Vector Algebra
Cross Product and Areas
Grade 12
Question:
<p>Let <strong>OA</strong> = <strong>a</strong>, <strong>OB</strong> = 10<strong>a</strong> + 2<strong>b</strong> and <strong>OC</strong> = <strong>b</strong>, where O, A and C are non-collinear points. Let p denote the area of quadrilateral OACB, and let q denote the area of parallelogram with <strong>OA</strong> and <strong>OC</strong> as adjacent sides. If p = kq, then k is equal to</p>
Step-by-Step Solution
Key Concept: Use the cross product property and break the quadrilateral into two triangles to relate areas p and q.
Step 1: Area of parallelogram with OA and OC as adjacent sides: \[q = |\mathbf{a} \times \mathbf{b}|\] Step 2: Area of quadrilateral OABC = Area of \(\triangle\)OAB + Area of \(\triangle\)OBC \[p = \frac{1}{2}|\mathbf{a} \times (10\mathbf{a} + 2\mathbf{b})| + \frac{1}{2}|(10\mathbf{a} + 2\mathbf{b}) \times \mathbf{b}|\] Step 3: Using properties of cross product (\(\mathbf{a} \times \mathbf{a} = 0\)): \[p = \frac{1}{2}|2\mathbf{a} \times \mathbf{b}| + \frac{1}{2}|10\mathbf{a} \times \mathbf{b}| = |\mathbf{a} \times \mathbf{b}| + 5|\mathbf{a} \times \mathbf{b}| = 6|\mathbf{a} \times \mathbf{b}|\] Step 4: Therefore, \(p = 6q\), so \(k = 6\)
Correct Answer: 6