JEE MAIN - Mathematics (2023 - 1st February Evening Shift - No. 13)
Let $$\vec{a}=2 \hat{i}-7 \hat{j}+5 \hat{k}, \vec{b}=\hat{i}+\hat{k}$$ and $$\vec{c}=\hat{i}+2 \hat{j}-3 \hat{k}$$ be three given vectors. If $$\overrightarrow{\mathrm{r}}$$ is a vector such that $$\vec{r} \times \vec{a}=\vec{c} \times \vec{a}$$ and $$\vec{r} \cdot \vec{b}=0$$, then $$|\vec{r}|$$ is equal to :
$$\frac{11}{7}$$
$$\frac{11}{5} \sqrt{2}$$
$$\frac{\sqrt{914}}{7}$$
$$\frac{11}{7} \sqrt{2}$$
Explanation
$\begin{aligned} & \vec{r} \times \vec{a}=\vec{c} \times \vec{a} \\\\ & \Rightarrow(\vec{r}-\vec{c}) \times \vec{a}=0 \Rightarrow \vec{r}-\vec{c}=\lambda \vec{a}((\vec{r}-\vec{c} ) \text{and} \overrightarrow{a} \text { are parallel }) \\\\ & \Rightarrow \vec{r}=\vec{c}+\lambda \vec{a} \\\\ & \Rightarrow \vec{r} \cdot \vec{b}=\vec{c} \cdot \vec{b}+\lambda \vec{a} \cdot \vec{b} \\\\ & 0=(1-3)+\lambda(2+5) \Rightarrow \lambda=\frac{2}{7} \\\\ & \text { Hence, } \vec{r}=\vec{c}+\frac{2 \vec{a}}{7} \\\\ & \vec{r} \Rightarrow \frac{11}{7} \hat{i}-\frac{11}{7} \hat{k} \\\\ & |\vec{r}|=\sqrt{\left(\frac{11}{7}\right)^2+\left(-\frac{11}{7}\right)^2} \Rightarrow r=\frac{11 \sqrt{2}}{7}\end{aligned}$
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