Question

Let \(\vec{a} = 3\hat{i} + \hat{j}\) and \(\vec{b} = \hat{i} + 2\hat{j} + \hat{k}\).
Let \(\vec{c}\) be a vector satisfying \(\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda \vec{c}\)
If \(\vec{b}\) and \(\vec{c}\) are non-parallel, then the value of \(λ\) is

• A. -5
• B. 5
• C. 1
• D. -1

Answer 1

The Correct Option is A

To find the value of \( \lambda \) in the expression \( \vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda \vec{c} \), we start with the given vectors:

• \(\vec{a} = 3\hat{i} + \hat{j}\) 
• \(\vec{b} = \hat{i} + 2\hat{j} + \hat{k}\)

Using the vector triple product identity, we have:

\[\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}\]

Plugging in the given expression, we need:

\[(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \vec{b} + \lambda \vec{c}\]

From the equation above, we equate coefficients:

\[(\vec{a} \cdot \vec{c})\vec{b} = \vec{b}\]

\[-(\vec{a} \cdot \vec{b})\vec{c} = \lambda \vec{c}\]

This implies:

1. \(\vec{a} \cdot \vec{c} = 1\)
2. \(-(\vec{a} \cdot \vec{b}) = \lambda\)

Next, calculate \(\vec{a} \cdot \vec{b}\):

\[\vec{a} \cdot \vec{b} = (3\hat{i} + \hat{j}) \cdot (\hat{i} + 2\hat{j} + \hat{k})\]

\[= 3 \times 1 + 1 \times 2 + 0 = 3 + 2 = 5\]

Thus, \( \vec{a} \cdot \vec{b} = 5 \). Plugging this into the equation:

\[-5 = \lambda\]

Therefore, the value of \( \lambda \) is \(-5\).

The correct answer is \(-5\).

Answer 2

\(\vec{a} = 3\hat{i} + \hat{j} \quad \text{and} \quad \vec{b} = \hat{i} + 2\hat{j} + \hat{k}\)
\(\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \vec{b} + \lambda \vec{c}\)
If \(\vec{b} \quad \text{and} \quad \vec{c}\)  are non-parallel, then
\(\vec{a} \cdot \vec{c} = 1 \quad \text{and} \quad \vec{a} \cdot \vec{b} = -\lambda\)
but \(\vec{a} \cdot \vec{b} = 5\)
\(⇒λ=−5\)
So, the correct option is (A): -5