P P is a point in the plane of the triangle ABC- Pv-s of A-B and C are a -b and c respectively with respect to P as the origin- If b - c -b - c - 0 and c - a -c - a - 0 then w-r-t- the Triangle ABC- P is its1CentroidQ If a -b -c are the position vectors of the three non collinear points A-B and C repectively such that the vector V - PA - PB - PC is a null vector then w-r-t- the - A BC- P is its2orthocentreRif P is a point inside the - A BC such that the vector R - BCPA - CAPB - ABPC is null vector then w-r-t- the - A BC- P is its3IncentreS If P is a point in the plane of the triangle ABC such that the scalar product PA -CB andPB -AC vanishes- then w-r-t- the - A BC- P is its4Circumcentre

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Byju's Answer

Standard XII

Mathematics

Geometrical Representation of a Complex Number

P P is a poin...

Question

P) P is a point in the plane of the triangle ABC. Pv's of A,B and C are $\to a, \to b a n d \to c$ respectively with respect to P as the origin. If $(\to b + \to c) . (\to b - \to c) = 0 a n d (\to c + \to a) . (\to c - \to a) = 0$ then w.r.t. the Triangle ABC, P is its 1)Centroid
Q) If $\to a, \to b, \to c a r e$ the position vectors of the three non collinear points A,B and C repectively such that the vector $\to V = - - \to P A + - - \to P B + - - \to P C$ is a null vector then w.r.t. the $Δ A B C$ , P is its 2)orthocentre
R)if P is a point inside the $Δ A B C$ such that the vector $\to R = (B C) (- - \to P A) + C A (- - \to P B) + (A B) (- - \to P C)$ is null vector then w.r.t. the $Δ A B C$ , P is its 3)Incentre
S) If P is a point in the plane of the triangle ABC such that the scalar product $- - \to P A . - - \to C B a n d - - \to P B . - - \to A C$ vanishes, then w.r.t. the $Δ A B C$ . P is its 4)Circumcentre

P-4,Q-1,R-2,S-3

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P-1,Q-4,R-3,S-2

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P-4,Q-1,R-3,S-2

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P-4,Q-1,R-2,S-2

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Solution

The correct option is C P-4,Q-1,R-3,S-2

Case (P) - Refer diagram under P) in attached image

Given,
$(\to b + \to c) . (\to b - \to c) = 0$
$\Rightarrow b^{2} - c^{2} = 0$
$\Rightarrow ∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣$ --- (i)

Similarly given,
$(\to c + \to a) . (\to c - \to a) = 0$
$\Rightarrow ∣ ∣ \to c ∣ ∣ = ∣ ∣ \to a ∣ ∣$ --- (ii)

On comparing (i) & (ii), we get:
$\Rightarrow ∣ ∣ \to a ∣ ∣ = ∣ ∣ ∣ \to b ∣ ∣ ∣ = ∣ ∣ \to c ∣ ∣$

Thus P is the Circumcentre.
So the match for P) is 4)Circumcentre

Case (Q) - Refer diagram under Q) in attached image
As
$- - \to P A + - - \to P B + - - \to P C = 0$
$\to a + \to b + \to c = 0$

As centroid is given by:
$\frac{- - \to P A + - - \to P B + - - \to P C}{3}$

$\Rightarrow$ Centroid is the origin.
i.e. P is the Centroid

So the match for Q) is 1)Centroid

Case (R) - Refer diagram under R) in attached image

We know that Incentre is given by
$\to I = \frac{B C (- - \to P A) + C A (- - \to P B) + A B (- - \to P C)}{A B + B C + C A}$

and $B C (- - \to P A) + C A (- - \to P B) + A B (- - \to P C) = 0$ (Given)

$\Rightarrow \to I = 0$ (I is the origin)

i.e. P is the Incentre.
So the match for R) is 3)Incentre

Case (S) - Refer diagram under S) in attached image

If $\to P$ is the orthocentre:

$- - \to P A ⊥ - - \to B C$

$- - \to P B ⊥ - - \to A C$

$- - \to P C ⊥ - - \to A B$

Given
$- - \to P A . - - \to C B = 0$

$\Rightarrow - - \to P A ⊥ - - \to C B$

and

$- - \to P B . - - \to A C = 0$

$\Rightarrow - - \to P B ⊥ - - \to A C$

i.e. P is the Orthocentre.
So the match for S) is 2)Orthocentre

Hence the correct answer is Option C (P-4,Q-1,R-3,S-2)

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P) P is a point in the plane of the triangle ABC. Pv's of A,B and C are $\to a, \to b a n d \to c$ respectively with respect to P as the origin. If $(\to b + \to c) . (\to b - \to c) = 0 a n d (\to c + \to a) . (\to c - \to a) = 0$ then w.r.t. the Triangle ABC, P is its	1)Centroid
Q) If $\to a, \to b, \to c a r e$ the position vectors of the three non collinear points A,B and C repectively such that the vector $\to V = - - \to P A + - - \to P B + - - \to P C$ is a null vector then w.r.t. the $Δ A B C$ , P is its	2)orthocentre
R)if P is a point inside the $Δ A B C$ such that the vector $\to R = (B C) (- - \to P A) + C A (- - \to P B) + (A B) (- - \to P C)$ is null vector then w.r.t. the $Δ A B C$ , P is its	3)Incentre
S) If P is a point in the plane of the triangle ABC such that the scalar product $- - \to P A . - - \to C B a n d - - \to P B . - - \to A C$ vanishes, then w.r.t. the $Δ A B C$ . P is its	4)Circumcentre