Let $$\vec{a} = \alpha \hat{i} + \hat{j} + \beta \hat{k}$$ and $$\vec{b} = 3\hat{i} - 5\hat{j} + 4\hat{k}$$ be two vectors, such that $$\vec{a} \times \vec{b} = -\hat{i} + 9\hat{j} + 12\hat{k}$$. Then the projection of $$\vec{b} - 2\vec{a}$$ on $$\vec{b} + \vec{a}$$ is equal to

Given

$$\vec a=(\alpha,1,\beta)$$

and

$$\vec b=(3,-5,4)$$

Compute the cross product:

$$\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\ \alpha&1&\beta\\ 3&-5&4\end{vmatrix}$$

$$=(4+5\beta)\hat i-(4\alpha-3\beta)\hat j+(-5\alpha-3)\hat k$$

Comparing with

$$-\hat i+9\hat j+12\hat k$$

we get

$$4+5\beta=-1$$

$$\beta=-1$$

Also,

$$-5\alpha-3=12$$

$$\alpha=-3$$

Hence

$$\vec a=-3\hat i+\hat j-\hat k$$

Now,

$$\vec b-2\vec a=(3,-5,4)-(-6,2,-2)=(9,-7,6)$$

and

$$\vec b+\vec a=(3,-5,4)+(-3,1,-1)=(0,-4,3)$$

The scalar projection of

$$\vec b-2\vec a$$

on

$$\vec b+\vec a$$

is

$$\frac{(\vec b-2\vec a)\cdot(\vec b+\vec a)}{|\vec b+\vec a|}$$

Now,

$$ (9,-7,6)\cdot(0,-4,3)=0+28+18=46 $$

and

$$ |\vec b+\vec a|=\sqrt{0^2+(-4)^2+3^2}=5 $$

Therefore the required projection is

$$\frac{46}{5}$$

Final Answer :

$$\frac{46}{5}$$