Verified Solution • 2025–26 Session moderate

QJEEM23Ma6a47: JEE Main 2023 Physics - Class 11 JEEM23Ma6a47 Solution: Let and be three given vectors. If is a vector such that and ,…

Let a⃗=2i^−7j^+5k^,b⃗=i^+k^\vec{a}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \vec{b}=\hat{i}+\hat{k} and c⃗=i^+2j^−3k^\vec{c}=\hat{i}+2 \hat{j}-3 \hat{k} be three given vectors. If r→\overrightarrow{\mathrm{r}} is a vector such that r⃗×a⃗=c⃗×a⃗\vec{r} \times \vec{a}=\vec{c} \times \vec{a} and r⃗⋅b⃗=0\vec{r} \cdot \vec{b}=0 , then ∣r⃗∣|\vec{r}| is equal to :

Multiple Choice Options

  • A 117\frac{11}{7}
  • B 1152\frac{11}{5} \sqrt{2}
  • C 9147\frac{\sqrt{914}}{7}
  • D 1172\frac{11}{7} \sqrt{2}

Step-by-Step Verified Solution

amp;r⃗×a⃗=c⃗×a⃗amp;⇒(r⃗−c⃗)×a⃗=0⇒r⃗−c⃗=λa⃗((r⃗−c⃗)anda→ are parallel )amp;⇒r⃗=c⃗+λa⃗amp;⇒r⃗⋅b⃗=c⃗⋅b⃗+λa⃗⋅b⃗amp;0=(1−3)+λ(2+5)⇒λ=27amp; Hence, r⃗=c⃗+2a⃗7amp;r⃗⇒117i^−117k^amp;∣r⃗∣=(117)2+(−117)2⇒r=1127\begin{aligned} & \vec{r} \times \vec{a}=\vec{c} \times \vec{a} \\\\ & \Rightarrow(\vec{r}-\vec{c}) \times \vec{a}=0 \Rightarrow \vec{r}-\vec{c}=\lambda \vec{a}((\vec{r}-\vec{c} ) \text{and} \overrightarrow{a} \text { are parallel }) \\\\ & \Rightarrow \vec{r}=\vec{c}+\lambda \vec{a} \\\\ & \Rightarrow \vec{r} \cdot \vec{b}=\vec{c} \cdot \vec{b}+\lambda \vec{a} \cdot \vec{b} \\\\ & 0=(1-3)+\lambda(2+5) \Rightarrow \lambda=\frac{2}{7} \\\\ & \text { Hence, } \vec{r}=\vec{c}+\frac{2 \vec{a}}{7} \\\\ & \vec{r} \Rightarrow \frac{11}{7} \hat{i}-\frac{11}{7} \hat{k} \\\\ & |\vec{r}|=\sqrt{\left(\frac{11}{7}\right)^2+\left(-\frac{11}{7}\right)^2} \Rightarrow r=\frac{11 \sqrt{2}}{7}\end{aligned}

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