Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Vectors

Question:

If $\vec b$ and $\vec c$ are any two non-collinear vectors, and $\vec a$ is any vector, then $(\vec a. \vec b) \vec b + (\vec a.\vec c) \vec c +\frac{\vec a(\vec b×\vec c)}{|\vec b×\vec c|^2}(\vec b×\vec c)=$

Options:

$\vec a$

$\vec b$

$\vec c$

none of these

Correct Answer:

$\vec a$

Explanation:

If $α,β,γ$ are three non-coplanar vectors, then any vector $\vec r$ can be written as

$\vec r=\frac{(\vec r.\vec α)}{|\vec α|^2}\vec α+\frac{(\vec r.\vec β)}{|\vec β|^2}\vec β+\frac{(\vec r.\vec γ)}{|\vec γ|^2}\vec γ$

Since $\vec b$ and $\vec c$ are non-collinear vectors. Therefore, $\vec b, \vec c$ and $\vec b×\vec c$ are three non-coplanar vectors.

$∴\vec a=\frac{(\vec a.\vec b)}{|\vec b|^2}\vec b+\frac{(\vec a.\vec c)}{|\vec c|^2}\vec c+\frac{\vec a.(\vec b×\vec c)}{|\vec b×\vec c|^2}(\vec b×\vec c)$

$⇒\vec a=(\vec a.\vec b)\vec b+(\vec a.\vec c)\vec c+\frac{\vec a(\vec b×\vec c)}{|\vec b×\vec c|^2}(\vec b×\vec c)$