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Composite Function

Assertion :A ...

Question

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 $\to R$ and $\to A$ as,
$\to R = (\frac{\to R . \to A}{\to A . \to A}) \to A - (\frac{1}{\to A . \to A}) (\to A \times (\to A \times \to R))$ Reason: If $\to a, \to b, \to c$ are coplanar then $\to a + \to b, \to b + \to c, \to c + \to a$ are also coplanar.

A

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

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B

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

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C

Assertion is correct but Reason is incorrect

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D

Assertion is incorrect but Reason is correct

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Solution

The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
The vector $A \times (A \times R)$ is coplaner with $R$ and $A$ is perpendicular to $A$ .
now any vector $R$ can be expressed as linear combination of a vector $A$
and another vector perpendicular to $A$ and coplaner with $R$ and $A$ .
So that $R = x A + y A \times (A \times R)$ ...(1)
Multiplying both sides of (1) scalorly with $A$ , we get
$R . A = x (A . A) + y [A \times (A \times R)] . A$
$= x (A . A) + y [A, A \times R, A]$ $= x (A . A)$ ,
Since $[A, A + R, A] = 0$ $∴ x = \frac{R . A}{A . A}$
Again Multiplying both sides of (1) vertically with $A$ , we get
$R \times A = x (A \times A) + y [A \times (A \times R)] \times A$ $= y [(A . R) A - (A . A) R] \times A$
Since $A \times A = 0$ $= y (A . R) A \times A - y (A . A) R \times A = - y (A . A) R \times A$
$∴ y (A . A) = - 1$ or $y = - \frac{1}{(A . A)}$
Putting the values of $x$ and $y$ in (1), we get
$R = \frac{R . A}{A . A} A - \frac{1}{A . A} A \times (A \times R)$ $= \frac{R . A}{A . A} + \frac{1}{A . A} (A \times R) \times A$

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Similar questions

\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) =

\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)

\to a, \to b, \to c

are three noncoplanar nonzero vectors and

\to r

is any vector in space 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)

\to a, \to b, \to c

are non-coplanar vectors and

\to α = \to b \times \to c, \to β = \to c \times \to a, \to r = \to a \times \to b

∣ ∣ ∣ ∣ ∣ \begin{matrix} \to α . \to a & \to α . \to b & \to α . \to c \to β . \to a & \to β . \to b & \to β . \to c \to r . \to a & \to r . \to b & \to r . \to c \end{matrix} ∣ ∣ ∣ ∣ ∣ =

Q. Assertion :If

\to r \times \to a = \to b, \to r . \to a = 1,

\to r = \to a + \to a \times \to b .

Reason: The solution of the

\to r \times \to a = \to b

\to r = \to a \times \to b

.

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