Question:

If \(|\bar{a}|=\sqrt{14},|\bar{b}|=\sqrt{10},|\bar{a}-\bar{b}|=\sqrt{24}\ and\ \theta\) is angle between \(\bar{a}\ and\ \bar{b}\), then \(\cos\theta=\)

Updated On: Apr 7, 2025
  • \(\frac{\sqrt{35}}{70}\)
  • \(\frac{\sqrt{6}}{12}\)
  • \(\frac{\sqrt{15}}{60}\)
  • \(\frac{\sqrt{210}}{35}\)
  • 0
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is

Approach Solution - 1

We are given the magnitudes of the vectors $\vec{a}$ and $\vec{b}$, and the magnitude of their difference. We can use the following identity for the magnitude of the difference between two vectors: \[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2|\vec{a}||\vec{b}|\cos \theta \] Substituting the given values: \[ |\vec{a} - \vec{b}|^2 = 24, \quad |\vec{a}| = \sqrt{14}, \quad |\vec{b}| = \sqrt{10} \] We have: \[ 24 = (\sqrt{14})^2 + (\sqrt{10})^2 - 2(\sqrt{14})(\sqrt{10})\cos \theta \] Simplifying: \[ 24 = 14 + 10 - 2\sqrt{140} \cos \theta \] \[ 24 = 24 - 2\sqrt{140} \cos \theta \] Subtract 24 from both sides: \[ 0 = -2\sqrt{140} \cos \theta \] \[ \cos \theta = 0 \]

The correct option is (E) : 0

Was this answer helpful?
0
1
Hide Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

We are given \(|\bar{a}| = \sqrt{14}\), \(|\bar{b}| = \sqrt{10}\), and \(|\bar{a} - \bar{b}| = \sqrt{24}\). We want to find \(\cos\theta\), where \(\theta\) is the angle between \(\bar{a}\) and \(\bar{b}\).

We know that \(|\bar{a} - \bar{b}|^2 = (\bar{a} - \bar{b}) \cdot (\bar{a} - \bar{b}) = \bar{a}\cdot\bar{a} - 2\bar{a}\cdot\bar{b} + \bar{b}\cdot\bar{b} = |\bar{a}|^2 - 2\bar{a}\cdot\bar{b} + |\bar{b}|^2\).

Therefore, \(|\bar{a} - \bar{b}|^2 = |\bar{a}|^2 + |\bar{b}|^2 - 2|\bar{a}||\bar{b}|\cos\theta\).

Substituting the given values, we have:

\((\sqrt{24})^2 = (\sqrt{14})^2 + (\sqrt{10})^2 - 2(\sqrt{14})(\sqrt{10})\cos\theta\)

\(24 = 14 + 10 - 2\sqrt{140}\cos\theta\)

\(24 = 24 - 2\sqrt{140}\cos\theta\)

\(0 = -2\sqrt{140}\cos\theta\)

This implies that \(\cos\theta = 0\).

Was this answer helpful?
0
0