# Vectors → a and → b are inclined at angle θ = 120°. If | → a | = 1 , ∣ ∣ → b ∣ ∣ = 2 , then [ ( → a + 3 → b ) × ( 3 → a − → b ) ] 2 is equal to - Mathematics

MCQ

Vectors $\vec{a} \text{ and } \vec{b}$ are inclined at angle θ = 120°. If $\left| \vec{a} \right| = 1, \left| \vec{b} \right| = 2,$ then  $\left[ \left( \vec{a} + 3 \vec{b} \right) \times \left( 3 \vec{a} - \vec{b} \right) \right]^2$  is equal to

• 300

•  325

•  275

•  225

#### Solution

$\left( \vec{a} + 3 \vec{b} \right) \times \left( 3 \vec{a} - \vec{b} \right)$

$= 3 \left( \vec{a} \times \vec{a} \right) - \vec{a} \times \vec{b} + 9 \left( \vec{b} \times \vec{a} \right) - 3 \left( \vec{b} \times \vec{b} \right)$

$= 3 \left( 0 \right) - \vec{a} \times \vec{b} - 9 \left( \vec{a} \times \vec{b} \right) - 3 \left( 0 \right)$

$= - 10 \left( \vec{a} \times \vec{b} \right)$

$\text{ Now } ,$

$\left| \left( \vec{a} \times 3 \vec{b} \right) \times \left( 3 \vec{a} - \vec{b} \right) \right| {}^2$

$= \left| - 10 \left( \vec{a} \times \vec{b} \right) \right|^2$

$= 100 \left| \left( \vec{a} \times \vec{b} \right) \right|^2$

$= 100 \left| \vec{a} \right|^2 \left| \vec{b} \right|^2 \sin^2 120$

$= 100 \left( 1 \right)^2 \left( 2 \right)^2 \left( \frac{\sqrt{3}}{2} \right)^2$

$= 400 \times \frac{3}{4}$

$= 300$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 25 Vector or Cross Product
MCQ | Q 6 | Page 35