Let $$\vec{a} = 4\hat{i} - \hat{j} + 3\hat{k}$$, $$\vec{b} = 10\hat{i} + 2\hat{j} - \hat{k}$$ and a vector $$\vec{c}$$ be such that $$2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{c}) = \vec{0}$$. If $$\vec{a} \cdot \vec{c} = 15$$, then $$\vec{c} \cdot (\hat{i} + \hat{j} - 3\hat{k})$$ is equal to :

To solve this efficiently, we use the properties of vector cross products.

Given: $$2(\vec{a} \times \vec{b}) + 3(\vec{b} \times \vec{c}) = \vec{0}$$

Rearrange: $$3(\vec{b} \times \vec{c}) = -2(\vec{a} \times \vec{b})$$

Using the property $$\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$$, we get:

$$3(\vec{b} \times \vec{c}) = 2(\vec{b} \times \vec{a})$$

$$\vec{b} \times 3\vec{c} = \vec{b} \times 2\vec{a}$$

This implies that $$3\vec{c} - 2\vec{a}$$ is parallel to $$\vec{b}$$:

$$3\vec{c} = 2\vec{a} + \lambda\vec{b}$$

$$\vec{c} = \frac{2\vec{a} + \lambda\vec{b}}{3}$$

Use the condition $$\vec{a} \cdot \vec{c} = 15$$:

$$\vec{a} \cdot \left( \frac{2\vec{a} + \lambda\vec{b}}{3} \right) = 15$$

$$2|\vec{a}|^2 + \lambda(\vec{a} \cdot \vec{b}) = 45$$

• $$|\vec{a}|^2 = 4^2 + (-1)^2 + 3^2 = 16 + 1 + 9 = 26$$
• $$\vec{a} \cdot \vec{b} = (4)(10) + (-1)(2) + (3)(-1) = 40 - 2 - 3 = 35$$
• $$\vec{a} \cdot \vec{d} = (4)(1) + (-1)(1) + (3)(-3) = 4 - 1 - 9 = -6$$
• $$\vec{b} \cdot \vec{d} = (10)(1) + (2)(1) + (-1)(-3) = 10 + 2 + 3 = 15$$

Substitute:

$$2(26) + \lambda(35) = 45$$

$$52 + 35\lambda = 45$$

$$35\lambda = -7 \implies \lambda = -\frac{1}{5}$$

We need $$\vec{c} \cdot \vec{d}$$ where $$\vec{d} = \hat{i} + \hat{j} - 3\hat{k}$$.

$$\vec{c} \cdot \vec{d} = \frac{2(\vec{a} \cdot \vec{d}) + \lambda(\vec{b} \cdot \vec{d})}{3}$$

$$\vec{c} \cdot \vec{d} = \frac{2(-6) + (-\frac{1}{5})(15)}{3}$$

$$\vec{c} \cdot \vec{d} = \frac{-12 - 3}{3} = \frac{-15}{3} = -5$$

Correct Answer: B (-5)