JEE Main 2024 — Mathematics Vectors & 3D Geometry

Given,

$\vec{a}=3\hat{i}+2\hat{j}+\hat{k},\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$a=3i^+2j^​+k^,b=2i^−j^​+3k^ and $\vec{c}$c be a vector such that $(\vec{a}+\vec{b})\times \vec{c}=2(\vec{a}\times \vec{b})+24\hat{j}-6\hat{k}$(a+b)×c=2(a×b)+24j^​−6k^

Now, finding $\vec{a}\times \vec{b}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 1 \\ 2 & -1 & 3\end{matrix}|=7\hat{i}-7\hat{j}-7\hat{k}$a×b=∣i^32​j^​2−1​k^13​∣=7i^−7j^​−7k^

Now, solving

$$(\vec{a}+\vec{b})\times \vec{c}=2(\vec{a}\times \vec{b})+24\hat{j}-6\hat{k}$$(a+b)×c=2(a×b)+24j^​−6k^

$$\Rightarrow (5\hat{i}+\hat{j}+4\hat{k})\times \vec{c}=2(7\hat{i}-7\hat{j}-7\hat{k})+24\hat{j}-6\hat{k}$$⇒(5i^+j^​+4k^)×c=2(7i^−7j^​−7k^)+24j^​−6k^

$$\Rightarrow |\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z\end{matrix}|=14\hat{i}+10\hat{j}-20\hat{k}$$⇒∣i^5x​j^​1y​k^4z​∣=14i^+10j^​−20k^

$$\Rightarrow \hat{i}(z-4y)-\hat{j}(5z-4x)+\hat{k}(5y-x)=14\hat{i}+10\hat{j}-20\hat{k}$$⇒i^(z−4y)−j^​(5z−4x)+k^(5y−x)=14i^+10j^​−20k^

Now, on comparing both side,

$$z-4y=14,4x-5z=10,5y-x=-20....(i)$$z−4y=14,4x−5z=10,5y−x=−20....(i)

Now, solving

$$(\vec{a}-\vec{b}+\hat{i})\cdot \vec{c}=-3$$(a−b+i^)⋅c=−3

$$\Rightarrow (2\hat{i}+3\hat{j}-2\hat{k})\cdot \vec{c}=-3$$⇒(2i^+3j^​−2k^)⋅c=−3

$$\Rightarrow 2x+3y-2z=-3.......(ii)$$⇒2x+3y−2z=−3.......(ii)

Now, solving the equation (i)&(ii) we get,

$$\therefore x=5,y=-3,z=2$$∴x=5,y=−3,z=2

$$\Rightarrow {|\vec{c}|}^{2}=25+9+4=38$$⇒∣c∣2=25+9+4=38