Q39 / 75 · JEE Main 2026 (05 April Shift 1) · Mathematics

Let $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{j} + 2\hat{k}$. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$, then $|3\vec{r}|^2$ is equal to:

  • Correct answer
    A
    $44$
  • B
    $54$
  • C
    $86$
  • D
    $132$

Solution

Given $\vec{r} \times \vec{a} + \vec{a} \times \vec{b} = \vec{0}$

$\Rightarrow \vec{r} \times \vec{a} - \vec{b} \times \vec{a} = \vec{0}$

$\Rightarrow (\vec{r} - \vec{b}) \times \vec{a} = \vec{0}$

$\Rightarrow \vec{r} - \vec{b} = \lambda \vec{a}$

$\Rightarrow \vec{r} = \vec{b} + \lambda \vec{a}$

Taking dot product with $\vec{a}$ on both sides:

$\vec{r} \cdot \vec{a} = \vec{b} \cdot \vec{a} + \lambda |\vec{a}|^2$

Since $\vec{r} \cdot \vec{a} = 0$, we get:

$\lambda = -\dfrac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}$

We have $\vec{a} = \sqrt{7}\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{j} + 2\hat{k}$.

$\vec{b} \cdot \vec{a} = (0)(\sqrt{7}) + (1)(1) + (2)(-1) = -1$

$|\vec{a}|^2 = (\sqrt{7})^2 + (1)^2 + (-1)^2 = 9$

$|\vec{b}|^2 = (1)^2 + (2)^2 = 5$

Substituting these values:

$\lambda = -\dfrac{-1}{9} = \dfrac{1}{9}$

Therefore, $\vec{r} = \vec{b} + \dfrac{1}{9}\vec{a}$

Squaring both sides:

$|\vec{r}|^2 = |\vec{b}|^2 + \dfrac{1}{81}|\vec{a}|^2 + \dfrac{2}{9}(\vec{b} \cdot \vec{a})$

$|\vec{r}|^2 = 5 + \dfrac{1}{81}(9) + \dfrac{2}{9}(-1)$

$|\vec{r}|^2 = 5 + \dfrac{1}{9} - \dfrac{2}{9} = \dfrac{44}{9}$

Finally, $|3\vec{r}|^2 = 9|\vec{r}|^2 = 9 \times \dfrac{44}{9} = 44$

Answer: $44$

Hence, the correct answer is option: 'A'.

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