Mathematics - Vector Question with Solution | TestHub

Let, $\vec{c}=a\hat{i}+b\hat{j}+c\hat{k}$ Given, $\vec{c}\times (\hat{i}+2\hat{j}+5\hat{k})=\overrightarrow{0}$ $$\begin{array}{rl}& \Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a& b& c\\ 1& 2& 5\end{array}\right|=\overrightarrow{0}\\ & \Rightarrow (5b-2c)\hat{i}-(5a-c)\hat{j}+(2a-b)\hat{k}\\ & =0\hat{i}+0\hat{j}+0\hat{k}\end{array}$$ Comparing both sides, we get $5b-2c=0;5a-c=0;2a-b=0$ or $5b=2c;5a=c;2a=b$ 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 $$\begin{array}{rl}& a^2+(2a{)}^2+(5a{)}^2=60\\ & \Rightarrow a^2+4a^2+25a^2=60\Rightarrow 30a^2=60\\ & a^2=2\\ & a=\pm \sqrt{2};b=2\sqrt{2};c=5\sqrt{2}\end{array}$$ 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 $\vec{c}\cdot (-7\hat{i}+2\hat{j}+3\hat{k})$ is $$\begin{array}{rl}& (\sqrt{2}\hat{i}+2\sqrt{2}\hat{j}+5\sqrt{2}\hat{k})\cdot (-7\hat{i}+2\hat{j}+3\hat{k})\\ & =-7\sqrt{2}+4\sqrt{2}+15\sqrt{2}=12\sqrt{2}\end{array}$$