JEE Main 2023 — Mathematics Vectors & 3D Geometry

Given,

$|\vec{a}|=\sqrt{31},4|\vec{b}|=|\vec{c}|=2$∣a∣=31​,4∣b∣=∣c∣=2, $2(\vec{a}\times \vec{b})=3(\vec{c}\times \vec{a})$2(a×b)=3(c×a) and angle between \vec{b}&\vec{c} is given as $\frac{2\pi }{3}$32π​

Now solving, $3(\vec{c}\times \vec{a})+2(\vec{b}\times \vec{a})=0$3(c×a)+2(b×a)=0

$$\Rightarrow (3\vec{c}\times 2\vec{b})\times \vec{a}=0$$⇒(3c×2b)×a=0

Means (3\vec{c}\times 2\vec{b})&\vec{a} are parallel vector,

So, let $3\vec{c}\times 2\vec{b}=\lambda \vec{a}$3c×2b=λa

Now squaring both sides we get,

$$9{|\vec{c}|}^{2}+4{|\vec{b}|}^{2}+12(\vec{b}\cdot \vec{c})={\lambda }^{2}{|\vec{a}|}^{2}$$9∣c∣2+4∣b∣2+12(b⋅c)=λ2∣a∣2

$$\Rightarrow 36+1+12\times \frac{1}{2}\times 2(\mathrm{cos}(2\frac{\pi }{3}))={\lambda }^{2}(31)$$⇒36+1+12×21​×2(cos(23π​))=λ2(31)

$$\Rightarrow {\lambda }^{2}=1$$⇒λ2=1

$$\Rightarrow \lambda =\pm 1$$⇒λ=±1

Now putting the value of $\lambda$λ in $(3\vec{c}\times 2\vec{b})=\lambda \vec{a}$(3c×2b)=λa we get,

$$3\vec{c}+2\vec{b}=\pm \vec{a}\ldots (1)$$3c+2b=±a…(1)

Now taking dot product with $\vec{b}$b in above equation we get,

$$3(\vec{b}\cdot \vec{c})+2(\vec{b}\cdot \vec{b})=\pm \vec{a}\cdot \vec{b}$$3(b⋅c)+2(b⋅b)=±a⋅b

$$\Rightarrow \vec{a}\cdot \vec{b}=\pm (-\frac{3}{2}+\frac{1}{2})=\pm (-1)$$⇒a⋅b=±(−23​+21​)=±(−1)

$$\Rightarrow {(\vec{a}\cdot \vec{b})}^{2}=1$$⇒(a⋅b)2=1

Again taking $3(\vec{c}\times \vec{a})=2(\vec{a}\times \vec{b})$3(c×a)=2(a×b) and sqauring both side,

$$\Rightarrow {(\vec{c}\times \vec{a})}^{2}=\frac{4}{9}{(\vec{a}\times \vec{b})}^{2}$$⇒(c×a)2=94​(a×b)2

$$\Rightarrow {(\vec{c}\times \vec{a})}^{2}=\frac{4}{9}[{|\vec{a}|}^{2}{|\vec{b}|}^{2}-{(\vec{a}\cdot \vec{b})}^{2}]$$⇒(c×a)2=94​[∣a∣2∣b∣2−(a⋅b)2]

$$\Rightarrow {(\vec{c}\times \vec{a})}^{2}=\frac{4}{9}[\frac{31}{4}-(1)]$$⇒(c×a)2=94​[431​−(1)]

$$\Rightarrow {(\vec{c}\times \vec{a})}^{2}=\frac{4}{9}\times \frac{27}{4}=3$$⇒(c×a)2=94​×427​=3

Hence, the value of ${(\frac{\vec{a}\times \vec{c}}{\vec{a}\cdot \vec{b}})}^{2}=\frac{3}{1}=3$(a⋅ba×c​)2=13​=3.