If a- b- c- are non-coplanar vectors- then show that the four points 2a-b- a-2b-c- 4a-2b-c- and 3a-4b-5c- are coplanar-

[ A( 2a⃗ +b⃗ #160; ) B( a⃗ +2b+2c⃗ #160; ) C( 4a⃗ 2b c⃗ #160; ) D( 3a⃗ +4b⃗ 5c⃗ #160; ) ...

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Condition for Coplanarity of Four Points

If a⃗, b⃗, ...

Question

If $\to a, \to b, \to c$ are non-coplanar vectors, then show that the four points $2 \to a + \to b, \to a + 2 \to b + \to c, 4 \to a - 2 \to b - \to c$ and $3 \to a + 4 \to b - 5 \to c$ are coplanar.

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2 \to a + 3 \to b - \to c, \to a - 2 \to b + 3 \to c, 3 \to a + 4 \to b - 2 \to c a n d \to a - 6 \to b + 6 \to c

\to a, \to b a n d \to c

\to a, \to b, \to c

\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]}

(2 \to a + 3 \to b + 4 \to c) . \to p + (2 \to b + 3 \to c + 4 \to a) . \to q +

(2 \to c + 3 \to a + 4 \to b) . \to r =

\to a, \to b, \to c

_{1} = \to a - \to b + \to c,_{2} = \to b + \to c - \to a

_{3} = \to c + \to a - \to b, \to r = 2 \to a - 2 \to b + 4 \to c

\to r = x_{1}_{1} + x_{2}_{2} + x_{3}_{3}

\to a, \to b, \to c

[\to a \to b \to c] = \frac{4}{7}

[2 \to a - \to b, 2 \to b - \to c, 2 \to c - \to a]

\to a, \to b, \to c

\to p = \to a - 2 \to b + 3 \to c

\to q = 2 \to a - 3 \to b + 4 \to c

\to r = 3 \to a - 4 \to b + 5 \to c

\to s = 7 \to a - 11 \to b + 15 \to c .

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