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Question: Answered & Verified by Expert
If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$ are unit vectors and $|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=1$, then $|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|$ is equal to
MathematicsVector AlgebraKCETKCET 2008
Options:
  • A $\sqrt{2}$
  • B 1
  • C $\sqrt{5}$
  • D $\sqrt{3}$
Solution:
2835 Upvotes Verified Answer
The correct answer is: $\sqrt{3}$
Given, $\quad|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=1$
$\therefore \quad|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|^{2}=1^{2}$
$\Rightarrow|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}+2|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|=1$
$\Rightarrow \quad 2|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|=1-(1+1)$
$\Rightarrow \quad 2|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|=-1$
Now, $|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|^{2}=|\overrightarrow{\mathbf{a}}|^{2}+|\overrightarrow{\mathbf{b}}|^{2}-2|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|$ $=1^{2}+1^{2}-(-1)=3$ [from Eq. (i)] $\Rightarrow \quad|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|=\sqrt{3}$

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