Search any question & find its solution
Question:
Answered & Verified by Expert
If the sum of two unit vectors is a unit vector, then the magnitude of their difference is
Options:
Solution:
1729 Upvotes
Verified Answer
The correct answer is:
$\sqrt{3}$ units
$\begin{aligned} &\text { Let a and } \mathbf{b} \text { are two unit vectors. }\\ &\text { Then, } \quad|\mathbf{a}+\mathbf{b}|=1\\ &\Rightarrow \quad|\mathbf{a}+\mathbf{b}|^{2}=1 \end{aligned}$
$\Rightarrow \quad(a+b) \cdot(a+b)=1$
$\Rightarrow|a|^{2}+|b|^{2}+2 a \cdot b=1$
$\Rightarrow \quad 1+1+2 \mathbf{a} \cdot \mathbf{b}=1$
$\Rightarrow \quad 2 \mathbf{a} \cdot \mathbf{b}=-1$
Again, $|\hat{\mathbf{a}}-\mathbf{b}|^{2}=(\mathbf{a}-\hat{\mathbf{b}}) \cdot(\mathbf{a}-\hat{\mathbf{b}})$
$=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}-2 \mathbf{a} \cdot \mathbf{b}$
$=1+1-(-1)$
$=1+1+1=3$
$|\hat{a}-\hat{b}|=\sqrt{3}$
$\Rightarrow \quad(a+b) \cdot(a+b)=1$
$\Rightarrow|a|^{2}+|b|^{2}+2 a \cdot b=1$
$\Rightarrow \quad 1+1+2 \mathbf{a} \cdot \mathbf{b}=1$
$\Rightarrow \quad 2 \mathbf{a} \cdot \mathbf{b}=-1$
Again, $|\hat{\mathbf{a}}-\mathbf{b}|^{2}=(\mathbf{a}-\hat{\mathbf{b}}) \cdot(\mathbf{a}-\hat{\mathbf{b}})$
$=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}-2 \mathbf{a} \cdot \mathbf{b}$
$=1+1-(-1)$
$=1+1+1=3$
$|\hat{a}-\hat{b}|=\sqrt{3}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.