Let P be an interior point of - ABC such that 4 PA -3 PB -5 PC -0- If Area - ABC - k - Area - APC - then k is

Name: JEE- Grade 12- Mathematics- Vector Algebra- Session 11- W08
Uploaded: 2022-07-04T10:36:42+05:30
Description: Let a=PA, b=PB, c=PC 4a+3b+5c=0 ⋯(1) Area(Δ APC) nbsp;= 12|c×a| Area(Δ ABC)=Area(Δ APB)+Area(Δ BPC)+Area(Δ APC) ⇒Area(Δ ABC)=12|a×b+b×c+c×a| Taking ...

Let a=PA, b=PB, c=PC 4a+3b+5c=0 ⋯(1) Area(Δ APC) nbsp;= 12|c×a| Area(Δ ABC)=Area(Δ APB)+Area(Δ BPC)+Area(Δ APC) ⇒Area(Δ ABC)=12|a×b+b×c+c×a| Taking ...

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Standard XII

Mathematics

Cross Product of Two Vectors

Let P be an i...

Question

Let $P$ be an interior point of $Δ A B C$ such that $4 - - \to P A + 3 - - \to P B + 5 - - \to P C = 0$ . If $Area (Δ A B C) = k \times Area (Δ A P C)$ , then $k$ is

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Solution

Let $\to a = - - \to P A, \to b = - - \to P B, \to c = - - \to P C$
$4 \to a + 3 \to b + 5 \to c = 0 \dots (1)$

$Area (Δ A P C) = \frac{1}{2} ∣ ∣ \to c \times \to a ∣ ∣$
$Area (Δ A B C) = Area (Δ A P B) + Area (Δ B P C) + Area (Δ A P C)$
$\Rightarrow Area (Δ A B C) = \frac{1}{2} ∣ ∣ ∣ \to a \times \to b + \to b \times \to c + \to c \times \to a ∣ ∣ ∣$

Taking cross product of equation $(1)$ with $\to a$ and $\to c$ one at a time, we get
$3 \to a \times \to b = 5 \to c \times \to a$
and $3 \to b \times \to c = 4 \to c \times \to a$
$\Rightarrow \to a \times \to b = \frac{5}{3} (\to c \times \to a)$
and $\to b \times \to c = \frac{4}{3} (\to c \times \to a)$

$∴ Area (Δ A B C) = \frac{4}{2} ∣ ∣ \to c \times \to a ∣ ∣ = 4 \times Area (Δ A P C)$
Hence, $k = 4$

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Cross Product of Two Vectors

Standard XII Mathematics

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Let P be an interior point of ΔABC such that 4−−→PA+3−−→PB+5−−→PC=0. If Area(ΔABC)=k×Area(ΔAPC), then k is

Let $P$ be an interior point of $Δ A B C$ such that $4 - - \to P A + 3 - - \to P B + 5 - - \to P C = 0$ . If $Area (Δ A B C) = k \times Area (Δ A P C)$ , then $k$ is