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Test for Coplanarity

If a,b,c are ...

Question

If $\to a, \to b, \to c$ are three non coplanar vectors and $\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]},$ then $(2 \to a + 3 \to b + 4 \to c) \cdot \to p + (2 \to b + 3 \to c + 4 \to a) \cdot \to q + (2 \to c + 3 \to a + 4 \to b) \cdot \to r =$

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Solution

Given : $\to p = \frac{\to b \times \to c}{[\to a \to b \to c]}, \to q = \frac{\to c \times \to a}{[\to a \to b \to c]}, \to r = \frac{\to a \times \to b}{[\to a \to b \to c]},$
$\Rightarrow \to a \cdot \to p = \frac{\to a \cdot (\to b \times \to c)}{[\to a \to b \to c]} = 1, \to b \cdot \to p = \frac{\to b \cdot (\to b \times \to c)}{[\to a \to b \to c]} = 0, \to c \cdot \to p = \frac{\to c \cdot (\to b \times \to c)}{[\to a \to b \to c]} = 0$
( $∵$ scalar triple product is zero if two vectors are same.)
similarly,
$\to a \cdot \to q = \to c \cdot \to q = 0, \to b \cdot \to q = 1$ and
$\to a \cdot \to r = \to b \cdot \to r = 0, \to c \cdot \to r = 1$
Now,
$(2 \to a + 3 \to b + 4 \to c) \cdot \to p + (2 \to b + 3 \to c + 4 \to a) \cdot \to q + (2 \to c + 3 \to a + 4 \to b) \cdot \to r = (2 + 0 + 0) + (2 + 0 + 0) + (2 + 0 + 0) = 6$

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Test for Coplanarity

Standard XII Mathematics

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