Question

Let $\vec{a} = 4\hat{i} - \hat{j} + \hat{k}$, $\vec{b} = 11\hat{i} - \hat{j} + \hat{k}$, and $\vec{c}$ be a vector such that $(\vec{a} + \vec{b}) \times \vec{c} = \vec{c} \times (-2\vec{a} + 3\vec{b}).$ If $(2\vec{a} + 3\vec{b}) \cdot \vec{c} = 1670$, then $|\vec{c}|^2$ is equal to:

• A. 1627
• B. 1618
• C. 1600
• D. 1609

Answer 1

Answer: (B)

1618

Answer 2

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

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

This implies that: $\vec{c} = \lambda (4\vec{b} - \vec{a}).$

We substitute the values of $\vec{a}$ and $\vec{b}$: $\vec{c} = \lambda \left( 4(11\hat{i} - \hat{j} + \hat{k}) - (4\hat{i} - \hat{j} + \hat{k}) \right),$ $\vec{c} = \lambda (40\hat{i} - 3\hat{j} + 3\hat{k}).$

Given that: $(2\vec{a} + 3\vec{b}) \cdot \vec{c} = 1670.$

Calculating $2\vec{a} + 3\vec{b}$: $2\vec{a} + 3\vec{b} = 2(4\hat{i} - \hat{j} + \hat{k}) + 3(11\hat{i} - \hat{j} + \hat{k}),$ $2\vec{a} + 3\vec{b} = (8 + 33)\hat{i} + (-2 - 3)\hat{j} + (2 + 3)\hat{k},$ $2\vec{a} + 3\vec{b} = 41\hat{i} - 5\hat{j} + 5\hat{k}.$

Now, we compute: $(2\vec{a} + 3\vec{b}) \cdot \vec{c} = (41\hat{i} - 5\hat{j} + 5\hat{k}) \cdot \lambda (40\hat{i} - 3\hat{j} + 3\hat{k}),$ $(2\vec{a} + 3\vec{b}) \cdot \vec{c} = \lambda (41 \cdot 40 + (-5) \cdot (-3) + 5 \cdot 3),$ $1670 = \lambda (1640 + 15 + 15),$ $1670 = 1670\lambda \implies \lambda = 1.$

Thus: $\vec{c} = 40\hat{i} - 3\hat{j} + 3\hat{k}.$

Calculating $|\vec{c}|^2$: $|\vec{c}|^2 = 40^2 + (-3)^2 + 3^2,$ $|\vec{c}|^2 = 1600 + 9 + 9,$ $|\vec{c}|^2 = 1618.$

Therefore: $1618.$