If $\overrightarrow{A B}=2 \hat{i}+3 \hat{j}-6 \hat{k} ; \overrightarrow{B C}=6 \hat{i}-2 \hat{j}+3 \hat{k}$…

If $\overrightarrow{A B}=2 \hat{i}+3 \hat{j}-6 \hat{k} ; \overrightarrow{B C}=6 \hat{i}-2 \hat{j}+3 \hat{k}$ are the vectors along two sides of a triangle ABC , then perimeter of triangle ABC is:
  1. 21
  2. $\sqrt{74}+14$
  3. $\sqrt{74}+19$
  4. $\sqrt{74}+3$

Solution

$\overrightarrow{A B}=2 \hat{i}+3 \hat{j}-6 \hat{k} ; \overrightarrow{B C}=6 \hat{i}-2 \hat{j}+3 \hat{k}$
$\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{B C}=8 \hat{i}+\hat{j}-3 \hat{k}$
Perimeter of $\triangle A B C$ $=|\overrightarrow{A B}|+|\overrightarrow{B C}|+|\overrightarrow{A C}|=14+\sqrt{74}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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