Want to ask us a question? Click here
Browse Questions
 Ad
0 votes

If $\overrightarrow{p} ,\overrightarrow{q}$ and $\overrightarrow{p}+\overrightarrow{q}$ are vectors of magnitude $\lambda$ than the magnitude of $|\overrightarrow{p}-\overrightarrow{q}|$ is

$\begin {array}{1 1}(1)2\lambda &(2)\sqrt{3}\lambda\\(3)\sqrt{2}\lambda&(4)1\end{array}$

Can you answer this question?

1 Answer

0 votes
Given $|\overrightarrow{p}|= \lambda$
$|\overrightarrow{q}|= \lambda$
$|\overrightarrow{p}+\overrightarrow{q}|= \lambda$
$|\overrightarrow{p}+\overrightarrow{q}|^2= \lambda^2$
$(\overrightarrow{p}+\overrightarrow{q})^2= \lambda^2$
$\overrightarrow{p^2}+\overrightarrow{q^2}+ 2.\overrightarrow{p}.\overrightarrow{q}= \lambda^2$
$\overrightarrow{|p|^2}+\overrightarrow{|q|^2}+ 2.\overrightarrow{p}.\overrightarrow{q}= \lambda^2$
$\lambda ^2+ \lambda ^2+ 2.\overrightarrow{p}.\overrightarrow{q}= \lambda^2$
$2.\overrightarrow{p}.\overrightarrow{q}= -\lambda^2$
$|\overrightarrow{p}+\overrightarrow{q}|^2=(\overrightarrow{p}+\overrightarrow{q})^2$
$\qquad=\overrightarrow{|p|^2}+\overrightarrow{|q|^2}+ 2.\overrightarrow{p}.\overrightarrow{q}$
$\qquad=\overrightarrow{|p|^2}+\overrightarrow{|q|^2} - (- \lambda)$
$\qquad= \lambda^2+\lambda^2+\lambda^2$
$\qquad=3 \lambda^2$
$|\overrightarrow{p}+\overrightarrow{q}| = \sqrt 3 \lambda$
Hence 2 is the correct answer
answered May 5, 2014 by

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer

0 votes
1 answer