If the vertices A,B,C of a triangle ABC are (1, 1, 3), (-1, 0, 0),(0, 1, 2) respectively, then determine ∠ABC. (∠ABC is the angle between the vectors \(\overrightarrow {BA} \) and \(\overrightarrow {BC} \))

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Question

This question was previously asked in

AAI ATC Junior Executive 25 March 2021 Official Paper (Shift 1)

Answer 1

Given:

The vertices A,B,C of a triangle ABC are (1, 1, 3), (-1, 0, 0), (0, 1, 2) respectively.

Concept:

\(\vec{PQ}\) = p.v.(\(\vec{Q}\)) - p.v.(\(\vec{P}\))

Formula:

The angle between two vectors \(\vec{P}\) and \(\vec{Q}\) is given by :

\(\theta = cos^{-1}(\frac{\vec{P}.\vec{Q}}{|\vec{P}||\vec{Q}|})\)

Solution:

\(\overrightarrow {BA} \) = (1, 1, 3) - (-1, 0, 0)

= 2î + ĵ + 3k̂

\(\overrightarrow {BC} \) = (0, 1, 2) - (-1, 0, 0) = î + ĵ + 2k̂

∴ \(\theta = cos^{-1}(\frac{\vec{BA}.\vec{BC}}{BA.BC})\)

\(\Rightarrow \theta = cos^{-1}(\frac{9}{\sqrt 14 \sqrt 6})\))

\(\Rightarrow \theta ={{\mathop{\rm cos}\nolimits} ^{ - 1}}(\frac{9}{{\sqrt 84}})\)