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Test for Coplanarity

Given: a⃗,b⃗ ...

Question

Given: $\to a, \to b a n d \to c$ are coplanar.
Vectors $\to a - 2 \to b + 3 c, - - \to 2 a + \to 3 b - \to 4 c a n d - \to b + \to 2 c$ are non-coplanar vectors.

A

True

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B

False

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Solution

The correct option is B False
We will check if the given 3 vectors are linearly dependent or not. If yes, then they are coplanar. Else not.
So
Let $α = \to a - 2 \to b + 3 c$
$β = 2 \to a + 3 \to b - 4 \to c$
$γ = - \to b + 2 \to c$
So if $α, β, a n d γ$ are linearly dependent then,
for $α = x \to β + y \to γ$
There should be values of x and y which are not simultaneously zero.
So let’s find out.
$\to a - 2 \to b + 3 \to c = x (- 2 \to a + 3 \to b - 4 \to c)$
$+ y (- \to b + 2 \to c)$
Equating the coefficients of $\to a, \to b a n d \to c$ . We get
-2x + 0.y = 1 (equating coefficient of $\to a$ )
3x – y = -2 (equating coefficient of $\to b$ )
-4x + 2y = 3 (equating coefficient of $\to c$ )
Solving the system of equations, we get $x = - \frac{1}{2} y = \frac{1}{2}$
These values satisfy all 3 equations and thus the system of equations is consistent and has a non-zero solution.
So $α, β a n d γ$ are linearly dependent and hence coplanar.

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