Mathematics - Vector Question with Solution | TestHub

Let $\overrightarrow{a}=\alpha \hat{i}+\hat{j}+\beta \hat{k},\overrightarrow{b}=3\hat{i}-5\hat{j}+4\hat{k}$

Now finding $\overrightarrow{a}\times \overrightarrow{b}=-\hat{i}+9\hat{j}+12\hat{k}$

$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \alpha & 1& \beta \\ 3& -5& 4\end{array}\right|$

$\Rightarrow \left(4+5\beta \right)\hat{i}+\left(3\beta -4\alpha \right)\hat{j}+\left(-5\alpha -3\right)\hat{k}$

$=-\hat{i}+9\hat{j}+12\hat{k}$

Now on comparing side we get,

$4+5\beta =-1,3\beta -4\alpha =9,-5\alpha -3=12$

$\beta =-1, \alpha =-3$

So, $\overrightarrow{a}=-3\hat{i}+\hat{j}-\hat{k} \& \overrightarrow{b}=3\hat{i}-5\hat{j}+4\hat{k}$

So,$\overrightarrow{a}+\overrightarrow{b}=-4\hat{j}+3\hat{k}$

Now finding ${\left|\overrightarrow{a}\right|}^2=11 \&\hspace{0.17em}{\left|\overrightarrow{b}\right|}^2=50$

And $\overrightarrow{a}·\overrightarrow{b}=-9+\left(-5\right)-4=-18$

Now projection of $\left(\overrightarrow{b}-2\overrightarrow{a}\right)$ on $\overrightarrow{a}+\overrightarrow{b}$ will be 

$=\frac{\left(\overrightarrow{b}-2\overrightarrow{a}\right)·\left(\overrightarrow{a}+\overrightarrow{b}\right)}{\left|\overrightarrow{a}+\overrightarrow{b}\right|}$

$=\frac{{\left|\overrightarrow{b}\right|}^2-2{\left|\overrightarrow{a}\right|}^2-\left(\overrightarrow{a}·\overrightarrow{b}\right)}{\left|\overrightarrow{a}+\overrightarrow{b}\right|}=\frac{50-22-\left(-18\right)}{5}=\frac{46}{5}$