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Fundamental Theorem in 2D

Any vector r⃗...

Question

Any vector $\to r$ in the plane of two non-zero, non collinear vectors $\to a$ and $\to b$ can be expressed uniquely as a linear combination $x \to a + y \to b$ of $\to a a n d \to b,$ where $x \to a a n d y \to b$ are the components of vector $\to r$

A

True

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B

False

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Solution

The correct option is A True
Isn't the statement conveying the same what fundamental theorem in 2D conveys?
Anyways let's prove the statement.
Let $\to a a n d \to b$ be any two non-zero, non collinear vectors and $\to r$ be any vector coplanar with $\to a a n d \to b$ . Take any point O in the plane of $\to a a n d \to b$ .

Let $- - \to O A = \to a, - - \to O B = \to b a n d - - \to O P = \to r$
Clearly OA, OB and OP are coplanar. Through P, draw two lines PM and PN perpendicular to OA and OB respectively meeting OA and OB at M and N respectively. Refer to the figure shown
We have $- - \to O P = - - \to O M + - - \to M P$ (Triangular law for vector addition)
= $- - \to O M + - - \to O N$ [Since MP = ON and MP $|$ $|$ ON] $\dots$ (1)
Now $- - \to O M$ and $- - \to O A$ are collinear vectors,
$- - \to O M = x - - \to O A = x \to a,$ where x is a scalar
$- - \to O N = y - - \to O B = y \to b,$ where y is a scalar
Hence from (1)
$- - \to O P = x \to a + y \to b$ or $\to r = x \to a + y \to b$
Now this is only one part of the statement. We have to prove that this is the unique vector $\to r$ i.e., uniqueness has to be established as well
If possible let
$\to r = x \to a + y \to b$ and $\to r = x^{'} a + y^{'} b$ be two different ways of representing $\to r$
Then we have $- \to x a + \to y b = - \to x^{'} a + - - \to y^{'} b^{'}$
or (x - x') $\to a$ +(y-y') $\to b = \to O$
But $\to a$ and $\to b$ are not collinear vectors
So x - x' = 0 $\Rightarrow$ x = x $\dots$ (2)
and y - y' = 0 $\Rightarrow$ y = y $\dots$ (3)
(For two non collinear vectors $\to a a n d \to b z \to a + y \to b = \to o$ is possible only when z = y = 0)
So uniqueness is established. This means that there is only one unique way of representing a vector $\to r$ in terms of 2 other coplanar vectors $\to a a n d \to b$ .

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Similar questions

\to r

in the plane of two non-zero, non collinear vectors

\to a

\to b

can be expressed uniquely as a linear combination

x \to a + y \to b

\to a a n d \to b,

x \to a a n d y \to b

are the components of vector

\to r

\to a, \to b, \to c

are three non-coplanar non-zero vectors and

\to r

is any vector, then

(\to a \times \to b) \times (\to r \times \to c) + (\to b \times \to c) \times (\to r \times \to a) + (\to c \times \to a) \times (\to r \times \to b) =

Q. Assertion :A vector

\to R

can be expressed as a linear combination of a vector

\to A

and another vector perpendicular to

\to A

and coplanar with

\to R

\to A

\to R = (\frac{\to R . \to A}{\to A . \to A}) \to A - (\frac{1}{\to A . \to A}) (\to A \times (\to A \times \to R))

\to a, \to b, \to c

are coplanar then

\to a + \to b, \to b + \to c, \to c + \to a

are also coplanar.

\to a, \to b, \to c

are three non- coplanar vectors for which

[\to a \to b \to c] \neq

{\to a}^{'}, {\to b}^{'}, {\to c}^{'}

constitute the reciprocal system of vectors, then any vector

\to r

can be expressed as

\to a, \to b, \to c

be three non-coplanar vectors and

\to r

be any arbitrary vector, then

(\to a \times \to b) \times (\to r \times \to c) + (\to b \times \to c) \times (\to r \times \to a) + (\to c \times \to a) \times (\to r \times \to b)

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