JEE Mathematics 2023 Question with Solution

Given:

• $\mathbf{a} = \hat{i} + 2\hat{j} + \lambda \hat{k}$
• $\mathbf{b} = 3\hat{i} - 5\hat{j} - \lambda \hat{k}$
• $\mathbf{a} \cdot \mathbf{c} = 7$
• $2\mathbf{b} \cdot \mathbf{c} + 43 = 0$
• $\mathbf{a} \times \mathbf{c} = \mathbf{b} \times \mathbf{c}$

Find: $|\mathbf{a} \cdot \mathbf{b}|$

From

$$\mathbf{a} \times \mathbf{c} - \mathbf{b} \times \mathbf{c} = \mathbf{0}$$

we get

$$(\mathbf{a} - \mathbf{b}) \times \mathbf{c} = \mathbf{0}$$

Hence $\mathbf{a} - \mathbf{b}$ is parallel to $\mathbf{c}$. So let

$$\mathbf{a} - \mathbf{b} = \mu \mathbf{c}$$

where $\mu$ is a scalar.

Now

$$\mathbf{a} - \mathbf{b} = (\hat{i} + 2\hat{j} + \lambda \hat{k}) - (3\hat{i} - 5\hat{j} - \lambda \hat{k})$$ $$= -2\hat{i} + 7\hat{j} + 2\lambda \hat{k}$$

So

$$-2\hat{i} + 7\hat{j} + 2\lambda \hat{k} = \mu \mathbf{c}$$

Using $\mathbf{a} \cdot \mathbf{c} = 7$, the solution gives

$$2\lambda^2 + 12 = 7\mu$$

Also, from $2\mathbf{b} \cdot \mathbf{c} + 43 = 0$, we have $\mathbf{b} \cdot \mathbf{c} = -\frac{43}{2}$, which gives

$$4\lambda^2 + 82 = 43\mu$$

Solving these, we get

$$\mu = 2$$

and

$$\lambda^2 = 1$$

Now compute

$$\mathbf{a} \cdot \mathbf{b} = (1)(3) + (2)(-5) + (\lambda)(-\lambda)$$ $$= 3 - 10 - \lambda^2 = -7 - 1 = -8$$

Therefore,

$$|\mathbf{a} \cdot \mathbf{b}| = 8$$

So the required numerical value is $8$.

Given: $\mathbf{a} \times \mathbf{c} = \mathbf{b} \times \mathbf{c}$ with the stated dot-product conditions.

Find: $|\mathbf{a} \cdot \mathbf{b}|$

Since

$$\mathbf{a} \times \mathbf{c} = \mathbf{b} \times \mathbf{c}$$

we get

$$(\mathbf{a} - \mathbf{b}) \times \mathbf{c} = \mathbf{0}$$

Therefore $\mathbf{c}$ is parallel to $\mathbf{a} - \mathbf{b}$. Let

$$\mathbf{c} = k(\mathbf{a} - \mathbf{b})$$

Now

$$\mathbf{a} - \mathbf{b} = (1-3)\hat{i} + (2-(-5))\hat{j} + (\lambda-(-\lambda))\hat{k}$$ $$= -2\hat{i} + 7\hat{j} + 2\lambda \hat{k}$$

Hence

$$\mathbf{c} = k(-2\hat{i} + 7\hat{j} + 2\lambda \hat{k})$$

Use $\mathbf{a} \cdot \mathbf{c} = 7$:

$$(\hat{i} + 2\hat{j} + \lambda \hat{k}) \cdot k(-2\hat{i} + 7\hat{j} + 2\lambda \hat{k}) = 7$$ $$k(-2 + 14 + 2\lambda^2) = 7$$ $$k(12 + 2\lambda^2) = 7$$

Also,

$$2\mathbf{b} \cdot \mathbf{c} + 43 = 0$$

so

$$\mathbf{b} \cdot \mathbf{c} = -\frac{43}{2}$$

Now use $\mathbf{b} = 3\hat{i} - 5\hat{j} - \lambda \hat{k}$:

$$(3\hat{i} - 5\hat{j} - \lambda \hat{k}) \cdot k(-2\hat{i} + 7\hat{j} + 2\lambda \hat{k}) = -\frac{43}{2}$$ $$k(-6 - 35 - 2\lambda^2) = -\frac{43}{2}$$ $$k(41 + 2\lambda^2) = \frac{43}{2}$$

From these two relations, the extracted solution yields

$$\lambda^2 = 1$$

and hence

$$\mathbf{a} \cdot \mathbf{b} = 3 - 10 - 1 = -8$$

Therefore,

$$|\mathbf{a} \cdot \mathbf{b}| = 8$$

Thus the answer is $8$.