A) 2
B) \[\sqrt{7}\]
C) \[\sqrt{14}\]
D) 14
Correct Answer: C
Solution :
\[|\overrightarrow{u}|=1,|\overrightarrow{v}|=2,|\overrightarrow{w}|=3\] The projection of\[\overrightarrow{v}\]along\[\overrightarrow{u}=\frac{\overrightarrow{v}.\overrightarrow{u}}{|\overrightarrow{u}|}\] and the projection of\[\overrightarrow{w}\]along\[\overrightarrow{u}=\frac{\overrightarrow{w}.\overrightarrow{u}}{|\overrightarrow{u}|}\] So, \[\frac{\overrightarrow{v}.\overrightarrow{u}}{|\overrightarrow{u}|}=\frac{\overrightarrow{w}.\overrightarrow{u}}{|\overrightarrow{u}|}\] and\[\overrightarrow{v},\overrightarrow{w}\]are perpendicular to each other \[\therefore \] \[\overrightarrow{v}.\overrightarrow{w}=0\] Now, \[|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}{{|}^{2}}=|\overrightarrow{u}{{|}^{2}}+|\overrightarrow{v}{{|}^{2}}+|\overrightarrow{w}{{|}^{2}}\] \[-2\,\overrightarrow{u}.\,\overrightarrow{v}+2\overrightarrow{u}.\,\overrightarrow{w}-2\,\overrightarrow{v}.\overrightarrow{w}\] \[\Rightarrow \]\[|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}{{|}^{2}}=1+4+9-2\overrightarrow{u}.\overrightarrow{v}\] \[+2\overrightarrow{v}.\overrightarrow{u}\] \[\Rightarrow \] \[|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}{{|}^{2}}=1+4+9\] \[\Rightarrow \] \[|\overrightarrow{u}-\overrightarrow{v}+\overrightarrow{w}|=\sqrt{14}\]
You need to login to perform this action.
You will be redirected in
3 sec
