JEE Mathematics 2023 Question with Solution

Given: $|\vec{a}| = \sqrt{31}, \ |\vec{b}| = 4, \ |\vec{c}| = 2$ and

$$2(\vec{a}\times\vec{b})=3(\vec{c}\times\vec{a})$$

Find: $\left(\frac{\vec{a}\times\vec{c}}{\vec{a}\cdot\vec{b}}\right)^2$

From the given relation,

$$2(\vec{a}\times\vec{b})=3(\vec{c}\times\vec{a})$$

Using $\vec{c}\times\vec{a}=-(\vec{a}\times\vec{c})$, we get

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

So,

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

Hence,

$$\vec{a}=\lambda(2\vec{b}+3\vec{c})$$

for some scalar $\lambda$.

Now,

$$|\vec{a}|^2=\lambda^2|2\vec{b}+3\vec{c}|^2$$

That is,

$$|\vec{a}|^2=\lambda^2\left(4|\vec{b}|^2+9|\vec{c}|^2+12\,\vec{b}\cdot\vec{c}\right)$$

Since the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2\pi}{3}$,

$$\vec{b}\cdot\vec{c}=|\vec{b}||\vec{c}|\cos\frac{2\pi}{3}=4\times2\times\left(-\frac12\right)=-4$$

Therefore,

$$|2\vec{b}+3\vec{c}|^2=4(16)+9(4)+12(-4)=64+36-48=52$$

So,

$$31=52\lambda^2$$

Thus,

$$\lambda^2=\frac{31}{52}$$

Next,

$$\vec{a}\times\vec{c}=\lambda(2\vec{b}+3\vec{c})\times\vec{c}=2\lambda(\vec{b}\times\vec{c})$$

Hence,

$$|\vec{a}\times\vec{c}|=2|\lambda|\,|\vec{b}\times\vec{c}|$$

Also,

$$\vec{a}\cdot\vec{b}=\lambda(2\vec{b}+3\vec{c})\cdot\vec{b}=\lambda\left(2|\vec{b}|^2+3\vec{c}\cdot\vec{b}\right)$$

So,

$$\vec{a}\cdot\vec{b}=\lambda(2\times16+3\times(-4))=20\lambda$$

Now,

$$|\vec{b}\times\vec{c}|=|\vec{b}||\vec{c}|\sin\frac{2\pi}{3}=4\times2\times\frac{\sqrt{3}}{2}=4\sqrt{3}$$

Therefore,

\frac{|\vec{a}\times\vec{c}|}{|\vec{a}\cdot\vec{b}|}=rac{2|\lambda|\cdot4\sqrt{3}}{|20\lambda|}=\frac{2\sqrt{3}}{5}

Hence,

$$\left(\frac{\vec{a}\times\vec{c}}{\vec{a}\cdot\vec{b}}\right)^2=\frac{12}{25}$$

the solution concludes with option 3, but the displayed working is inconsistent at multiple places. Following the working from the given data gives the value $\frac{12}{25}$.

The extracted solution contains internal inconsistencies:

1. It states

$$31=31\lambda^2 \Rightarrow \lambda=\pm1$$

but from the given values,

$$|2\vec{b}+3\vec{c}|^2=52$$

not $31$.

2. It writes a cross-product magnitude line in an incorrect form and then uses numerical substitutions that do not match $|\vec{b}|=4$ and $|\vec{c}|=2$.

3. The final statement "So, the correct answer is 3" refers to option number 3, not a numerical value answer derived consistently from the algebra.

Therefore, the defensible result from the visible mathematical working based on the question data is

$$\frac{12}{25}$$

and not $3$ or $4$.