JEE Mathematics 2023 Question with Solution

Given: $\vec{a} = \hat{i} + 2\hat{k}$, $\vec{b} = \hat{i} + \hat{j} + \hat{k}$, $\vec{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}$, with

$$\vec{r} \times \vec{b} + \vec{b} \times \vec{c} = 0$$

and $\vec{r} \cdot \vec{a} = 0$.

Find: $\vec{r} \cdot \vec{c}$.

From

$$\vec{r} \times \vec{b} + \vec{b} \times \vec{c} = 0$$

we get

$$\vec{r} \times \vec{b} = -\vec{b} \times \vec{c}$$

So, $\vec{r}$ can be written as

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

for some scalar $\lambda$.

Now use $\vec{r} \cdot \vec{a} = 0$:

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

Compute the dot products:

$$\vec{c} \cdot \vec{a} = (7)(1) + (-3)(0) + (4)(2) = 15$$ $$\vec{b} \cdot \vec{a} = (1)(1) + (1)(0) + (1)(2) = 3$$

Substituting,

$$15 + 3\lambda = 0$$ $$\lambda = -5$$

Hence,

$$\vec{r} = \vec{c} + \lambda \vec{b} = (7\hat{i} - 3\hat{j} + 4\hat{k}) - 5(\hat{i} + \hat{j} + \hat{k})$$ $$\vec{r} = 2\hat{i} - 8\hat{j} - \hat{k}$$

Now calculate:

$$\vec{r} \cdot \vec{c} = (2)(7) + (-8)(-3) + (-1)(4)$$ $$\vec{r} \cdot \vec{c} = 14 + 24 - 4 = 34$$

Therefore, $\vec{r} \cdot \vec{c} = 34$ and the correct option is A.

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

Find: $\vec{r} \cdot \vec{c}$.

Notice that if two vectors have the same cross product with $\vec{b}$, then they differ by a vector parallel to $\vec{b}$. Therefore,

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

Using the orthogonality condition,

$$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0$$ $$15 + 3\lambda = 0$$ $$\lambda = -5$$

So,

$$\vec{r} = \vec{c} - 5\vec{b}$$

Then

$$\vec{r} \cdot \vec{c} = (\vec{c} - 5\vec{b}) \cdot \vec{c}$$

Using the computed vector from the working,

$$\vec{r} = 2\hat{i} - 8\hat{j} - \hat{k}$$

Hence,

$$\vec{r} \cdot \vec{c} = 34$$

Therefore, the correct option is A.