What are the values of λ and μ?

μ=3 (from foot on line condition: (1−4)/1=(μ−9)/2=(2−5)/1=−3). λ=2 (from perpendicularity: vector (1−λ,1,−1)·(1,2,1)=0).

What is the distance between the two lines?

√146/7. The lines are parallel with direction (2,3,6). Using AB=(1,1,−1) and AB×d=(9,−8,1), distance=√146/7.

Foot of perpendicular from (λ,2,3) on line (x−4)/1=(y−9)/2=(z−5)/1 is (1,μ,2) — distance between lines

JEERANKERS

Math3D Geometry

QMCQ3D Geometry

Let the foot of perpendicular from the point $(\lambda,2,3)$ on the line $\dfrac{x-4}{1}=\dfrac{y-9}{2}=\dfrac{z-5}{1}$ be the point $(1,\mu,2)$. Then the distance between the lines $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z+4}{6}$ and $\dfrac{x-\lambda}{2}=\dfrac{y-\mu}{3}=\dfrac{z+5}{6}$ is:

A) $\dfrac{12}{7}$ B) $\dfrac{\sqrt{145}}{7}$ C) $\dfrac{\sqrt{146}}{7}$ D) $\dfrac{\sqrt{143}}{7}$

✅ Correct Answer

C) √146/7

✓

Solution

Find μ: foot (1,μ,2) lies on the given line

$\dfrac{1-4}{1}=\dfrac{\mu-9}{2}=\dfrac{2-5}{1}=-3$

$\Rightarrow\quad\mu=9+2(-3)=9-6=3$

Find λ using perpendicularity

Vector from $(\lambda,2,3)$ to foot $(1,3,2)$: $\vec{v}=(1-\lambda,\ 1,\ -1)$.

This must be perpendicular to direction $(1,2,1)$:

$(1-\lambda)(1)+(1)(2)+(-1)(1)=0$

$1-\lambda+2-1=0\quad\Rightarrow\quad2-\lambda=0\quad\Rightarrow\quad\lambda=2$

Find distance between the two parallel lines

Both lines have direction $\vec{d}=(2,3,6)$, $|\vec{d}|=\sqrt{4+9+36}=7$.

Points on the lines: $A=(1,2,-4)$ on $L_1$ and $B=(\lambda,\mu,-5)=(2,3,-5)$ on $L_2$.

$\overrightarrow{AB}=(2-1,\ 3-2,\ -5-(-4))=(1,\ 1,\ -1)$

$\overrightarrow{AB}\times\vec{d}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&-1\\2&3&6\end{vmatrix}$

$=(6+3)\hat{i}-(6+2)\hat{j}+(3-2)\hat{k}=9\hat{i}-8\hat{j}+\hat{k}$

$|\overrightarrow{AB}\times\vec{d}|=\sqrt{81+64+1}=\sqrt{146}$

$$d=\frac{|\overrightarrow{AB}\times\vec{d}|}{|\vec{d}|}=\frac{\sqrt{146}}{7}$$

📘 Distance Between Parallel Lines in 3D

For two parallel lines with direction $\vec{d}$ passing through points $A$ and $B$: Distance $=\dfrac{|\overrightarrow{AB}\times\vec{d}|}{|\vec{d}|}$. First verify the lines are parallel (same direction vector), then choose one point on each line, form $\overrightarrow{AB}$, compute the cross product with $\vec{d}$, and divide by $|\vec{d}|$.

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Important Concepts

Foot on Line Condition$(1,\mu,2)$ lies on $(x-4)/1=(y-9)/2=(z-5)/1=t$. At $(1,\mu,2)$: $t=(1-4)/1=-3$. So $\mu=9+2(-3)=3$ and $z=5+(-3)=2$ ✓.

Perpendicularity ConditionThe foot-of-perpendicular vector must be perpendicular to the line direction. Vector from source to foot: $(1-\lambda,1,-1)$. Dot with direction $(1,2,1)=0$: $(1-\lambda)+2-1=0$, giving $\lambda=2$.

Distance Formula for Parallel Lines$d=|\overrightarrow{AB}\times\vec{d}|/|\vec{d}|$. The cross product $\overrightarrow{AB}\times\vec{d}$ gives a vector perpendicular to $\vec{d}$ with magnitude equal to $|\overrightarrow{AB}|\sin\theta$ where $\theta$ is the angle between them.

Verify Parallel LinesBoth lines have direction $(2,3,6)$. $|(2,3,6)|=\sqrt{4+9+36}=7$. The lines are parallel. Distance $=\sqrt{146}/7$.

FAQs

How to verify foot (1,3,2) is correct?

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Substitute into line: $(1-4)/1=-3$, $(3-9)/2=-3$, $(2-5)/1=-3$ — all equal ✓. The foot lies on the line.

Why is |(2,3,6)|=7?

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$\sqrt{2^2+3^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7$.

How to compute the cross product (1,1,−1)×(2,3,6)?

⌄

Using the determinant: $\hat{i}(6-(-3))-\hat{j}(6-(-2))+\hat{k}(3-2)=9\hat{i}-8\hat{j}+1\hat{k}$.

Can I verify |(1,1,−1)×(2,3,6)|=√146?

⌄

$\sqrt{9^2+(-8)^2+1^2}=\sqrt{81+64+1}=\sqrt{146}$ ✓

Is this from JEE Main 2026?

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Yes, this question appeared in JEE Main 2026.

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