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
Let $\vec{a}$, $\vec{b}$, $\vec{c}$ have magnitudes 1, 2, 3 respectively, satisfying $|(\vec{a}\times\vec{b})\cdot\vec{c}|=6$. If $\vec{d}$ is a unit vector coplanar with $\vec{b}$ and $\vec{c}$ such that $\vec{b}\cdot\vec{d}=1$, then $|(\vec{a}\times\vec{c})\cdot\vec{d}|^2+|(\vec{a}\times\vec{c})\times\vec{d}|^2$ is
Let $\vec{a},\vec{b},\vec{c},\vec{d}$ be position vectors of $A,B,C,D$ with $|\vec{a}|=1$, $|\vec{b}|=2$, $|\vec{c}|=3$, $|\vec{d}|=4$, $(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})=24$, $\vec{a}\cdot\vec{c}=\frac{3}{2}$, $\vec{b}\cdot\vec{c}>0$. Area of $\triangle ABC=\frac{1}{4}(2\alpha+\beta\sqrt{3})$ and area of $\triangle BCD=\frac{1}{2}(m-n\sqrt{3})$ where $\alpha,\beta,m,n\in\mathbb{N}$. Match P)$\alpha$, Q)$\beta$, R)$m$, S)$n$ with 1)1, 2)3, 3)4, 4)9.
Let $\vec{a},\vec{b},\vec{c},\vec{d}$ be position vectors of $A,B,C,D$ with $|\vec{a}|=1$, $|\vec{b}|=2$, $|\vec{c}|=3$, $|\vec{d}|=4$, $(\vec{a}\times\vec{b})\cdot(\vec{c}\times\vec{d})=24$, $\vec{a}\cdot\vec{c}=\frac{3}{2}$, $\vec{b}\cdot\vec{c}>0$. Area of $\triangle ABC=\frac{1}{4}(2\alpha+\beta\sqrt{3})$ and area of $\triangle BCD=\frac{1}{2}(m-n\sqrt{3})$ where $\alpha,\beta,m,n\in\mathbb{N}$. Match P)$\alpha$, Q)$\beta$, R)$m$, S)$n$ with 1)1, 2)3, 3)4, 4)9.
Let A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3), D(x4, y4, z4) be the vertices of a tetrahedron. If E is the centroid of face BCD and G is the centroid of ABCD, then find the value of K such that AG = K(AE).
A tetrahedron has vertices at \(O(0, 0, 0)\), \(A(1, 2, 1)\), \(B(2, 1, 3)\) and \(C(-1, 1, 2)\). Then, the angle between the faces \(OAB\) and \(ABC\) will be
Let P be the plane, passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of P from the point $(3, -2, 2)$ is
If the length of the perpendicular from point $P(a, 4, 2)$, $a>0$, to the line $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z-1}{-1}$ is $2\sqrt{6}$ units, and $Q(\alpha_1,\alpha_2,\alpha_3)$ is the image of $P$ on this line, then $a+\displaystyle\sum_{i=1}^3 \alpha_i$ equals
A ray of light is coming along the line $\frac{x-2}{3} = \frac{y-1}{4} = \frac{z-6}{5}$ and strikes the plane mirror kept along the plane through the points $(2, 1, 0)$, $(5, 0, 1)$ and $(4, 1, 1)$. Then the equation of reflected ray is:
The distance of the point $(7, -3, -4)$ from the plane passing through the points $(2, -3, 1)$, $(-1, 1, -2)$ and $(3, -4, 2)$ is: