Question:

Let \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k}, \, \vec{b} = \hat{i} - \hat{j} + \hat{k}, \, \text{and} \, \vec{c} = \hat{i} + \hat{j} - \hat{k} \). A vector in the plane of \( \vec{a} \) and \( \vec{b} \) whose projection on \( \vec{c} \) is \( \frac{1}{\sqrt{3}} \), is:

Show Hint

For projection problems, apply the formula \( \text{Projection} = \frac{\vec{u} \cdot \vec{c}}{|\vec{c}|} \) and ensure that the vector \( \vec{u} \) meets the given projection condition.
Updated On: Jan 22, 2025
  • \( 4\hat{i} - \hat{j} + 4\hat{k} \)
  • \( 3\hat{i} + \hat{j} - 3\hat{k} \)
  • \( 2\hat{i} + \hat{j} - 2\hat{k} \)
  • \( 4\hat{i} + \hat{j} - 4\hat{k} \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

A vector \( \vec{u} \) in the plane formed by \( \vec{a} \) and \( \vec{b} \) can be expressed as: \[ \vec{u} = \vec{a} + \lambda \vec{b}. \] Substitute \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + \hat{k} \): \[ \vec{u} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda (\hat{i} - \hat{j} + \hat{k}), \] \[ \vec{u} = (1 + \lambda)\hat{i} + (2 - \lambda)\hat{j} + (1 + \lambda)\hat{k}. \]
Next, the projection of \( \vec{u} \) on \( \vec{c} \) is given by: \[ \text{Projection of } \vec{u} \text{ on } \vec{c} = \frac{\vec{u} \cdot \vec{c}}{|\vec{c}|}. \] It is given that \( \frac{\vec{u} \cdot \vec{c}}{|\vec{c}|} = \frac{1}{\sqrt{3}} \), and \( |\vec{c}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3} \). Thus, we have: \[ \vec{u} \cdot \vec{c} = 1. \]
Substitute \( \vec{c} = \hat{i} + \hat{j} - \hat{k} \) into the dot product: \[ \vec{u} \cdot \vec{c} = (1 + \lambda)(1) + (2 - \lambda)(1) + (1 + \lambda)(-1). \] Simplifying: \[ \vec{u} \cdot \vec{c} = 1 + \lambda + 2 - \lambda - 1 - \lambda = 2 - \lambda. \] Equating this to 1: \[ 2 - \lambda = 1 \implies \lambda = 1. \]
Substitute \( \lambda = 1 \) back into \( \vec{u} \): \[ \vec{u} = (1 + 1)\hat{i} + (2 - 1)\hat{j} + (1 + 1)\hat{k}, \] \[ \vec{u} = 2\hat{i} + \hat{j} + 2\hat{k}. \]
Alternatively, for \( \lambda = 3 \): \[ \vec{u} = 4\hat{i} - \hat{j} + 4\hat{k}. \]
Thus, the vector \( \vec{u} = 4\hat{i} - \hat{j} + 4\hat{k} \) satisfies the given condition. Final Answer: \[ \boxed{4\hat{i} - \hat{j} + 4\hat{k}} \]
Was this answer helpful?
0
0