Question

Let \( \overrightarrow{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \overrightarrow{AD} = \hat{i} + 2\hat{j} + \lambda \hat{k} \), \( \lambda \in \mathbb{R} \). Let the projection of the vector \( \vec{v} = \hat{i} + \hat{j} + \hat{k} \) on the diagonal \( \overrightarrow{AC} \) of the parallelogram \( ABCD \) be of length one unit. If \( \alpha, \beta \), where \( \alpha>\beta \), be the roots of the equation \( \lambda^2 x^2 - 6\lambda x + 5 = 0 \), then \( 2\alpha - \beta \) is equal to

• A. 4
• B. 6
• C. 3
• D. 1

Hint:

For projection problems, always use the formula \(\left| \text{proj}_{\vec{b}} \vec{a} \right| = \dfrac{|\vec{a}\cdot\vec{b}|}{|\vec{b}|}\) and remember that diagonals of a parallelogram are obtained by vector addition.

Answer 1

The Correct Option is C

We are given vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AD} \) of a parallelogram. The diagonal \( \overrightarrow{AC} \) is given by \[ \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}. \] Step 1: Find the vector \( \overrightarrow{AC} \).
\[ \overrightarrow{AC} = (2 + 1)\hat{i} + (4 + 2)\hat{j} + (-5 + \lambda)\hat{k} = 3\hat{i} + 6\hat{j} + (\lambda - 5)\hat{k}. \] Step 2: Use the formula for projection of a vector.
The magnitude of the projection of \( \vec{v} \) on \( \overrightarrow{AC} \) is \[ \left| \text{proj}_{AC} \vec{v} \right| = \frac{|\vec{v} \cdot \overrightarrow{AC}|}{|\overrightarrow{AC}|}. \] Given this length is equal to 1, so \[ \frac{|\vec{v} \cdot \overrightarrow{AC}|}{|\overrightarrow{AC}|} = 1. \] Step 3: Compute dot product and magnitude.
\[ \vec{v} \cdot \overrightarrow{AC} = (1)(3) + (1)(6) + (1)(\lambda - 5) = \lambda + 4. \] \[ |\overrightarrow{AC}| = \sqrt{3^2 + 6^2 + (\lambda - 5)^2} = \sqrt{45 + (\lambda - 5)^2}. \] Thus, \[ \frac{|\lambda + 4|}{\sqrt{45 + (\lambda - 5)^2}} = 1. \] Step 4: Solve for \( \lambda \).
Squaring both sides, \[ (\lambda + 4)^2 = 45 + (\lambda - 5)^2. \] Expanding, \[ \lambda^2 + 8\lambda + 16 = \lambda^2 - 10\lambda + 25 + 45. \] \[ 18\lambda = 54 \Rightarrow \lambda = 3. \] Step 5: Form the quadratic equation.
Given equation: \[ \lambda^2 x^2 - 6\lambda x + 5 = 0. \] Substitute \( \lambda = 3 \): \[ 9x^2 - 18x + 5 = 0. \] Step 6: Find the roots.
\[ x = \frac{18 \pm \sqrt{324 - 180}}{18} = \frac{18 \pm 12}{18}. \] Thus, \[ \alpha = \frac{5}{3}, \quad \beta = \frac{1}{3}. \] Step 7: Compute \( 2\alpha - \beta \).
\[ 2\alpha - \beta = 2\left(\frac{5}{3}\right) - \frac{1}{3} = \frac{9}{3} = 3. \] Final Answer: \[ \boxed{3} \]