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Condition for Coplanarity of Four Points

If a⃗, b⃗, ...

Question

If $\to a, \to b, \to c$ be any three non-zero, non-coplanar vectors, then find the linear relation between the following four vectors $\to p = \to a - 2 \to b + 3 \to c$ , $\to q = 2 \to a - 3 \to b + 4 \to c$ , $\to r = 3 \to a - 4 \to b + 5 \to c$ , $\to s = 7 \to a - 11 \to b + 15 \to c .$

A

\to s = \to p + 2 \to q + \to z

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B

\to s = 2 \to p + \to q + \to z

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C

\to s = 2 \to p + 3 \to q - \to z

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D

\to s = \to p + 3 \to q

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Solution

The correct option is D $\to s = \to p + 3 \to q$
Let $s = x p + y q + z r$

$\Rightarrow 7 a - 11 b + 15 c = x (a - 2 b + 3 c) + y (2 a - 3 b + 4 c) + z (3 a - 4 b + 5 c)$

$\Rightarrow 7 a - 11 b + 15 c = x a - 2 x b + 3 x c + 2 a y - 3 b y + 4 c y + 3 a z - 4 b z + 5 c z$

$\Rightarrow 7 a - 11 b + 15 c = a (x + 2 y + 3 z) + b (- 2 x - 3 y - 4 z) + c (3 x + 4 y + 5 z)$

Comparing the coefficients of $a, b$ and $c$ on both sides we get,

$x + 2 y + 3 z = 7 \dots (1), - 2 x - 3 y - 4 z = - 11 \dots (2), 3 x + 4 y + 5 z = 15 \dots (3)$

Solving these equation we get, $∴ x = 1, y = 3, z = 0.$
Hence, $s = p + 3 q$

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