The vector $$\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$$ is rotated through a right angle, passing through the y-axis in its way and the resulting vector is $$\vec{b}$$. Then the projection of $$3\vec{a} + \sqrt{2}\vec{b}$$ on $$\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$$ is

We are given the vector $$\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$$, which is rotated through a right angle in a plane that passes through the y-axis to yield $$\vec{b}$$, and we wish to find the projection of $$3\vec{a} + \sqrt{2}\vec{b}$$ onto $$\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$$.

Since the rotation passes through the y-axis ($$\hat{j}$$ direction), it occurs in the plane spanned by $$\vec{a}$$ and $$\hat{j}$$, we write $$\vec{b} = \lambda\vec{a} + \mu\hat{j}$$.

Enforcing perpendicularity between $$\vec{a}$$ and $$\vec{b}$$ gives $$\vec{a} \cdot \vec{b} = \lambda|\vec{a}|^2 + \mu(\vec{a} \cdot \hat{j}) = 0$$. Since $$|\vec{a}|^2 = (-1)^2 + 2^2 + 1^2 = 6$$ and $$\vec{a} \cdot \hat{j} = 2$$, it follows that $$6\lambda + 2\mu = 0 \implies \mu = -3\lambda$$.

Substituting into $$\vec{b}$$ yields $$\vec{b} = \lambda(-1, 2, 1) + (-3\lambda)(0, 1, 0) = (-\lambda, 2\lambda - 3\lambda, \lambda) = (-\lambda, -\lambda, \lambda)$$. Imposing $$|\vec{b}| = |\vec{a}|$$ leads to $$|\vec{b}|^2 = \lambda^2 + \lambda^2 + \lambda^2 = 3\lambda^2 = 6 \implies \lambda = \pm\sqrt{2}$$.

To select the correct sign, note that during the $$90°$$ rotation from $$\vec{a} = (-1, 2, 1)$$ the path must pass through the $$+\hat{j}$$ direction. For $$\lambda = -\sqrt{2}$$ one finds $$\vec{b} = (\sqrt{2}, \sqrt{2}, -\sqrt{2})$$ and $$\sqrt{2}\vec{b} = (2, 2, -2)$$, whereas for $$\lambda = \sqrt{2}$$ one finds $$\vec{b} = (-\sqrt{2}, -\sqrt{2}, \sqrt{2})$$ and $$\sqrt{2}\vec{b} = (-2, -2, 2)$$. Only the choice $$\lambda = -\sqrt{2}$$ gives a positive $$y$$-component in $$\vec{b}$$, so we take $$\lambda = -\sqrt{2}$$.

Therefore, it follows that $$3\vec{a} = 3(-1, 2, 1) = (-3, 6, 3)$$ and, with this choice of $$\lambda$$, $$\sqrt{2}\vec{b} = (2, 2, -2)$$, so $$3\vec{a} + \sqrt{2}\vec{b} = (-3 + 2, 6 + 2, 3 - 2) = (-1, 8, 1)$$. Since $$|\vec{c}| = \sqrt{25 + 16 + 9} = 5\sqrt{2}$$, the projection of $$3\vec{a} + \sqrt{2}\vec{b}$$ on $$\vec{c}$$ is $$\frac{(3\vec{a} + \sqrt{2}\vec{b}) \cdot \vec{c}}{|\vec{c}|} = \frac{(-1)(5) + (8)(4) + (1)(3)}{5\sqrt{2}} = \frac{-5 + 32 + 3}{5\sqrt{2}} = \frac{30}{5\sqrt{2}} = \frac{6}{\sqrt{2}} = 3\sqrt{2}$$.

Therefore, the projection is $$\mathbf{3\sqrt{2}}$$.