MHT CET · Maths · Vector Algebra
If the area of the parallelogram with \(\vec{a}\) and \(\vec{b}\) as two adjacent sides is 15 sq. units, then the area of the parallelogram having \(3 \vec{a}+\vec{b}\) and \(\vec{a}+3 \vec{b}\) as two adjacent sides, in square units, is
- A \(135\)
- B 90
- C 150
- D 120
Answer & Solution
Correct Answer
(D) 120
Step-by-step Solution
Detailed explanation
Area of parallelogram \(=|\vec{a} \times \vec{b}|=15\) [given]
Area of second parallelogram \(=|(3 \vec{a}+\vec{b}) \times(\vec{a}+3 \vec{b})|\)
\(\begin{aligned} & =|3 \vec{a} \times \vec{a}+9 \vec{a} \times \vec{b}+\vec{b} \times \vec{a}+3 \vec{b} \times \vec{b}| \\ & =|0+9 \vec{a} \times \vec{b}-\vec{a} \times \vec{b}+0| \\ & =|8 \vec{a} \times \vec{b}|=8|\vec{a} \times \vec{b}|=8 \times 15=120\end{aligned}\)
Area of second parallelogram \(=|(3 \vec{a}+\vec{b}) \times(\vec{a}+3 \vec{b})|\)
\(\begin{aligned} & =|3 \vec{a} \times \vec{a}+9 \vec{a} \times \vec{b}+\vec{b} \times \vec{a}+3 \vec{b} \times \vec{b}| \\ & =|0+9 \vec{a} \times \vec{b}-\vec{a} \times \vec{b}+0| \\ & =|8 \vec{a} \times \vec{b}|=8|\vec{a} \times \vec{b}|=8 \times 15=120\end{aligned}\)
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- If in triangle \(A B C\), with usual notations \(\sin \frac{A}{2} \cdot \sin \frac{C}{2}=\sin \frac{B}{2}\) and \(2 s\) is the perimeter of the triangle, then the value of \(s\) isMHT CET 2025 Medium
- \(\int \frac{\tan ^4 \sqrt{x} \cdot \sec ^2 \sqrt{x}}{\sqrt{x}} d x=\)MHT CET 2021 Medium
- A spherical rain drop evaporates at a rate proportional to its surface area. If initially its radius is 3 mm and after 1 second it is reduced to 2 mm , then at any time t its radius is (where \(0 \leq \mathrm{t} \lt 3\) )MHT CET 2024 Hard
- If \(\cos ^{-1} x-\cos ^{-1} \frac{y}{3}=\alpha\), where \(-1 \leq x \leq 1\), \(-3 \leq y \leq 3, x \leq \frac{y}{3}\), then for all \(x, y\), \(9 x^2-6 x y \cos \alpha+y^2\) is equal toMHT CET 2023 Hard
- \(\int_{0}^{\pi} \frac{x d x}{1+\cos \alpha \sin x},(0 < \alpha < \pi)\) is equal toMHT CET 2011 Hard
- The number of values of \(x\) in the interval \([0,3 \pi]\) satisfying the equation \(2 \sin ^2 x+5 \sin x-3=0\) isMHT CET 2025 Medium
More PYQs from MHT CET
- A conductivity cell shows resistance of \(600 \mathrm{ohm}\). If conductivity of \(0.01 \mathrm{M} \mathrm{KCl}\) is \(0.0015 \Omega^{-1} \mathrm{~cm}^{-1}\), what is cell constant?MHT CET 2021 Medium
- \(\mathrm{ABCD}\) is a parallelogram, \(\mathrm{P}\) is the mid-point of \(\mathrm{AB}\). If \(\mathrm{R}\) is the point of intersection
of \(\mathrm{AC}\) and \(\mathrm{DP}\), then \(\mathrm{R}\) divides \(\mathrm{AC}\) internally in the ratioMHT CET 2020 Medium - The different types of 3-C compounds formed during glycolysis of one glucose molecules areMHT CET 2023 Hard
- Identify substrate ' \(A\) ' in the following reaction.
\(\mathrm{A}+\mathrm{CO}_2 \xrightarrow[6 \mathrm{~atm}]{398 \mathrm{~K}}\) Sodium salicylate \(\xrightarrow{\mathrm{H}_3 \mathrm{O}^{+}}\)Salicylic acidMHT CET 2025 Easy - The \(\mathrm{H}-\mathrm{C}-\mathrm{H}\) bond angle in \(\mathrm{CH}_4\) molecule isMHT CET 2022 Easy
- For the differential equation \(\left[1-\left(\frac{d y}{d x}\right)^2\right]^{5 / 2}=8 \frac{d^2 y}{d x^2}\) has the order and degreeMHT CET 2022 Easy