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: Understanding the Concept:
In any triangle, the sum of the vectors representing the sides taken in order is the zero vector. We are given the vectors for the sides and information about their magnitudes and dot products. We can use the vector sum property to relate the vectors and then use the dot product properties to find the required value.
Step 2: Key Formula or Approach:
1. For triangle ABC, the sum of side vectors is zero: \(\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}\). 2. Substitute the given vector names: \(\vec{c} + \vec{a} + \vec{b} = \vec{0}\). 3. We can rearrange this to express one vector in terms of the others, for example, \(\vec{a} = -(\vec{b} + \vec{c})\). 4. Use the dot product property: \(\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta\) and \(\vec{u} \cdot \vec{u} = |\vec{u}|^2\).
Step 3: Detailed Explanation:
From the triangle law of vector addition, we have:
\[ \vec{a} + \vec{b} + \vec{c} = \vec{0} \] From this, we can write \(\vec{a} = -(\vec{b} + \vec{c})\).
We are given \(|\vec{a}| = 3\) and \(|\vec{b}| = 3\).
Let's find \(|\vec{c}|\). From \(\vec{c} = -(\vec{a}+\vec{b})\), we have \(|\vec{c}|^2 = |- (\vec{a}+\vec{b})|^2 = (\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})\). This path seems complicated. Let's use the given dot product \(\vec{b} \cdot \vec{c} = -9\).
We know \(|\vec{b}| = 3\). Let's find \(|\vec{c}|\).
From \(\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos(\pi - A)\), where A is the angle at vertex A. This also seems complex. Let's try a different approach. From \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), let's rearrange to \(\vec{c} = -(\vec{a}+\vec{b})\).
Now substitute this into the given dot product:
\[ \vec{b} \cdot (-(\vec{a}+\vec{b})) = -9 \] \[ -\vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b} = -9 \] \[ -\vec{a} \cdot \vec{b} - |\vec{b}|^2 = -9 \] We are given \(|\vec{b}| = 3\), so \(|\vec{b}|^2 = 9\).
\[ -\vec{a} \cdot \vec{b} - 9 = -9 \] \[ -\vec{a} \cdot \vec{b} = 0 \] \[ \vec{a} \cdot \vec{b} = 0 \] Let's check for consistency. If \(\vec{a} \cdot \vec{b} = 0\), the angle between \(\vec{BC}\) and \(\vec{CA}\) is 90\(^\circ\). This means angle C is 90\(^\circ\). The triangle is a right-angled isosceles triangle with sides \(|\vec{a}|=3\) and \(|\vec{b}|=3\). The hypotenuse would be \(|\vec{c}| = \sqrt{3^2+3^2} = 3\sqrt{2}\). The angle between \(\vec{b}\) and \(\vec{c}\) (vectors \(\vec{CA}\) and \(\vec{AB}\)) is the exterior angle at A, which is \(180-45 = 135^\circ\). So \(\vec{b}\cdot\vec{c} = |\vec{b}||\vec{c}|\cos(135^\circ) = 3 \cdot 3\sqrt{2} \cdot (-\frac{1}{\sqrt{2}}) = -9\). This is consistent with the given information. Step 4: Final Answer:
The value of \(\vec{a} \cdot \vec{b}\) is 0. This corresponds to option (D).