JEE Main 2024 (Online) 27th January Evening Shift | 3D Geometry Question 40 | Mathematics

Let the image of the point $$(1,0,7)$$ in the line $$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$$ be the point $$(\alpha, \beta, \gamma)$$. Then which one of the following points lies on the line passing through $$(\alpha, \beta, \gamma)$$ and making angles $$\frac{2 \pi}{3}$$ and $$\frac{3 \pi}{4}$$ with $$y$$-axis and $$z$$-axis respectively and an acute angle with $$x$$-axis ?

$$(1,-2,1+\sqrt{2})$$

$$(3,-4,3+2 \sqrt{2})$$

$$(3,4,3-2 \sqrt{2})$$

$$(1,2,1-\sqrt{2})$$

JEE Main 2024 (Online) 27th January Morning Shift

MCQ (Single Correct Answer)

-1

The distance, of the point $(7,-2,11)$ from the line

$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is :

JEE Main 2024 (Online) 27th January Morning Shift

MCQ (Single Correct Answer)

-1

If the shortest distance between the lines

$\frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3}$ and $\frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5}$ is $\frac{6}{\sqrt{5}}$, then the sum of all possible values of $\lambda$ is :

JEE Main 2023 (Online) 15th April Morning Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

Let the foot of perpendicular of the point $P(3,-2,-9)$ on the plane passing through the points $(-1,-2,-3),(9,3,4),(9,-2,1)$ be $Q(\alpha, \beta, \gamma)$. Then the distance of $Q$ from the origin is :

$\sqrt{38}$

$\sqrt{29}$

$\sqrt{42}$

$\sqrt{35}$

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