Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

If $\vec{a}, \vec{b}$ and $\vec{c}$ are three non-coplanar vectors, then the length of projection of vector $\vec{a}$ in the plane of the vectors $\vec{b}$ and $\vec{c}$ may be given by

Options:

$\frac{|\vec{a}. (\vec{b}×\vec{c})|}{|\vec{b}×\vec{c}|}$

$\frac{|\vec{a}×(\vec{b}×\vec{c})|}{|\vec{b}×\vec{c}|}$

$\frac{\vec{a} \vec{b}\vec{c}}{(\vec{b}.\vec{c})}$

none of these

Correct Answer:

$\frac{|\vec{a}×(\vec{b}×\vec{c})|}{|\vec{b}×\vec{c}|}$

Explanation:

A vector normal to the plane containing $\vec{b} $ and $\vec{c}$ is $\vec{n}= \vec{b}× \vec{c}$

We have,

BN = Projection of $\vec{a}$ on $\vec{n}= \vec{a}. \hat{n}$

∴ Projection of $\vec{a}$ on the plane containing $\vec{b}$ and $\vec{c}$

$= LM = AN = \sqrt{AB^2 -BN^2}$

$= \sqrt{|\vec{a}|^2-(\vec{a}. \hat{n})^2}= \sqrt{|\vec{a}|^2 |\vec{n}|^2-(\vec{a}. \hat{n})^2}$

$= \sqrt{|\vec{a}. \hat{n}|^2}= |\vec{a}. \hat{n}|= \frac{\vec{a}× (\vec{b}×\vec{c})}{|\vec{b}×\vec{c}|}$