Question

Let \( \vec{a}, \vec{b} \) and \( \vec{c} \) be the sides of a triangle \( ABC \) such that \( \overrightarrow{BC}=\vec{a}, \overrightarrow{CA}=\vec{b} \) and \( \overrightarrow{AB}=\vec{c} \). If \( BC=AC=3 \) and \( \vec{b}\cdot\vec{c}=-9 \), then \( \vec{a}\cdot\vec{b} \) is equal to

• A. \( 27 \)
• B. \( 9 \)
• C. \( 3\sqrt{3} \)
• D. \( 0 \)
• E. \( -3\sqrt{3} \)

Hint:

For vectors representing the sides of a triangle taken in order, always use \( \vec{a}+\vec{b}+\vec{c}=0 \). It is the key identity in most triangle vector problems.

Answer 1

The Correct Option is D

Step 1: Use the triangle side relation.
Since \[ \overrightarrow{BC}=\vec{a}, \qquad \overrightarrow{CA}=\vec{b}, \qquad \overrightarrow{AB}=\vec{c} \] and these are the directed sides of triangle \( ABC \), we have the standard vector relation \[ \vec{a}+\vec{b}+\vec{c}=0 \] because \[ \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=0 \] Equivalently, \[ \vec{c}=-(\vec{a}+\vec{b}) \]

Step 2: Interpret the given lengths.
We are given \[ BC=AC=3 \] Now, \[ |\vec{a}|=|\overrightarrow{BC}|=BC=3 \] and \[ |\vec{b}|=|\overrightarrow{CA}|=CA=AC=3 \] Thus, \[ |\vec{a}|=|\vec{b}|=3 \]

Step 3: Use the given dot product \( \vec{b}\cdot\vec{c}=-9 \).
From \[ \vec{c}=-(\vec{a}+\vec{b}) \] we get \[ \vec{b}\cdot\vec{c} = \vec{b}\cdot\bigl(-(\vec{a}+\vec{b})\bigr) \] \[ = -(\vec{b}\cdot\vec{a}+\vec{b}\cdot\vec{b}) \] Since dot product is commutative, \[ \vec{b}\cdot\vec{a}=\vec{a}\cdot\vec{b} \] and \[ \vec{b}\cdot\vec{b}=|\vec{b}|^2=3^2=9 \] Hence, \[ \vec{b}\cdot\vec{c}=-(\vec{a}\cdot\vec{b}+9) \]

Step 4: Substitute the given value.
It is given that \[ \vec{b}\cdot\vec{c}=-9 \] So, \[ -(\vec{a}\cdot\vec{b}+9)=-9 \]

Step 5: Solve for \( \vec{a}\cdot\vec{b} \).
Multiply both sides by \( -1 \): \[ \vec{a}\cdot\vec{b}+9=9 \] \[ \vec{a}\cdot\vec{b}=0 \]

Step 6: Interpret the result geometrically.
Since \[ \vec{a}\cdot\vec{b}=0 \] the vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular.
That means the sides \( BC \) and \( CA \) are mutually perpendicular, so the angle at \( C \) is a right angle.

Step 7: Final conclusion.
Therefore, \[ \boxed{\vec{a}\cdot\vec{b}=0} \] Hence, the correct option is \[ \boxed{(4)\ 0} \]