$A B C D$ is a tetrahedron and $O$ is any point. If the line joining $O$ to the vertices meets the opposite faces at $P, Q, R$ and $S$ , then the value of
$\frac{O P}{A P} + \frac{O Q}{B Q} + \frac{O R}{C R} + \frac{O S}{D S}$ is

Open in App

Solution

Here, $A B C D$ is the tetrahedron.
Let $O$ be the origin and position vectors of $A, B, C$ and $D$ be $\to a, \to b, \to c$ and $\to d$ respectively.
$4$ linearly dependent variable can be expressed as
$x \to a + y \to b + z \to c + t \to d = 0$
where $x, y, z, t$ are scalars.
$\Rightarrow y \to b + z \to c + t \to d = - x \to a \Rightarrow \frac{y \to b + z \to c + t \to d}{y + z + t} = - \frac{x \to a}{y + z + t}$
Here, L.H.S. represents the position vector of a point in $B C D$ plane and R.H.S. is the point on $- - \to A O$
$∴$ There must be a point common to both the plane and the straight line, that is
$- - \to O P = - \frac{x \to a}{y + z + t} - - \to A P = - - \to O P - - - \to O A$
$= - \frac{x \to a}{y + z + t} - \to a = - \frac{x + y + z + t}{y + z + t} (\to a) \Rightarrow \frac{O P}{A P} = \frac{x}{x + y + z + t}$

Simillarly, $\frac{O Q}{B Q} = \frac{y}{x + y + z + t}$ ,
$\frac{O R}{C R} = \frac{z}{x + y + z + t}$ ,
and $\frac{O S}{D S} = \frac{t}{x + y + z + t}$

$∴$ Required value is $1$

Alternate:
Let $O$ be the origin
And the position vectors of $A, B, C, D$ be $a, b, c, d$ respectively,
Any one of these vectors may be expressed in terms of the other three, so
$x \to a + y \to b + z \to c + t \to d = 0 \dots (1)$

Let the equation of the line $A P$ be,
$\to r = - u \to a$
$u$ is positive when it lies on the oposite side of the origin w.r.t. $A$
Using equation $(1)$ ,
$x \to r - u (y \to b + z \to c + t \to d) = 0 \dots (2)$
The point $P$ , which lies on the plane $B C D$ ,
So $\to r, \to b, \to c, \to d$ are coplanar.
Hence, the sum of the coefficients in equation $(2)$ is zero,
$u = \frac{x}{(y + z + t)}$ ,
$\frac{O P}{A P} = \frac{u}{1 + u} = \frac{x}{x + y + z + t}$
Simillarly,
$\frac{O Q}{B Q} = \frac{y}{x + y + z + t}$ ,
$\frac{O R}{C R} = \frac{z}{x + y + z + t}$ ,
$\frac{O S}{D S} = \frac{t}{x + y + z + t}$
Therefore,
$\frac{O P}{A P} + \frac{O Q}{B Q} + \frac{O R}{C R} + \frac{O S}{D S} = 1$

Suggest Corrections

Similar questions

If a circle and rectangular hyperbola $x y = c^{2}$ meets in 4 points $P, Q, R a n d S$ then $O P^{2} + O Q^{2} + O R^{2} + O S^{2} =$ ______ where r is the radius of the circle. O is the origin.

Q. In figure, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS then prove that PQRS is a parallelogram

Q. The centre of a circle is at O and its diameter is of length 8 cm. Which of the points P, Q, R and S lies on the circle if OP = 8 cm; OQ = 4 cm; OR = 5 cm and OS = 6 cm?

Q. Let

O

be the origin and let

P Q R

be an arbitrary triangle. The point

S

is such that

¯ ¯¯¯¯¯¯ ¯ O P . ¯ ¯¯¯¯¯¯¯ ¯ O Q + ¯ ¯¯¯¯¯¯ ¯ O R . ¯ ¯¯¯¯¯¯ ¯ O S = ¯ ¯¯¯¯¯¯ ¯ O R . ¯ ¯¯¯¯¯¯ ¯ O P + ¯ ¯¯¯¯¯¯¯ ¯ O Q . ¯ ¯¯¯¯¯¯ ¯ O S = ¯ ¯¯¯¯¯¯¯ ¯ O Q . ¯ ¯¯¯¯¯¯ ¯ O R + ¯ ¯¯¯¯¯¯ ¯ O P . ¯ ¯¯¯¯¯¯ ¯ O S

. Then the

Δ

PQR has

S

as its

Q. Through the vertex O of the parabola

y^{2} = 4 a x, a

perpendicular is drawn to any tangent meeting it at P and the parabola T Q find the value of OP .OQ?

Join BYJU'S Learning Program

ABCD is a tetrahedron and O is any point. If the line joining O to the vertices meets the opposite faces at P,Q,R and S, then the value of OPAP+OQBQ+ORCR+OSDS is

$A B C D$ is a tetrahedron and $O$ is any point. If the line joining $O$ to the vertices meets the opposite faces at $P, Q, R$ and $S$ , then the value of
$\frac{O P}{A P} + \frac{O Q}{B Q} + \frac{O R}{C R} + \frac{O S}{D S}$ is