Search any question & find its solution
Question:
Answered & Verified by Expert
If the magnitudes of $\mathbf{a}, \mathbf{b}$ and $\mathbf{a}+\mathbf{b}$ are respectively 3,4 and 5 , then the magnitude of $(\mathbf{a}-\mathbf{b})$ is
Options:
Solution:
1733 Upvotes
Verified Answer
The correct answer is:
5
We have, $|\mathbf{a}|=3,|\mathbf{b}|=4$ and $|\mathbf{a}+\mathbf{b}|=5$
Since,
$\begin{gathered}
|\mathbf{a}+\mathbf{b}|=5 \\
|\mathbf{a}+\mathbf{b}|^2=25
\end{gathered}$
$\begin{aligned}
& \therefore \quad|\mathbf{a}+\mathbf{b}|^2=25 \\
& \Rightarrow \quad(\mathbf{a}+\mathbf{b}\rangle \cdot(\mathbf{a}+\mathbf{b}\rangle=25 \\
& \Rightarrow \mathbf{a} \cdot \mathbf{a}+\mathbf{a} \cdot \mathbf{b}+\mathbf{b}+\mathbf{a}+\mathbf{b} \cdot \mathbf{b}=25 \\
& \Rightarrow \quad|\mathbf{a}|^2+2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2=25 \quad[\because \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{a}] \\
& \Rightarrow \quad 9+2 \mathbf{a} \cdot \mathbf{b}+16=25 \\
& \Rightarrow \quad \mathbf{a} \cdot \mathbf{b}=0
\end{aligned}$
Now, consider $|\mathbf{a}-\mathbf{b}|^2=(\mathbf{a}-\mathbf{b})(\mathbf{a}-\mathbf{b})$
$\begin{aligned}
& =\mathbf{a} \cdot \mathbf{a}-\mathbf{a} \cdot \mathbf{b}-\mathbf{b} \cdot \mathbf{a}+\mathbf{b} \cdot \mathbf{b} \\
& =|\mathbf{a}|^2-2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2 \quad[\because \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{a}] \\
& =9-0+16=25 \\
\Rightarrow \mid \mathbf{a}-\mathbf{b} & =5
\end{aligned}$
Since,
$\begin{gathered}
|\mathbf{a}+\mathbf{b}|=5 \\
|\mathbf{a}+\mathbf{b}|^2=25
\end{gathered}$
$\begin{aligned}
& \therefore \quad|\mathbf{a}+\mathbf{b}|^2=25 \\
& \Rightarrow \quad(\mathbf{a}+\mathbf{b}\rangle \cdot(\mathbf{a}+\mathbf{b}\rangle=25 \\
& \Rightarrow \mathbf{a} \cdot \mathbf{a}+\mathbf{a} \cdot \mathbf{b}+\mathbf{b}+\mathbf{a}+\mathbf{b} \cdot \mathbf{b}=25 \\
& \Rightarrow \quad|\mathbf{a}|^2+2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2=25 \quad[\because \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{a}] \\
& \Rightarrow \quad 9+2 \mathbf{a} \cdot \mathbf{b}+16=25 \\
& \Rightarrow \quad \mathbf{a} \cdot \mathbf{b}=0
\end{aligned}$
Now, consider $|\mathbf{a}-\mathbf{b}|^2=(\mathbf{a}-\mathbf{b})(\mathbf{a}-\mathbf{b})$
$\begin{aligned}
& =\mathbf{a} \cdot \mathbf{a}-\mathbf{a} \cdot \mathbf{b}-\mathbf{b} \cdot \mathbf{a}+\mathbf{b} \cdot \mathbf{b} \\
& =|\mathbf{a}|^2-2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2 \quad[\because \mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{a}] \\
& =9-0+16=25 \\
\Rightarrow \mid \mathbf{a}-\mathbf{b} & =5
\end{aligned}$
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.