JEE Main 2024 — Mathematics Vectors & 3D Geometry

Given $|\vec{a}|=1,|\vec{b}|=4,\vec{a}\cdot \vec{b}=2$∣a∣=1,∣b∣=4,a⋅b=2

Also, $\vec{c}=2(\vec{a}\times \vec{b})-3\vec{b}...(i)$c=2(a×b)−3b...(i)

Applying dot product with $\vec{a}$a on both sides of equation $(i)$(i)

$$\Rightarrow \vec{c}\cdot \vec{a}=-6...(ii)$$⇒c⋅a=−6...(ii)

$${\text{as}(\vec{a}\times \vec{b})\cdot \vec{a}=(\vec{a}\times \vec{b})\cdot \vec{b}=0}$$as(a×b)⋅a=(a×b)⋅b=0

Again, applying dot product with $\vec{b}$b on both sides of equation $(i)$(i)

$$\Rightarrow \vec{b}\cdot \vec{c}=-48...(iii)$$⇒b⋅c=−48...(iii)

Now, using equation $(i)$(i)

$$\Rightarrow {|\vec{c}|}^{2}=4|\vec{a}\times \vec{b}{|}^{2}+9|\vec{b}{|}^{2}-12(\vec{a}\times \vec{b})\cdot \vec{b}$$⇒∣c∣2=4∣a×b∣2+9∣b∣2−12(a×b)⋅b

We know that, ${(\vec{a}\times \vec{b})}^{2}+{(\vec{a}.\vec{b})}^{2}={|\vec{a}|}^{2}{|\vec{b}|}^{2}$(a×b)2+(a.b)2=∣a∣2∣b∣2

$$\Rightarrow |\vec{c}{|}^{2}=4[|\vec{a}{|}^{2}|\vec{b}{|}^{2}-(\vec{a}\cdot \vec{b}{)}^{2}]+9|\vec{b}{|}^{2}$$⇒∣c∣2=4[∣a∣2∣b∣2−(a⋅b)2]+9∣b∣2

$$\Rightarrow |\vec{c}{|}^{2}=4[(1)(4{)}^{2}-(4)]+9(16)$$⇒∣c∣2=4[(1)(4)2−(4)]+9(16)

$$\Rightarrow |\vec{c}{|}^{2}=4\times 12+144$$⇒∣c∣2=4×12+144

$$\Rightarrow |\vec{c}{|}^{2}=48+144$$⇒∣c∣2=48+144

$$\Rightarrow |\vec{c}{|}^{2}=192$$⇒∣c∣2=192

Now, $\mathrm{cos}\theta =\frac{\vec{b}\cdot \vec{c}}{|\vec{b}||\vec{c}|}$cosθ=∣b∣∣c∣b⋅c​

$$\Rightarrow \mathrm{cos}\theta =\frac{-48}{\sqrt{192}\times 4}$$⇒cosθ=192​×4−48​

$$\Rightarrow \mathrm{cos}\theta =\frac{-48}{8\sqrt{3}.4}$$⇒cosθ=83​.4−48​

$$\Rightarrow \mathrm{cos}\theta =\frac{-3}{2\sqrt{3}}$$⇒cosθ=23​−3​

$$\Rightarrow \mathrm{cos}\theta =\frac{-\sqrt{3}}{2}$$⇒cosθ=2−3​​

$$\Rightarrow \theta ={\mathrm{cos}}^{-1}(\frac{-\sqrt{3}}{2})$$⇒θ=cos−1(2−3​​)