Question

$ABC$ is a right-angled triangle in which $\max \{AB,BC,AC\}$ $=BC$. 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
$\overrightarrow{AB}\cdot \overrightarrow{AC}+\overrightarrow{BA}\cdot \overrightarrow{BC}+\overrightarrow{CA}\cdot \overrightarrow{CB}=$

• A. 28
• B. 29
• C. 27
• D. 25

Answer 1

Answer: (B)

Given, $B=3\hat{i}-2\hat{j}+\hat{k}$
$C=5\hat{i}+\hat{j}-3\hat{k}$
$BC=2\hat{i}+3\hat{j}-4\hat{k}$
$\max \{AB,BC,AC\}=BC$
$\therefore BC$ is hypotenuse of $△ABC$

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