Question

Let $ \vec{a} = \hat{i} + 4\hat{j} + 3\hat{k} $ and $ \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} $, and $ \vec{c} $ is a vector perpendicular to $ \vec{a} $ and lies in the plane of $ \vec{a} $, $ \vec{b} $. Then $ \vec{c} $ is equal to:

• A. \( \hat{i} + \hat{j} + \hat{k} \)
• B. \( -\hat{i} + \hat{j} - \hat{k} \)
• C. \( \hat{i} - \hat{j} + \hat{k} \)
• D. \( -\hat{i} - \hat{j} - \hat{k} \)

Hint:

For problems involving vectors perpendicular to multiple vectors, use the cross product to find a vector perpendicular to both.

Answer 1

The Correct Option is B

Given vectors \( \vec{a} = \hat{i} + 4\hat{j} + 3\hat{k} \) and \( \vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k} \). We need to determine vector \( \vec{c} \), which is orthogonal to \( \vec{a} \) and coplanar with \( \vec{a} \) and \( \vec{b} \).
Condition 1: Orthogonality As \( \vec{c} \) is orthogonal to \( \vec{a} \), their dot product is zero:\[\vec{a} \cdot \vec{c} = 0\]
Condition 2: Coplanarity Since \( \vec{c} \) lies in the plane formed by \( \vec{a} \) and \( \vec{b} \), it can be expressed as a linear combination of these vectors:\[\vec{c} = \lambda \vec{a} + \mu \vec{b}\]where \( \lambda \) and \( \mu \) are scalar coefficients.
Applying Dot Product for Orthogonality Substitute the expression for \( \vec{c} \) into the orthogonality condition:\[\vec{a} \cdot (\lambda \vec{a} + \mu \vec{b}) = 0\]Expand the equation:\[\lambda (\vec{a} \cdot \vec{a}) + \mu (\vec{a} \cdot \vec{b}) = 0\]Calculate the dot products: \( \vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1^2 + 4^2 + 3^2 = 1 + 16 + 9 = 26 \) and \( \vec{a} \cdot \vec{b} = (1)(2) + (4)(3) + (3)(4) = 2 + 12 + 12 = 26 \).This yields the equation:\[\lambda (26) + \mu (26) = 0 \quad \Rightarrow \quad \lambda + \mu = 0\]
Determining Vector \( \vec{c} \) From \( \lambda + \mu = 0 \), we get \( \mu = -\lambda \). Substitute this back into the expression for \( \vec{c} \):\[\vec{c} = \lambda \vec{a} + (-\lambda) \vec{b} = \lambda (\vec{a} - \vec{b})\]Calculate the difference \( \vec{a} - \vec{b} \):\[\vec{a} - \vec{b} = (\hat{i} + 4\hat{j} + 3\hat{k}) - (2\hat{i} + 3\hat{j} + 4\hat{k}) = (\hat{i} - 2\hat{i}) + (4\hat{j} - 3\hat{j}) + (3\hat{k} - 4\hat{k}) = -\hat{i} + \hat{j} - \hat{k}\]Therefore, \( \vec{c} = \lambda (-\hat{i} + \hat{j} - \hat{k}) \). For simplicity, we can choose \( \lambda = 1 \).Thus, the resulting vector is \( \vec{c} = -\hat{i} + \hat{j} - \hat{k} \).