JEE Main 2023 — Mathematics Vectors & 3D Geometry

Given,

The vector $\vec{a}=-\hat{i}+2\hat{j}+\hat{k}$a=−i^+2j^​+k^ is rotated through a right angle, passing through the $y$y-axis in its way and the resulting vector is $\vec{b}$b

Now taking, $\vec{b}=\lambda \vec{a}\times (\vec{a}\times \hat{j})$b=λa×(a×j^​) {note here we have taken cross vector with $\hat{j}$j^​ as it is rotated along $y-$y−axis}

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

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

$$\Rightarrow \vec{b}=\lambda |\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ -1 & 0 & -1\end{matrix}|$$⇒b=λ∣i^−1−1​j^​20​k^1−1​∣

$$\Rightarrow \vec{b}=\lambda (-2\hat{i}-2\hat{j}+2\hat{k})$$⇒b=λ(−2i^−2j^​+2k^)

Also magnitude of both vector will be equal,

So,$|\vec{b}|=|\vec{a}|\Rightarrow \sqrt{6}=\sqrt{12}|\lambda |$∣b∣=∣a∣⇒6​=12​∣λ∣

$$\Rightarrow \lambda =\pm \frac{1}{\sqrt{2}}$$⇒λ=±2​1​

(Now $\lambda =\frac{1}{\sqrt{2}}$λ=2​1​ rejected $\because \vec{b}$∵b makes acute angle with y axis )

So, $\vec{b}=-\sqrt{2}(-\hat{i}-\hat{j}+\hat{k})$b=−2​(−i^−j^​+k^)

Now projection of $3\vec{a}+\sqrt{2}\vec{b}$3a+2​b on $\vec{c}$c is given by,

$$\frac{(3\vec{a}+\sqrt{2b})\cdot \vec{c}}{|\vec{c}|}=3\sqrt{2}$$∣c∣(3a+2b​)⋅c​=32​