In a ABC- P be an interior point such that PA - 2PB - 3PC - 0- The ratio of the area of ABC to that of APC is

PA + 2PB + 3PC = 0 Let a=PA, b=PB, c=PC ⇒a+2b+3c=0 Taking cross product of above equation with a, we get a×( a+2b+3c)=0 ⇒ 2a×b=3c×a ⇒a×b=32(c×a) ...(1) ...

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Byju's Answer

Standard XI

Mathematics

Applications of Cross Product

In a ABC, P b...

Question

In a $△ A B C,$ $P$ be an interior point such that $- - \to P A + 2 - - \to P B + 3 - - \to P C = 0.$ The ratio of the area of $△ A B C$ to that of $△ A P C$ is

3 : 1

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3 : 2

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5 : 2

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2 : 1

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Solution

The correct option is A $3 : 1$
$- - \to P A + 2 - - \to P B + 3 - - \to P C = 0$

Let $\to a = - - \to P A, \to b = - - \to P B, \to c = - - \to P C$
$\Rightarrow \to a + 2 \to b + 3 \to c = 0$

Taking cross product of above equation with $\to a$ , we get
$\to a \times (\to a + 2 \to b + 3 \to c) = 0$
$\Rightarrow 2 \to a \times \to b = 3 \to c \times \to a$
$\Rightarrow \to a \times \to b = \frac{3}{2} (\to c \times \to a) . . . (1)$

Similarly, $\to c \times (\to a + 2 \to b + 3 \to c) = 0$
$\Rightarrow 2 \to b \times \to c = \to c \times \to a$
$\Rightarrow \to b \times \to c = \frac{1}{2} (\to c \times \to a) . . . (2)$

Now, area of $△ A P C = \frac{1}{2} ∣ ∣ \to c \times \to a ∣ ∣$

$Area (△ A B C) = Area (△ A P B) + Area (△ B P C) + Area (△ A P C)$
$\Rightarrow Area (△ A B C) = \frac{1}{2} ∣ ∣ ∣ \to a \times \to b + \to b \times \to c + \to c \times \to a ∣ ∣ ∣$
From equation $(1)$ and $(2),$ we have
$Area (△ A B C) = \frac{3}{2} ∣ ∣ \to c \times \to a ∣ ∣$

$∴ \frac{Area (△ A B C)}{Area (△ A P C)} = \frac{\frac{3}{2} ∣ ∣ \to c \times \to a ∣ ∣}{\frac{1}{2} ∣ ∣ \to c \times \to a ∣ ∣} = 3$

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In a △ABC, P be an interior point such that −−→PA+2−−→PB+3−−→PC=0. The ratio of the area of △ABC to that of △APC is

In a $△ A B C,$ $P$ be an interior point such that $- - \to P A + 2 - - \to P B + 3 - - \to P C = 0.$ The ratio of the area of $△ A B C$ to that of $△ A P C$ is