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 solve this problem, we are given two vectors \(\vec{a} = 3\hat{i} + \hat{j}\) and \(\vec{b} = \hat{i} + 2\hat{j} + \hat{k}\), and we need to find the value of \(\lambda\) in the equation:

\(\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \lambda \vec{c}\)

Here, \(\vec{b}\) and \(\vec{c}\) are non-parallel vectors.

First, apply the vector triple product identity:

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

Equating this to the given identity:

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

Matching coefficients, we get two equations:

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

Next, compute the dot products:

1. Calculate \((\vec{a} \cdot \vec{b})\):

\(\vec{a} \cdot \vec{b} = (3\hat{i} + \hat{j}) \cdot (\hat{i} + 2\hat{j} + \hat{k}) = 3(1) + 1(2) + 0 = 3 + 2 = 5\)

Substitute the result to find \(\lambda\):

\(-\vec{a} \cdot \vec{b} = -(5) = -5\)

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

Since our calculation aligns with one of the options given, we conclude:

The correct answer is -5.