JEE Main 2023 (Online) 30th January Evening Shift | 3D Geometry Question 127 | Mathematics

A vector $\vec{v}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$, to the $y$-axis at 45 and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2},-1,1)$ and $(a, b, c)$, is normal to $\vec{v}$, then :

$a+b+\sqrt{2} c=1$

$\sqrt{2} a+b+c=1$

$\sqrt{2} a-b+c=1$

$a+\sqrt{2} b+c=1$

JEE Main 2023 (Online) 30th January Evening Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

If a plane passes through the points $(-1, k, 0),(2, k,-1),(1,1,2)$ and is parallel to the line $\frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}$, then the value of $\frac{k^2+1}{(k-1)(k-2)}$ is :

$\frac{17}{5}$

$\frac{6}{13}$

$\frac{13}{6}$

$\frac{5}{17}$

JEE Main 2023 (Online) 30th January Morning Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

The line $$l_1$$ passes through the point (2, 6, 2) and is perpendicular to the plane $$2x+y-2z=10$$. Then the shortest distance between the line $$l_1$$ and the line $$\frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$$ is :

$$\frac{19}{3}$$

$$\frac{13}{3}$$

JEE Main 2023 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

-1

Out of Syllabus

The plane $$2x-y+z=4$$ intersects the line segment joining the points A ($$a,-2,4)$$ and B ($$2,b,-3)$$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $$\sqrt5$$. If $$ab < 0$$ and P is the point $$(a-b,b,2b-a)$$ then CP$$^2$$ is equal to :

$$\frac{17}{3}$$

$$\frac{97}{3}$$

$$\frac{16}{3}$$

$$\frac{73}{3}$$

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