3, 5)$. If the line $L_3$ is parallel to the vector $-3\hat{i} + 5\hat{j} + 16\hat{k}$ and intersects the lines $L_1$ and $L_2$ at the points $C$ and $D$, respectively, then $|\vec{CD}|^2$ is equal to:
A290 ✓
B171
C89
✅ Correct Answer: (A) 290
Step-by-Step Solution
1
Define Equations of Lines $L_1$ and $L_2$
General point $C$ on $L_1$:
$C = (2 – 3\lambda, 6 + 2\lambda, 7 + 4\lambda)$
General point $D$ on $L_2$:
$D = (4 + 2\mu, 3 + \mu, 5 + 3\mu)$
2
Form Vector $\vec{CD}$
$\vec{CD} = D – C$
$\vec{CD} = (2 + 2\mu + 3\lambda)\hat{i} + (-3 + \mu – 2\lambda)\hat{j} + (-2 + 3\mu – 4\lambda)\hat{k}$
3
Apply Parallelism to $L_3$
Since $L_3$ is parallel to $\vec{v} = -3\hat{i} + 5\hat{j} + 16\hat{k}$, the components of $\vec{CD}$ must be proportional to $\vec{v}$:
$\frac{2 + 2\mu + 3\lambda}{-3} = \frac{-3 + \mu – 2\lambda}{5} = \frac{-2 + 3\mu – 4\lambda}{16} = k$
4
Solve for $\lambda$ and $\mu$
Using the first two ratios: $5(2 + 2\mu + 3\lambda) = -3(-3 + \mu – 2\lambda)$
$10 + 10\mu + 15\lambda = 9 – 3\mu + 6\lambda \implies 9\lambda + 13\mu = -1$ …(i)
Using second and third: $16(-3 + \mu – 2\lambda) = 5(-2 + 3\mu – 4\lambda)$
$-48 + 16\mu – 32\lambda = -10 + 15\mu – 20\lambda \implies -12\lambda + \mu = 38$ …(ii)
Solving (i) and (ii), we get $\mu = 2$ and $\lambda = -3$.
5
Calculate $|\vec{CD}|^2$
Substitute $\mu=2, \lambda=-3$ into the ratio $k$:
$k = \frac{2 + 2(2) + 3(-3)}{-3} = \frac{2+4-9}{-3} = \frac{-3}{-3} = 1$
So, $\vec{CD} = 1 \cdot (-3\hat{i} + 5\hat{j} + 16\hat{k})$
$|\vec{CD}|^2 = (-3)^2 + 5^2 + 16^2 = 9 + 25 + 256 = 290$
✅ Correct Option: (A)
Related Theory: Lines in 3D Space
📌 Equations of a Line
In three-dimensional space, a line is uniquely determined by a point through which it passes and its direction.
1. Vector Form: $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is the direction vector.
2. Cartesian Form: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, where $(x_1, y_1, z_1)$ is a point on the line and $(a, b, c)$ are the direction ratios.
1. Vector Form: $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\vec{a}$ is the position vector of a point on the line and $\vec{b}$ is the direction vector.
2. Cartesian Form: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, where $(x_1, y_1, z_1)$ is a point on the line and $(a, b, c)$ are the direction ratios.
📌 Intersection of Lines
To find if two lines intersect, we equate their general points. If a common set of parameters exists, they intersect. In this specific problem, $L_3$ acts as a “transversal” intersecting both $L_1$ and $L_2$. This means the vector joining point $C$ (on $L_1$) and $D$ (on $L_2$) must be collinear with the direction vector of $L_3$.
📌 Distance Between Points in 3D
The distance $d$ between points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by the formula:
$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$
In vector terms, if $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$, then $|\vec{v}|^2 = a^2 + b^2 + c^2$.
📌 Collinearity and Parallelism
Two vectors $\vec{u}$ and $\vec{v}$ are parallel if $\vec{u} = k\vec{v}$ for some scalar $k$. This leads to the proportionality of their components:
$\frac{u_x}{v_x} = \frac{u_y}{v_y} = \frac{u_z}{v_z}$. This property is used to solve for unknown parameters in intersection problems.
📌 Important Results for JEE
Shortest Distance between Skew Lines
Angle between two lines
Condition for Coplanarity
Projection of a vector
Frequently Asked Questions
1. What is the direction vector of L3?
The direction vector is $-3\hat{i} + 5\hat{j} + 16\hat{k}$, as given in the problem.
2. Why did we assume $\vec{CD}$ is parallel to $L_3$?
Because the problem states that line $L_3$ passes through points $C$ and $D$ and is parallel to the given vector.
3. Can we solve this using Cartesian form?
Yes, but vector notation is generally faster for intersection and parallelism problems in JEE.
4. What if the lines L1 and L2 were parallel?
If they were parallel, the approach remains similar, but the system of equations might have different properties. Here they are skew lines.
5. How to find $|\vec{CD}|$ instead of its square?
Take the square root of 290, which is approximately 17.03.
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