Question

Let \(\vec{a}, \vec{b}\) be two vectors, and let \(P, Q\) and \(R\) be the points with position vectors \(\vec{a}, \vec{b}\) and \(\vec{a} + \vec{b}\), respectively, with respect to the origin \(O\). If \[ |\vec{a} + \vec{b}| = \sqrt{21}, \qquad |\vec{a} - \vec{b}| = 3, \] and \(\vec{a}\) and \((\vec{a} - \vec{b})\) are perpendicular to each other, then the area of the triangle \(OPR\) is:

• A. $\sqrt{3}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{3\sqrt{3}}{2}$
• D. $\frac{3}{2}$

Hint:

For any triangle with vertices $A, B, C$, the area can be calculated using the cross product of any two vectors formed by the vertices, such as $\frac{1}{2} |\vec{AB} \times \vec{AC}|$. Remember that $\vec{a} \times \vec{a} = 0$ simplifies such expressions significantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem involves vector geometry where we need to find the area of a triangle formed by the origin and two specific position vectors. We are given magnitudes of the sum and difference of vectors $\vec{a}$ and $\vec{b}$, along with a perpendicularity condition.
Step 2: Key Formula or Approach:

Use the parallelogram law for magnitudes: $|\vec{u} + \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + 2\vec{u} \cdot \vec{v}$.

Area of triangle $OPR$ with vertices $O(\vec{0}), P(\vec{a}), R(\vec{a} + \vec{b})$ is given by $\frac{1}{2} |\vec{OP} \times \vec{OR}|$.

Dot product of perpendicular vectors is zero.

Step 3: Detailed Explanation:

Given $|\vec{a} + \vec{b}| = \sqrt{21} \implies |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} = 21$. (Equation 1)

Given $|\vec{a} - \vec{b}| = 3 \implies |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} = 9$. (Equation 2)

Subtracting Equation 2 from Equation 1: $4\vec{a} \cdot \vec{b} = 12 \implies \vec{a} \cdot \vec{b} = 3$.

Since $\vec{a}$ and $(\vec{a} - \vec{b})$ are perpendicular, $\vec{a} \cdot (\vec{a} - \vec{b}) = 0$.
\[ \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} = 0 \implies |\vec{a}|^2 = \vec{a} \cdot \vec{b} = 3 \]
From Equation 2: $3 + |\vec{b}|^2 - 2(3) = 9 \implies |\vec{b}|^2 - 3 = 9 \implies |\vec{b}|^2 = 12$.

The area of triangle $OPR$ is: \[ \text{Area} = \frac{1}{2} |\vec{a} \times (\vec{a} + \vec{b})| = \frac{1}{2} |(\vec{a} \times \vec{a}) + (\vec{a} \times \vec{b})| = \frac{1}{2} | \vec{a} \times \vec{b} | \]
Using the identity $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$: \[ |\vec{a} \times \vec{b}|^2 = (3)(12) - (3)^2 = 36 - 9 = 27 \implies |\vec{a} \times \vec{b}| = \sqrt{27} = 3\sqrt{3} \]
Therefore, Area $= \frac{1}{2} (3\sqrt{3}) = \frac{3\sqrt{3}}{2}$.

Step 4: Final Answer:
The area of triangle $OPR$ is $\frac{3\sqrt{3}}{2}$ square units.