KCET · Maths · Vector Algebra
If the area of the parallelogram with \(\mathbf{a}\) and \(\mathbf{b}\) as two adjacent sides is 15 sq units, then the area of the parallelogram having \(3 \mathbf{a}+2 \mathbf{b}\) and \(\mathbf{a}+3 \mathbf{b}\) as two adjacent sides in sq units is
- A 45
- B 75
- C 105
- D 120
Answer & Solution
Correct Answer
(C) 105
Step-by-step Solution
Detailed explanation
Area of parallelogram having \(\mathbf{a}\) and \(\mathbf{b}\) as its adjacent sides is 15 sq units.
\(\Rightarrow \quad|\mathbf{a} \times \mathbf{b}|=15\)
Area of parallelogram having \((3 \mathbf{a}+2 \mathbf{b})\) and
\((\mathbf{a}+3 \mathbf{b})\) as two adjacent side
\(=|(3 \mathbf{a}+2 \mathbf{b}) \times(\mathbf{a}+3 \mathbf{b})|\)
\(=|3 \mathbf{a} \times \mathbf{a}+2 \mathbf{b} \times \mathbf{a}+9 \mathbf{a} \times \mathbf{b}+6 \mathbf{b} \times \mathbf{b}|\)
\(=|2 \mathbf{b} \times \mathbf{a}+9 \mathbf{a} \times \mathbf{b}|\)
\([\therefore \mathbf{a} \times \mathbf{a}=\mathbf{b} \times \mathbf{b}=0]\)
\(=7|\mathbf{a} \times \mathbf{b}|\)
\([\therefore \mathbf{b} \times \mathbf{a}=-\mathbf{a} \times \mathbf{b}]\)
\(=7 \times 15\)
\(=105\)
\(\Rightarrow \quad|\mathbf{a} \times \mathbf{b}|=15\)
Area of parallelogram having \((3 \mathbf{a}+2 \mathbf{b})\) and
\((\mathbf{a}+3 \mathbf{b})\) as two adjacent side
\(=|(3 \mathbf{a}+2 \mathbf{b}) \times(\mathbf{a}+3 \mathbf{b})|\)
\(=|3 \mathbf{a} \times \mathbf{a}+2 \mathbf{b} \times \mathbf{a}+9 \mathbf{a} \times \mathbf{b}+6 \mathbf{b} \times \mathbf{b}|\)
\(=|2 \mathbf{b} \times \mathbf{a}+9 \mathbf{a} \times \mathbf{b}|\)
\([\therefore \mathbf{a} \times \mathbf{a}=\mathbf{b} \times \mathbf{b}=0]\)
\(=7|\mathbf{a} \times \mathbf{b}|\)
\([\therefore \mathbf{b} \times \mathbf{a}=-\mathbf{a} \times \mathbf{b}]\)
\(=7 \times 15\)
\(=105\)
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