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

To solve this problem, we need to analyze the given vectors and conditions. We have:

• \(\overrightarrow{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k}\)
• \(\overrightarrow{AD} = \hat{i} + 2\hat{j} + \lambda \hat{k}\)
• The projection of the vector \(\vec{v} = \hat{i} + \hat{j} + \hat{k}\) on \(\overrightarrow{AC}\) is of length 1 unit.

Let's find the vector \(\overrightarrow{AC}\) using vector addition:

• \(\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}\)
• \(= (2\hat{i} + 4\hat{j} - 5\hat{k}) + (\hat{i} + 2\hat{j} + \lambda \hat{k})\)
• \(= 3\hat{i} + 6\hat{j} + (\lambda - 5)\hat{k}\)

The length of the projection of \(\vec{v}\) on \(\overrightarrow{AC}\) is 1, thus:

\(\text{Projection of } \vec{v} \text{ on } \overrightarrow{AC} = \frac{\vec{v} \cdot \overrightarrow{AC}}{\|\overrightarrow{AC}\|}\)

Given that this length is 1, we write:

\(\left|\frac{\vec{v} \cdot \overrightarrow{AC}}{\|\overrightarrow{AC}\|}\right| = 1\)

Calculate the dot product:

• \(\vec{v} \cdot \overrightarrow{AC} = (1\hat{i} + 1\hat{j} + 1\hat{k}) \cdot (3\hat{i} + 6\hat{j} + (\lambda - 5)\hat{k})\)
• \(= 1 \cdot 3 + 1 \cdot 6 + 1 \cdot (\lambda - 5)\)
• \(= 9 + \lambda - 5 = \lambda + 4\)

Now calculate the magnitude of \(\overrightarrow{AC}\):

\(\|\overrightarrow{AC}\| = \sqrt{3^2 + 6^2 + (\lambda - 5)^2} = \sqrt{9 + 36 + (\lambda - 5)^2}\)

Simplifying role:

• \(= \sqrt{45 + \lambda^2 - 10\lambda + 25}\)
• \(= \sqrt{\lambda^2 - 10\lambda + 70}\)

Use the projection equation:

\(\left|\frac{\lambda + 4}{\sqrt{\lambda^2 - 10\lambda + 70}}\right| = 1\)

Implying:

• \(\lambda + 4 = \pm \sqrt{\lambda^2 - 10\lambda + 70}\)

Squaring both sides:

• \((\lambda + 4)^2 = \lambda^2 - 10\lambda + 70\)
• \(=\lambda^2 + 8\lambda + 16 = \lambda^2 - 10\lambda + 70\)
• \(=18\lambda = 54\)
• \(\lambda = 3\)

Substitute back to equation:

• \(\lambda^2x^2 - 6\lambda x + 5 = 0\)
  Find the roots using the quadratic formula:
• \(x = \frac{18 \pm \sqrt{18^2 - 4 \cdot 9 \cdot 5}}{2 \cdot 9}\)
• \(= \frac{18 \pm \sqrt{324 - 180}}{18}\)
• \(= \frac{18 \pm \sqrt{144}}{18}\)
• \(= \frac{18 \pm 12}{18}\)
• Roots: \(\alpha = \frac{30}{18} = \frac{5}{3}\)and \(\beta = \frac{6}{18} = \frac{1}{3}\)

Finally, calculate \(2\alpha - \beta\):

• \(= 2 \cdot \frac{5}{3} - \frac{1}{3}\)
• \(= \frac{10}{3} - \frac{1}{3}\)
• = \(\frac{9}{3} = 3\)

Thus, the value of \(2\alpha - \beta\) is 3.