Introduction to...

A plane meets the co-ordinate axes in A,B,C such that the centroid of the triangle ABC is the point $$(p,q,r)$$. The equation of the plane is 

If a point $$P$$ from where line drawn cuts coordinates axes at $$A$$ and $$B$$ (with $$A$$ on $$x-$$axis and $$B$$ on $$y-$$axis ) satisfies $$\alpha \dfrac{x^{2}}{PB^{2}}+\beta \dfrac{y^{2}}{PA^{2}}=1$$, then $$\alpha+\beta$$ is

The distance between the points $$(\cos \theta, \sin \theta)$$ and $$(\sin \theta - \cos \theta)$$ is

The point equidistant from the point $$O(0, 0, 0), A(a, 0, 0), B(0, b, 0)$$ and $$C(0, 0, c)$$ has the coordinates

$$A$$ point $$C$$ with position vector $$\frac{{3a + 4b - 5c}}{3}$$ (where a,b and c are non co-planar vectors) divides the line joining $$A$$ and $$B$$ in the ratio $$2:1$$. If the position vector of $$A$$ is $$a-2b+3c$$, then the position vector of $$B$$ is

If the distance between a point P and the point (1, 1, 1) on the line $$\frac{{x\, - \,1}}{3}\, = \,\frac{{y - \,1}}{4}\, = \,\frac{{z\, - 1}}{{12}}$$ is 13, then the coordinates of P are

The xy-plane divides the line joining the points (-1, 3, 4) and (2,-5,6).

Let $$O$$ be the origin and $$P$$ be the point at a distance $$3$$ units from origin. If d.x.s' of OP are 1, - 2, - 2, then coordinates of P is given by 

The vector $$\vec{AF}$$, is given by?

The position vector of the vertices of a triangle $$ABC$$ are $$\hat { i } ,\hat { j } ,\hat { k } $$ then the position vector of its orthocentre is