Given that \( \vec{u} = x\hat{i} + y\hat{j} + z\hat{k} \) is a unit vector, it satisfies: \(x^2 + y^2 + z^2 = 1\)
Step 1. Using the angle conditions:
- The angle between \( \vec{u} \) and \( \frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}} \) is \( \frac{\pi}{2} \):
\(\vec{u} \cdot \left( \frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}} \right) = 0 \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = 0\)
\(x + y = 0\) ---(1)
- The angle between \( \vec{u} \) and \( \frac{\hat{i}}{\sqrt{2}} + \hat{j} + \frac{\hat{k}}{\sqrt{2}} \) is \( \frac{\pi}{3} \):
\(\vec{u} \cdot \left( \frac{\hat{i}}{\sqrt{2}} + \hat{j} + \frac{\hat{k}}{\sqrt{2}} \right) = \frac{1}{2} \implies \frac{x}{\sqrt{2}} + y + \frac{z}{\sqrt{2}} = \frac{1}{2}\)
\(x + \sqrt{2}y + z = \frac{\sqrt{2}}{2}\)
- The angle between \( \vec{u} \) and \( \frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}} + \hat{k} \) is \( \frac{\pi}{2} \):
\(\vec{u} \cdot \left( \frac{\hat{i}}{\sqrt{2}} + \frac{\hat{j}}{\sqrt{2}} + \hat{k} \right) = 0 \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} + z = 0\)
\(x + y + \sqrt{2}z = 0\)
Step 2. Solving the system of equations:** From equations (1), (2), and (3):
- Substitute \( z = -x \) in (2):
\(x + \sqrt{2}y - x = \frac{\sqrt{2}}{2} \implies y = \frac{1}{\sqrt{2}}\)
- Substitute \( y = \frac{1}{\sqrt{2}} \) and \( z = -x \) in (3):
\(x + \frac{1}{\sqrt{2}} + \sqrt{2}(-x) = 0 \implies x = -\frac{1}{2\sqrt{2}}, \, z = \frac{1}{2\sqrt{2}}\)
Step 3. Calculate \( |\vec{u} - \vec{v}|^2 \):
\(|\vec{u} - \vec{v}|^2 = \left( x - \frac{1}{\sqrt{2}} \right)^2 + \left( y - \frac{1}{\sqrt{2}} \right)^2 + \left( z - \frac{1}{\sqrt{2}} \right)^2\)
Substituting \( x = -\frac{1}{2\sqrt{2}}, \, y = \frac{1}{\sqrt{2}}, \, z = \frac{1}{2\sqrt{2}} \):
\(|\vec{u} - \vec{v}|^2 = \frac{5}{2}\)
The Correct answer is :\( \frac{5}{2} \).
Let \( \vec{a} \) and \( \vec{b} \) be two co-initial vectors forming adjacent sides of a parallelogram such that:
\[
|\vec{a}| = 10, \quad |\vec{b}| = 2, \quad \vec{a} \cdot \vec{b} = 12
\]
Find the area of the parallelogram.
A bob of mass \(m\) is suspended at a point \(O\) by a light string of length \(l\) and left to perform vertical motion (circular) as shown in the figure. Initially, by applying horizontal velocity \(v_0\) at the point ‘A’, the string becomes slack when the bob reaches at the point ‘D’. The ratio of the kinetic energy of the bob at the points B and C is: