JEE Mathematics 2023 Question with Solution

Given: $\vec{a} = 2\hat{i} - 7\hat{j} + 5\hat{k}$, $\vec{b} = \hat{i} + \hat{k}$, $\vec{c} = \hat{i} + 2\hat{j} - 3\hat{k}$.

Find: $|\vec{r}|$, given that $\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$ and $\vec{r} \cdot \vec{b} = 0$.

From

$$\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$$

we get

$$(\vec{r} - \vec{c}) \times \vec{a} = 0.$$

This implies that $\vec{r} - \vec{c}$ is parallel to $\vec{a}$, so

$$\vec{r} = \vec{c} + \lambda \vec{a}$$

for some scalar $\lambda$.

Using $\vec{r} \cdot \vec{b} = 0$,

$$(\vec{c} + \lambda \vec{a}) \cdot \vec{b} = 0$$

which gives

$$\vec{c} \cdot \vec{b} + \lambda (\vec{a} \cdot \vec{b}) = 0.$$

Now,

$$\vec{c} \cdot \vec{b} = (\hat{i} + 2\hat{j} - 3\hat{k}) \cdot (\hat{i} + \hat{k}) = 1 + 0 - 3 = -2$$

and

$$\vec{a} \cdot \vec{b} = (2\hat{i} - 7\hat{j} + 5\hat{k}) \cdot (\hat{i} + \hat{k}) = 2 + 0 + 5 = 7.$$

So,

$$-2 + 7\lambda = 0$$

which gives

$$\lambda = \frac{2}{7}.$$

Hence,

$$\vec{r} = (\hat{i} + 2\hat{j} - 3\hat{k}) + \frac{2}{7}(2\hat{i} - 7\hat{j} + 5\hat{k}).$$

Expanding,

$$\vec{r} = \hat{i} + 2\hat{j} - 3\hat{k} + \frac{4}{7}\hat{i} - 2\hat{j} + \frac{10}{7}\hat{k}.$$

Therefore,

$$\vec{r} = \left(1 + \frac{4}{7}\right)\hat{i} + (2 - 2)\hat{j} + \left(-3 + \frac{10}{7}\right)\hat{k}$$

so

$$\vec{r} = \frac{11}{7}\hat{i} - \frac{11}{7}\hat{k}.$$

Now,

$$|\vec{r}| = \sqrt{\left(\frac{11}{7}\right)^2 + \left(-\frac{11}{7}\right)^2}$$

which becomes

$$|\vec{r}| = \sqrt{\frac{121}{49} + \frac{121}{49}} = \sqrt{\frac{242}{49}} = \frac{11}{7}\sqrt{2}.$$

Therefore, the magnitude of $\vec{r}$ is $\frac{11}{7}\sqrt{2}$, so the correct option is A.

Given: $\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$ and $\vec{r} \cdot \vec{b} = 0$.

Find: $|\vec{r}|$.

Since

$$(\vec{r} - \vec{c}) \times \vec{a} = 0,$$

vector $\vec{r} - \vec{c}$ must be parallel to $\vec{a}$. So write directly

$$\vec{r} = \vec{c} + \lambda \vec{a}.$$

Now enforce the perpendicularity condition with $\vec{b}$:

$$(\vec{c} + \lambda \vec{a}) \cdot \vec{b} = 0.$$

That gives

$$\lambda = -\frac{\vec{c} \cdot \vec{b}}{\vec{a} \cdot \vec{b}} = -\frac{-2}{7} = \frac{2}{7}.$$

Hence,

$$\vec{r} = \vec{c} + \frac{2}{7}\vec{a} = \frac{11}{7}\hat{i} - \frac{11}{7}\hat{k}.$$

So,

$$|\vec{r}| = \sqrt{\left(\frac{11}{7}\right)^2 + \left(\frac{11}{7}\right)^2} = \frac{11}{7}\sqrt{2}.$$

This shortcut works because the cross product condition immediately restricts $\vec{r}$ to a line parallel to $\vec{a}$ through $\vec{c}$, and the dot product condition then determines the scalar uniquely. Therefore, the correct option is A.