What is the direction of L1 perpendicular to both L2 and L3?

Direction d1=d2×d3=(1,2,2)×(2,2,1)=(−2,3,−2). L1 passes through origin with this direction.

4. The integer point on L3 at distance √17 from (2,−3,2) is (−1,−1,0) (from s=−2). Sum=−2, square=4.

L1 through origin perpendicular to L2 and L3 — point on L3 at √17 from intersection of L1 and L2 — (a+b+c)²

JEERANKERS

Math3D Geometry

QInteger Type3D Geometry

Let a line $L_1$ pass through the origin and be perpendicular to the lines
$L_2:\vec{r}=(3+t)\hat{i}+(2t-1)\hat{j}+(2t+4)\hat{k}$ and
$L_3:\vec{r}=(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k},\quad t,s\in\mathbb{R}$.

If $(a,b,c)$, $a\in\mathbb{Z}$, is the point on $L_3$ at a distance of $\sqrt{17}$ from the point of intersection of $L_1$ and $L_2$, then $(a+b+c)^2$ is equal to __________.

✅ Correct Answer

✓

Solution

Find direction of L1

Direction of $L_2$: $\vec{d_2}=(1,2,2)$. Direction of $L_3$: $\vec{d_3}=(2,2,1)$.

$L_1\perp L_2$ and $L_1\perp L_3$, so direction of $L_1=\vec{d_2}\times\vec{d_3}$:

$\vec{d_1}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&2\\2&2&1\end{vmatrix}=(2-4)\hat{i}-(1-4)\hat{j}+(2-4)\hat{k}=-2\hat{i}+3\hat{j}-2\hat{k}$

$L_1$: passes through origin with direction $(-2,3,-2)$: $\vec{r}=\lambda(-2,3,-2)$.

Find intersection of L1 and L2

$L_1$: $(-2\lambda, 3\lambda, -2\lambda)$. $L_2$: $(3+t, 2t-1, 2t+4)$.

$-2\lambda=3+t\quad\cdots(1)$
$3\lambda=2t-1\quad\cdots(2)$
$-2\lambda=2t+4\quad\cdots(3)$

From (1) and (3): $3+t=2t+4\Rightarrow t=-1$. Substituting: $-2\lambda=3+(-1)=2\Rightarrow\lambda=-1$.

Check (2): $3(-1)=-3$, $2(-1)-1=-3$ ✓

Intersection point $= (-2(-1),\ 3(-1),\ -2(-1))=(2,\ -3,\ 2)$

Find point on L3 at distance √17 from (2,−3,2)

Point on $L_3$: $(3+2s,\ 3+2s,\ 2+s)$. Distance$^2=17$:

$(3+2s-2)^2+(3+2s+3)^2+(2+s-2)^2=17$

$(1+2s)^2+(6+2s)^2+s^2=17$

$1+4s+4s^2+36+24s+4s^2+s^2=17$

$9s^2+28s+37=17$

$9s^2+28s+20=0$

$(9s+20)(s+1)... \text{disc}=784-720=64$

$s=\dfrac{-28\pm8}{18}$: $s=-\dfrac{20}{9}$ or $s=-2$

Select integer point and compute (a+b+c)²

$s=-20/9$: gives non-integer coordinates. Rejected (since $a\in\mathbb{Z}$).

$s=-2$: point $=(3+2(-2),\ 3+2(-2),\ 2+(-2))=(-1,\ -1,\ 0)$.

$a=-1,\ b=-1,\ c=0$

$(a+b+c)^2=(-1-1+0)^2=(-2)^2=\boxed{4}$

📘 Perpendicular Line via Cross Product

A line perpendicular to two given lines has direction equal to the cross product of their direction vectors. Once the intersection point is found by solving the parametric equations simultaneously, the distance condition gives a quadratic in the parameter $s$, yielding two candidate points — select the one with integer coordinates as required.

💡

Important Concepts

Direction of L1 = d2×d3$\vec{d_1}=\vec{d_2}\times\vec{d_3}$ ensures $L_1$ is perpendicular to both $L_2$ and $L_3$. Here $\vec{d_1}=(-2,3,-2)$ — note this is the only direction (up to scaling) perpendicular to both.

Intersection L1∩L2Equate parametric forms: $(-2\lambda,3\lambda,-2\lambda)=(3+t,2t-1,2t+4)$. From equations 1 and 3: $3+t=2t+4\Rightarrow t=-1,\lambda=-1$. Intersection is $(2,-3,2)$.

Distance Equation on L3General point on $L_3$: $(3+2s,3+2s,2+s)$. Set distance from $(2,-3,2)$ equal to $\sqrt{17}$. This gives quadratic $9s^2+28s+20=0$ with solutions $s=-2$ and $s=-20/9$.

Integer Constraint Selects Solution$a\in\mathbb{Z}$ means we need integer $a$-coordinate. $s=-20/9$ gives $a=3+2(-20/9)=3-40/9=-13/9\notin\mathbb{Z}$. $s=-2$ gives $a=-1\in\mathbb{Z}$ ✓.

FAQs

How to find direction of L1?

⌄

L1 is perpendicular to both L2 and L3, so its direction is $\vec{d_2}\times\vec{d_3}=(1,2,2)\times(2,2,1)$. Using the determinant formula: $\hat{i}(2-4)-\hat{j}(1-4)+\hat{k}(2-4)=-2\hat{i}+3\hat{j}-2\hat{k}$.

How to find L1∩L2?

⌄

Set $L_1$ parameter equations equal to $L_2$ parameter equations. Solve the system for $\lambda$ and $t$. With $t=-1,\lambda=-1$: intersection is $(2,-3,2)$.

What is the distance formula used?

⌄

Euclidean: $\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}=\sqrt{17}$. Squaring: $(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2=17$.

Why is s=−20/9 rejected?

⌄

When $s=-20/9$: $a=3+2(-20/9)=-13/9\notin\mathbb{Z}$. The condition $a\in\mathbb{Z}$ forces $s=-2$.

Is this from JEE Main 2026?

⌄

Yes, this appeared in JEE Main 2026 Section B.

🔗