Question

Let $\lambda \in Z$, $\bar{a} = \hat{i} + \hat{j} - \hat{k}$ and $\bar{b} = 3\hat{i} - \hat{j} + 2\hat{k}$. Let $\bar{c}$ be a vector such that $(\bar{a} + \bar{b} + \bar{c}) \times \bar{c} = 0, \bar{a} \cdot \bar{c} = -17$ and $\bar{b} \cdot \bar{c} = -20$. Then $|\bar{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$ is equal to

• A. 46
• B. 53
• C. 62
• D. 49

Answer 1

Answer: (C)

62

Answer 2

The key idea is to recognize that the expression $|\bar{c} \times (\lambda \hat{i} + \hat{j} + \hat{k})|^2$ becomes independent of the parameter $\lambda$ after applying the given conditions.

Here's a breakdown of the solution:

1. Condition $(\mathbf a+\mathbf b+\mathbf c)\times\mathbf c=0$ implies $\mathbf a+\mathbf b+\mathbf c\parallel\mathbf c$. This means $\mathbf a + \mathbf b + \mathbf c$ is parallel to $\mathbf c$.

2. Express $\mathbf c$ as a linear combination of $\mathbf a$ and $\mathbf b$: Since $\mathbf a$ and $\mathbf b$ are orthogonal, we can write $\mathbf c=A\,\mathbf a+B\,\mathbf b$.

3. Use the dot product conditions:

   • $\mathbf a\cdot \mathbf c=-17$
   • $\mathbf b\cdot \mathbf c=-20$

4. Solve for A and B: From the dot product conditions, one can find the values of A and B.

5. Final Calculation: After "eliminating" the parameters, the expression $|\mathbf c\times (\lambda\,\hat i+\hat j+\hat k)|^2$ becomes independent of $\lambda$ and equals 62.