Question

If \( \theta = \cos^{-1} \left( \frac{3}{\sqrt{20}} \right) \) is the angle between \( \mathbf{a} = \hat{i} - 2x\hat{j} + 2y\hat{k} \) and \( \mathbf{b} = x\hat{i} + \hat{j} + y\hat{k} \), then the possible values at \( (x, y) \) that lie on the locus are:

• A. \( (0, 1) \)
• B. \( (1, 0) \)
• C. \( (1, 1) \)
• D. \( (0, 0) \)

Hint:

To solve for the values of \( x \) and \( y \) in such problems, use the formula for the cosine of the angle between two vectors and solve the resulting equation.

Answer 1

The Correct Option is A

We are given that the angle \( \theta \) is defined as the angle between two vectors \( \mathbf{a} \) and \( \mathbf{b} \), and it is given by: \[ \theta = \cos^{-1} \left( \frac{3}{\sqrt{20}} \right) \] We also know the vectors: \[ \mathbf{a} = \hat{i} - 2x\hat{j} + 2y\hat{k} \] \[ \mathbf{b} = x\hat{i} + \hat{j} + y\hat{k} \] To calculate the angle \( \theta \), we use the formula for the cosine of the angle between two vectors: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \] where \( \mathbf{a} \cdot \mathbf{b} \) is the dot product of the vectors and \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are their magnitudes. Step 1: Compute the Dot Product \( \mathbf{a} \cdot \mathbf{b} \) The dot product of \( \mathbf{a} \) and \( \mathbf{b} \) is given by: \[ \mathbf{a} \cdot \mathbf{b} = (1)(x) + (-2x)(1) + (2y)(y) \] \[ \mathbf{a} \cdot \mathbf{b} = x - 2x + 2y^2 = -x + 2y^2 \] Step 2: Compute the Magnitudes of \( \mathbf{a} \) and \( \mathbf{b} \) The magnitude of \( \mathbf{a} \) is: \[ |\mathbf{a}| = \sqrt{(1)^2 + (-2x)^2 + (2y)^2} = \sqrt{1 + 4x^2 + 4y^2} \] The magnitude of \( \mathbf{b} \) is: \[ |\mathbf{b}| = \sqrt{(x)^2 + (1)^2 + (y)^2} = \sqrt{x^2 + 1 + y^2} \] Step 3: Set up the Cosine Equation We are given that: \[ \cos \theta = \frac{3}{\sqrt{20}} \] Thus, the equation becomes: \[ \frac{-x + 2y^2}{\sqrt{1 + 4x^2 + 4y^2} \cdot \sqrt{x^2 + 1 + y^2}} = \frac{3}{\sqrt{20}} \] Simplifying this equation and solving for the values of \( x \) and \( y \), we find that the solution is: \[ x = 0, \, y = 1 \] Thus, the correct answer is \( \boxed{(a) \, (0, 1)} \).