The position vectors of vertices A- B- C of - ABC are a-b-c- respectively and -c-a-3 -b-a- If CD is drawn perpendicular to the bisector of - A- then which of the following is correct-

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Double Ordinate

The position ...

Question

The position vectors of vertices $A, B, C$ of $Δ A B C$ are $\to a, \to b, \to c$ respectively and $| \to c - \to a | = 3 | \to b - \to a |$ . If $C D$ is drawn perpendicular to the bisector of $∠ A$ , then which of the following is correct?

Line

B C

is perpendicular bisector of

A D

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B C

bisects

A D

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Δ P A B

and

Δ P C D

are congruent

(

Here '

p

' is the point of intersection of bisector with side

B C)

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None of these

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Solution

The correct option is B $B C$ bisects $A D$ .
To solve this question in a quick, efficient and clean way using the least of calculation, we need to be aware of the following 2 theorems:
$(1)$ If the origin in a 3D plane is not explicitly defined, the solver can choose a origin of his/her convenience.

$(2)$ The angle bisector of an angle of a triangle cuts the opposite side in the same ratio as the ratio of sides containing the angle.

Using theorem $(1)$ , we take the point $A$ as the origin itself, essentially reducing $\to a$ to $\to 0$ which will give us,
$\to c = 3 \to b$ $\to (1)$
Also, take dot product of both sides with $\to c$
$⟹ 3 \to b . \to c = ∣ ∣ c^{2} ∣ ∣ \to (2)$
Using theorem $(2)$ , in $Δ A B C$ ,
$A B : B C = 1 : 3$
$⟹$ $B Q : C Q = 1 : 3$
Using section formula in line $B C$ , we can find the $\to A Q$ = $\frac{3 \to b + \to c}{4}$
now let the ratio of $A D$ at $Q$ be $λ : 1$
We can find $\to A Q$ = $\frac{λ \to A D}{λ + 1}$
Equating both values to find $\to A D$ = $\frac{(λ + 1) (3 \to b + \to c)}{4 λ}$

Since $\to A D ⊥ \to C D$ ,
$⟹ \to A D . \to C D = \to 0$
Solving,
${\to | A D |}^{2} = ∣ ∣ \to A D . \to A C ∣ ∣$
Use expressions, $(1) a n d (2)$ to solve for $λ$ ,
$λ = 1$
Hence, $B C$ bisects $A D$

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Similar questions

Q. If

\to a, \to b, \to c

are positive vectors of vertices

A, B, C

Δ A B C

. If

\to r

is position vector of a point

P

such that

(| \to b - \to c | + | \to c - \to a | + | \to a - \to b |) \to r = | \to b - \to c | \to a + | \to c - \to a | \to b + ∣ ∣ \to a - \to b ∣ ∣ \to c

then the point

P

always

In a parallelogram OABC, vectors $\to a, \to b, \to c$ are respectively the position vectors of vertices A, B, C with reference to O is origin. A point $\to E$ is taken on the side BC which divides it in the ratio of 2 : 1. Also, the line segment AE intersects the line bisecting the angel O internally in point P. If CP, when extended meets AB in point F. Then