Question

If \(ABC\) is a right-angled triangle in which \(BC\) is the longest side and the position vector of \(B\) and \(C\) are respectively \(3\hat{i}-2\hat{j}+\hat{k}\) and \(5\hat{i}+\hat{j}-3\hat{k}\), then the value of \(\overline{AB}\cdot \overline{AC}+\overline{BA}\cdot \overline{BC}+\overline{CA}\cdot \overline{CB}\) is

• A. \(25\)
• B. \(27\)
• C. \(29\)
• D. \(31\)

Hint:

Since BC is the longest side, angle A is 90 degrees; the sum reduces to BC squared.

Answer 1

The Correct Option is C

Step 1: Use Squared Lengths:
Let $AB=u$ and $AC=w$ as vectors from A. The right angle gives $u\cdot w=0$.

Step 2: Expand Each Term:
$BA\cdot BC=(-u)\cdot(w-u)=u\cdot u-u\cdot w=|u|^2$. $CA\cdot CB=(-w)\cdot(u-w)=|w|^2-u\cdot w=|w|^2$.

Step 3: Total:
Sum $=0+|u|^2+|w|^2=|w-u|^2=|BC|^2=2^2+3^2+4^2=29$. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } 29} \]