JEE Main 2023 — Mathematics Vectors & 3D Geometry

Given that,

$\vec{a}=2\hat{i}+3\hat{j}+4\hat{k},\vec{b}=\hat{i}-2\hat{j}-2\hat{k}$a=2i^+3j^​+4k^,b=i^−2j^​−2k^ and $\vec{c}=-\hat{i}+4\hat{j}+3\hat{k}.$c=−i^+4j^​+3k^.

Let us find $\vec{b}\times \vec{c}$b×c

$$\Rightarrow \vec{b}\times \vec{c}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & -2 \\ -1 & 4 & 3\end{matrix}|$$⇒b×c=∣i^1−1​j^​−24​k^−23​∣

$$=(-6+8)\hat{i}-(3-2)\hat{j}+(4-2)\hat{k}$$=(−6+8)i^−(3−2)j^​+(4−2)k^

$$\Rightarrow \vec{b}\times \vec{c}=2\hat{i}-\hat{j}+2\hat{k}$$⇒b×c=2i^−j^​+2k^

Since $\vec{d}$d is perpendicular to both $\vec{b}\text{and}\vec{c}$bandc

$$\therefore \vec{d}=\lambda (2\hat{i}-\hat{j}+2\hat{k})$$∴d=λ(2i^−j^​+2k^)

Also $\vec{a}\cdot \vec{d}=18$a⋅d=18

$$\Rightarrow \lambda (2\hat{i}+3\hat{j}+4\hat{k}).(2\hat{i}-\hat{j}+2\hat{k})=18$$⇒λ(2i^+3j^​+4k^).(2i^−j^​+2k^)=18

$$\Rightarrow 9\lambda =18$$⇒9λ=18

$$\Rightarrow \lambda =2$$⇒λ=2

Now let us apply LaGrange's identity which is as follows,

$$\therefore |\vec{a}\times \vec{d}{|}^{2}=|\vec{a}{|}^{2}\cdot |\vec{d}{|}^{2}-(\vec{a}\cdot \vec{d}{)}^{2}$$∴∣a×d∣2=∣a∣2⋅∣d∣2−(a⋅d)2

$$={|2\hat{i}+3\hat{j}+4\hat{k}|}^{2}{|4\hat{i}-2\hat{j}+4\hat{k}|}^{2}-{(18)}^{2}$$=∣2i^+3j^​+4k^∣2∣4i^−2j^​+4k^∣2−(18)2

$$=720$$=720