First slide

Scalar triple product of vectors

Question

$Let G_{1}, G_{2}, G_{3} be the centroids of the triangular faces OBC, OCA, OAB$ $of a tetrahedron OABC . If V_{1} denote the volume of the tetrahedron OABC and V_{2} that of$ $the parallelopiped with O G_{1}, O G_{2}, O G_{3} as three$ $concurrent edges, then :$

Moderate

A

$4 V_{1} = 9 V_{2}$
B

$9 V_{1} = 4 V_{2}$
C

$3 V_{1} = 2 V_{2}$
D

$3 V_{2} = 2 V_{1}$

Solution

$Taking O as the origin,let the position vectors of A, B and$ $C be \vec{a}, \vec{b} and \vec{c} respectively. Then the position vectors$ of
$G_{1}, G_{2} and G_{3} are \frac{\vec{b} + \vec{c}}{3}, \frac{\vec{c} + \vec{a}}{3} and \frac{\vec{a} + \vec{b}}{3} respectively.$
$∴ V_{1} = \frac{1}{6} [\vec{a} \vec{b} \vec{c}]$
$and V_{2} = [O \vec{G_{1}} O \vec{G_{2}} O \vec{G_{3}}]$
$Now, V 2 = [O \vec{G_{1}} O \vec{G_{2}} O \vec{G_{3}}]$
$\Rightarrow V 2 = [\frac{\vec{b} + \vec{c}}{3} \frac{\vec{c} + \vec{a}}{3} \frac{\vec{a} + \vec{b}}{3}]$
$\Rightarrow V 2 = \frac{1}{27} [\vec{b} + \vec{c} \vec{c} + \vec{a} \vec{a} + \vec{b}]$
$\Rightarrow V 2 = \frac{2}{27} [\vec{a} \vec{b} \vec{c}]$
$\Rightarrow V 2 = \frac{2}{27} × 6 V_{1} \Rightarrow 9 V_{2} = 4 V_{1}$
$Hence, (a) is the correct answer.$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App