Question

$ABC$ is a right-angled triangle in which $\max \{A B, B C, A C\}$ $=B C$. If the position vectors of $B$ and $C$ are respectively $3 \hat{i}-2 \hat{j}+\hat{k}$ and $5 \hat{i}+\hat{j}-3 \hat{k}$ then
$\overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{B A} \cdot \overrightarrow{B C}+\overrightarrow{C A} \cdot \overrightarrow{C B}=$

• A. 28
• B. 29
• C. 27
• D. 25

Answer 1

Answer: (B)

Given, $B=3 \hat{i}-2 \hat{j}+\hat{k}$
$C=5 \hat{i}+\hat{j}-3 \hat{k}$
$B C=2 \hat{i}+3 \hat{j}-4 \hat{k}$
$\max \{A B, B C, A C\}=B C$
$\therefore B C$ is hypotenuse of $\triangle A B C$

$\angle A=90^{\circ}$
$\therefore A B \cdot A C=0$
$B A \cdot B C=|B A| B C \mid \cos B$
$C A \cdot C B=|C A| C B \mid \cos C$
$\therefore A B-A C+B A-B C+C A-C B$
$=0+|B C|(|B A| \cos B+|C B| \cos C)$
$=0+|B C| B C \mid[\because$ By projection formula]
$=|B C|^{2}-\left(\sqrt{(2)^{2}+3^{2}+4^{2}}\right)^{2}$
$=4+9 +16=29$