A) \[\frac{2}{3}\]
B) \[\frac{3}{2}\]
C) \[2\]
D) \[3\]
Correct Answer: B
Solution :
We have,\[\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{c}}=|\overrightarrow{\mathbf{c}}|\]and\[|\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}|=2\sqrt{2}\] \[\Rightarrow \]\[\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{c}}=|\overrightarrow{\mathbf{c}}|\]and\[|\overrightarrow{\mathbf{c}}{{|}^{2}}+|\overrightarrow{\mathbf{a}}{{|}^{2}}-2(\overrightarrow{\mathbf{a}}\cdot \overrightarrow{\mathbf{c}})=8\] \[\Rightarrow \] \[|\mathbf{\vec{c}}{{|}^{2}}+\,\,9-2|\mathbf{\vec{c}}|\,\,=8\] \[\Rightarrow \] \[{{(|\mathbf{\vec{c}}|-1)}^{2}}=0\] \[\Rightarrow \] \[|\mathbf{\vec{c}}|\,\,=1\] \[\Rightarrow \] \[|(\overrightarrow{\mathbf{a}}\times \overrightarrow{\mathbf{b}})\times \mathbf{\vec{c}}|=|\mathbf{\vec{a}}\times \mathbf{\vec{b}}||\mathbf{\vec{c}}|sin{{30}^{o}}\] \[=\frac{1}{2}|\mathbf{\vec{a}}\times \mathbf{\vec{b}}|\,\,=\frac{3}{2}\] \[[\because \,\,\mathbf{\vec{a}}\times \mathbf{\vec{b}}=2\widehat{\mathbf{i}}-2\mathbf{\hat{j}}+\mathbf{\hat{k}}]\]
You need to login to perform this action.
You will be redirected in
3 sec
