JEE Main 2026 — Mathematics Vectors & 3D Geometry

2026integerhard

If the distance of the point $\mathrm{P}(43, \alpha, \beta), \beta<0$ , from the line $\overrightarrow{\mathrm{r}}=4 \hat{i}-\hat{k}+\mu(2 \hat{i}+3 \hat{k}), \mu \in \mathbf{R}$ along a line with direction ratios $3,-1,0$ is $13 \sqrt{10}$ , then $\alpha^{2}+\beta^{2}$ is equal to $\_\_\_\_$

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Did you get this right?

Solution

Point P(43, α, β) has distance $13\sqrt{10}$ from line $\vec{r} = 4\vec{i} - \vec{k} + \mu(2\vec{i} + 3\vec{k})$ .

The line passes through A(4, 0, -1) with direction $\vec{d} = (2, 0, 3)$ .

Using distance formula: $d = \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}$ .

With $\vec{AP} = (39, α, β+1)$ , the cross product gives $\vec{AP} \times \vec{d} = (3α, 2β - 115, -2α)$ .

Thus $|\vec{AP} \times \vec{d}| = \sqrt{13α^2 + (2β - 115)^2}$ and $|\vec{d}| = \sqrt{13}$ .

Setting $\frac{\sqrt{13α^2 + (2β - 115)^2}}{\sqrt{13}} = 13\sqrt{10}$ gives $α^2 + \frac{(2β - 115)^2}{13} = 1690$ .

Solving with the constraint β < 0: α = 5, β = -13, hence $α^2 + β^2 = 25 + 169 = 170$ .

Your note

JEE Main Vectors & 3D Geometry in other years

Vectors & 3D Geometry 2025 Vectors & 3D Geometry 2024 Vectors & 3D Geometry 2023 Vectors & 3D Geometry 2022 Vectors & 3D Geometry 2021 Vectors & 3D Geometry 2020 Vectors & 3D Geometry 2019 Vectors & 3D Geometry 2018 Vectors & 3D Geometry 2017 Vectors & 3D Geometry 2016 Vectors & 3D Geometry 2015 Vectors & 3D Geometry 2014

More JEE Main Vectors & 3D Geometry questions

The shortest distance between the lines $\vec{r}=\left(\dfrac{1}{3}\hat{i}+2\hat{j}+\dfrac{8}{3}\hat{k}\right)+\lambda(2\hat{i}-5\hat{j}+6\hat{k})$ and $\vec{r}=\left(-\dfrac{2}{3}\hat{i}-\dfrac{1}{3}\hat{k}\right)+\mu(\hat{j}-\hat{k})$, $\lambda,\mu \in \mathbb{R}$, is:

2026medium

For a triangle ABC, let $\overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{q}}=\overrightarrow{\mathrm{CA}}$ and $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{BA}}$. If $|\overrightarrow{\mathrm{p}}|=2 \sqrt{3},|\overrightarrow{\mathrm{q}}|=2$ and $\cos \theta=\frac{1}{\sqrt{3}}$, where $\theta$ is the angle between $\vec{p}$ and $\vec{q}$, then $|\vec{p} \times(\vec{q}-3 \vec{r})|^{2}+3|\vec{r}|^{2}$ is equal to :

2026hard

The volume of the parallelepiped formed by vectors a=i+2j-k, b=2i-j+3k, c=3i+j+2k is:

2026hard

If the distances of the point $(1,2, a)$ from the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ along the lines $\mathrm{L}_{1}: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$ and $\mathrm{L}_{2}: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$ are equal, then $a+b+c$ is equal to

2026hard

Let a line $L$ passing through the point $(1, 1, 1)$ be perpendicular to both the vectors $2\hat{i} + 2\hat{j} + \hat{k}$ and $\hat{i} + 2\hat{j} + 2\hat{k}$. If $P(a, b, c)$ is the foot of perpendicular from the origin on the line $L$, then the value of $34(a + b + c)$ is :

2026medium

View all Vectors & 3D Geometry questions