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: Locate the Right Angle:
In a right-angled triangle the hypotenuse is the longest side. Since \(BC\) is the longest, the right angle is at \(A\). So \(\overline{AB}\cdot\overline{AC}=0\).

Step 2: Reduce the Other Terms:
\(\overline{BA}\cdot\overline{BC}=\overline{BA}\cdot(\overline{BA}+\overline{AC})=|BA|^2+0=|AB|^2\).
\(\overline{CA}\cdot\overline{CB}=\overline{CA}\cdot(\overline{CA}+\overline{AB})=|CA|^2+0=|AC|^2\).

Step 3: Add:
The sum is \(0+|AB|^2+|AC|^2=|BC|^2\) by Pythagoras.

Step 4: Compute BC:
\(\overline{BC}=(5-3)\hat i+(1+2)\hat j+(-3-1)\hat k=2\hat i+3\hat j-4\hat k\).
\[ |BC|^2=4+9+16=29 \]
So the value is 29. Options 25, 27 and 31 do not equal this sum of squares.

Final Answer:
The value is 29, option (C). \[ \boxed{\text{(C) } 29} \]