Vector Algebra
Cross Product and Area
Grade 12
Question:
<p>If <strong>a</strong> = 2<strong>i</strong> − 3<strong>j</strong> + <strong>k</strong>, <strong>b</strong> = −<strong>i</strong> + <strong>k</strong>, <strong>c</strong> = 2<strong>j</strong> − <strong>k</strong>, then the area (in sq units) of parallelogram with diagonals <strong>a</strong> + <strong>b</strong> and <strong>b</strong> + <strong>c</strong> will be</p>
<p>(a) \(\sqrt{21}\)</p>
<p>(b) \(2\sqrt{21}\)</p>
<p>(c) \(\frac{1}{2}\sqrt{21}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The area of a parallelogram with diagonals **d₁** and **d₂** is given by \(\frac{1}{2}|\mathbf{d_1} \times \mathbf{d_2}|\). Calculate the cross product of the two diagonal vectors and take half its magnitude.
Step 1: We have, a = 2 i − 3 j + k , b = − i + k , c = 2 j − k Step 2: Since ( a + b ) and ( b + c ) are the diagonals of the parallelogram Step 3: a + b = i − 3 j + 2 k Step 4: b + c = − i + 2 j Step 5: Area of parallelogram = \(\frac{1}{2}|( a + b ) \times ( b + c )|\) ∴ Answer is (c) \(\frac{1}{2}\sqrt{21}\)
Correct Answer: C