Consider the line $L$ passing through the points $(1,2,3)$ and $(2,3,5)$. The distance of the point $\left(\dfrac{11}{3},\dfrac{11}{3},\dfrac{19}{3}\right)$ from the line $L$ along the line $\dfrac{3x-11}{2}=\dfrac{3y-11}{1}=\dfrac{3z-19}{2}$ is equal to:
Let $Q$ be the cube with vertices $\{(x_1,x_2,x_3)\in\mathbb{R}^3:x_1,x_2,x_3\in\{0,1\}\}$. Let $F$ be the set of all 12 lines containing face diagonals and $S$ be the set of 4 main diagonals. For lines $l_1\in F$ and $l_2\in S$, let $d(l_1,l_2)$ denote shortest distance. Maximum of $d(l_1,l_2)$ is $\lambda$. Find $\lambda^{-2}$.
Let \(A_1, A_2, A_3, A_4\) be the areas of the triangular faces of a tetrahedron, and \(h_1, h_2, h_3, h_4\) be the corresponding altitude of the tetrahedron. If volume of tetrahedron is \(1/6\) cubic units, then find the minimum value of \((A_1 + A_2 + A_3 + A_4)(h_1 + h_2 + h_3 + h_4)\) (in cubic units).
The line $(x,y,z)=(2,-3,4)+\lambda(1,2,-3)$ intersects $2x+3y-z=13$ at $P$ and $yz$-plane at $Q$. If $PQ=a\sqrt{b}$, $a,b\in\mathbb{N}$, $a>3$, then $\dfrac{a+b}{3}$ equals
Statement-1: The point \(A(3, 1, 6)\) is the mirror image of the point \(B(1, 3, 4)\) in the plane \(x - y + z = 5\).Statement-2: The plane \(x - y + z = 5\) bisects the line segment joining \(A(3, 1, 6)\) and \(B(1, 3, 4)\).
Match List-I with List-II: (A) Line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-2k}{2}$ lies in plane $2x-4y+z=3$; (B) Lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, value of $k$; (C) Plane through $(1,1,1)$ with OA=OB=OC, volume of tetrahedron OABC; (D) Distance from $(-1,5/\sqrt{2},3/\sqrt{2})$ to plane $P$ through $(1,-2,1)$ perpendicular to $2x-2y+z=0$ and $x-y+2z=4$
ABC is a triangle with vertices A(0, 0, 6), B(0, 4, 0) and C(6, 0, 0). Let points D, E and F are the mid-points of BC, AC and AB, respectively. Find the length of median AD.
Let (l, 2, 1) be a point on the plane which passes through the point (4, -2, 2). If the plane is perpendicular to the line joining the points (-2, -21, 29) and (-1, -16, 23), then \(\left(\frac{l}{11}\right)^2 - \frac{4l}{11} - 4\) is equal to
Let \(A(2, 3, 5)\), \(B(-1, 3, 2)\) and \(C(\lambda, 5, \mu)\) be the vertices of a \(\triangle ABC\). If the median through \(A\) is equally inclined to the coordinate axes, then