Let $\vec{a}=\sqrt{7 } \hat{\imath}+\hat{\jmath}-\hat{k}$ and $\vec{b}=\hat{j}+2 \hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}+\vec{a} \times \vec{b}=\overrightarrow{0}$ and $\vec{r} \cdot \vec{a}=0$, then $|3 \vec{r}|^2$ is equal to: [JEE Main 2026, 5 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: a) $44$. $\vec{r} \times \vec{a}-\vec{b} \times \vec{a}=\overrightarrow{0}$ $(\vec{r}-\vec{b}) \times \vec{a}=\overrightarrow{0}$ $\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{b}}=\lambda \overrightarrow{\mathrm{a}}$ $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{b}}+\lambda \overrightarrow{\mathrm{a}}$ $ \vec{r} \cdot \vec{a}=0 \Rightarrow \vec{a} \cdot \vec{b}+\lambda|\vec{a}|^2=0$ $\lambda=-\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{a}}|^2}=-\frac{(1-2)}{9}=\frac{1}{9} $ $ \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{b}}+\frac{\overrightarrow{\mathrm{a}}}{9} $ $ |3 \vec{r}|^2=9|\vec{r}|^2=9\left(|\vec{b}|^2+\frac{|\vec{a}|^2}{81}+\frac{2(\vec{a} \cdot \vec{b})}{9}\right)=44$

Let \(\vec{a}=\sqrt{7 } \hat{\imath}+\hat{\jmath}-\hat{k}\) and \(\vec{b}=\hat{j}+2 \hat{k}\). If \(\vec{r}\) is a vecto…

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