JEE Main 2024 (Online) 4th April Evening Shift | 3D Geometry Question 18 | Mathematics

Let $$\mathrm{P}$$ be the point of intersection of the lines $$\frac{x-2}{1}=\frac{y-4}{5}=\frac{z-2}{1}$$ and $$\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-3}{2}$$. Then, the shortest distance of $$\mathrm{P}$$ from the line $$4 x=2 y=z$$ is

$$\frac{3 \sqrt{14}}{7}$$

$$\frac{5 \sqrt{14}}{7}$$

$$\frac{\sqrt{14}}{7}$$

$$\frac{6 \sqrt{14}}{7}$$

JEE Main 2024 (Online) 4th April Morning Shift

MCQ (Single Correct Answer)

-1

Let the point, on the line passing through the points $$P(1,-2,3)$$ and $$Q(5,-4,7)$$, farther from the origin and at a distance of 9 units from the point $$P$$, be $$(\alpha, \beta, \gamma)$$. Then $$\alpha^2+\beta^2+\gamma^2$$ is equal to :

150

155

160

165

JEE Main 2024 (Online) 1st February Evening Shift

MCQ (Single Correct Answer)

-1

Consider a $\triangle A B C$ where $A(1,3,2), B(-2,8,0)$ and $C(3,6,7)$. If the angle bisector of $\angle B A C$ meets the line $B C$ at $D$, then the length of the projection of the vector $\overrightarrow{A D}$ on the vector $\overrightarrow{A C}$ is :

$\frac{37}{2 \sqrt{38}}$

$\sqrt{19}$

$\frac{39}{2 \sqrt{38}}$

$\frac{\sqrt{38}}{2}$

JEE Main 2024 (Online) 1st February Evening Shift

MCQ (Single Correct Answer)

-1

Let $\mathrm{P}$ and $\mathrm{Q}$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of 6 units from the point $\mathrm{R}(1,2,3)$. If the centroid of the triangle PQR is $(\alpha, \beta, \gamma)$, then $\alpha^2+\beta^2+\gamma^2$ is :