Question

In a \( \triangle A B C, \quad \overrightarrow{A B}=6 \hat{i}+3 \hat{j}+3 \hat{k} \quad ; \quad \overrightarrow{A C}=3 \hat{i}-3 \hat{j}+6 \hat{k} \) \( D \) and \( D^{\prime} \) are points trisections of side \( B C \) Find \( \overline{ AD } \) and \( \overrightarrow{ AD } \).

Answer 1

https://www.sarthaks.com/?qa=blob&qa_blobid=13998160689201843963

Answer 2

We have, \(\overrightarrow{BC} = \overrightarrow{AC}- \overrightarrow{AB}\) (from triangle law)

or \(\overrightarrow{BC} =(3\hat i - 3\hat j + 6\hat k) - (6\hat i + 3\hat j + 3\hat k)\)

\(= -3\hat i + 6\hat j + 3\hat k\)

Now, according to the question,

\(\overrightarrow{BD} = \overrightarrow {D'C} = \frac 1 3 \overrightarrow {BC} = (-\hat i - 2\hat j + \hat k)\)

\(\therefore \overrightarrow{AD} = \overrightarrow{BD} - \overrightarrow{BA} = \overrightarrow{BD} + \overrightarrow{AB} = (5\hat i + \hat j + 4\hat k)\)

And \( \overrightarrow{AD'} = \overrightarrow{AC} - \overrightarrow{D'C} = (4\hat i - \hat j + 5\hat k)\)