A) \[4\sqrt{2}\]
B) 12
C) 24
D) 122
Correct Answer: D
Solution :
Let, \[\vec{c}=a\hat{i}+b\hat{j}+c\hat{k}\] Given, \[\vec{c}\times (\hat{i}+2\hat{j}+5\hat{k})=\vec{0}\] \[\Rightarrow \]\[\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ a & b & c \\ 1 & 2 & 3 \\ \end{matrix} \right|=\vec{0}\] \[\Rightarrow \]\[(5b-2c)\hat{i}-(5a-c)\hat{j}+(2a-b)\hat{k}=0\hat{i}+0\hat{j}+0\hat{k}\] Comparing both sides, we get \[5b-2c=0;5a-c=0;2a-b=0\] Also given\[|\vec{c}{{|}^{2}}=60\Rightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}=60\] Putting the value of b and c in above eqn., we get \[{{a}^{2}}+{{(2a)}^{2}}+{{(5a)}^{2}}=60\] \[\Rightarrow \]\[{{a}^{2}}+4{{a}^{2}}+25{{a}^{2}}=60\Rightarrow 30{{a}^{2}}=60\] \[{{a}^{2}}=2\] \[a=\pm \sqrt{2};b=2\sqrt{2};c=5\sqrt{2}\] Now, \[\vec{c}=a\hat{i}+b\hat{j}+c\hat{k}\] \[\therefore \]\[\vec{c}=\sqrt{2}\hat{i}+2\sqrt{2}\hat{j}+5\sqrt{2}\hat{k}\] Value of \[\vec{c}=.(-7\hat{i}+2\hat{j}+3\hat{k})\]is \[\left( \sqrt{2}\hat{i}+2\sqrt{2}\hat{j}+5\sqrt{2}\hat{k} \right).\left( -7\hat{i}+2\hat{j}+3\hat{k} \right)\] \[=-7\sqrt{2}+4\sqrt{2}+15\sqrt{2}=12\sqrt{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
