Let $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be three vectors such that $$\vec{a} = \vec{b} \times (\vec{b} \times \vec{c})$$. If magnitudes of the vectors $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ are $$\sqrt{2}, 1$$ and 2 respectively and the angle between $$\vec{b}$$ and $$\vec{c}$$ is $$\theta$$ $$(0 < \theta < \frac{\pi}{2})$$, then the value of $$1 + \tan \theta$$ is equal to:

We have the relation $$\vec a=\vec b\times(\vec b\times\vec c)$$.

First we recall the standard vector triple-product identity:

$$\vec p\times(\vec q\times\vec r)=\vec q(\vec p\cdot\vec r)-\vec r(\vec p\cdot\vec q).$$

Applying it with $$\vec p=\vec b,\;\vec q=\vec b,\;\vec r=\vec c$$ we get

$$\vec a=\vec b(\vec b\cdot\vec c)-\vec c(\vec b\cdot\vec b).$$

Now we substitute the given magnitudes. We know $$|\vec b|=1,\;|\vec c|=2,$$ and if the angle between $$\vec b$$ and $$\vec c$$ is $$\theta\;(0<\theta<\tfrac{\pi}{2})$$, then

$$\vec b\cdot\vec c=|\vec b||\vec c|\cos\theta=1\cdot2\cos\theta=2\cos\theta.$$

Also $$\vec b\cdot\vec b=|\vec b|^{2}=1^{2}=1.$$

Substituting these scalar products into the expression for $$\vec a$$ gives

$$\vec a=2\cos\theta\,\vec b-1\cdot\vec c=2\cos\theta\,\vec b-\vec c.$$

Next we compute the magnitude of $$\vec a$$. Let us denote

$$\vec u=2\cos\theta\,\vec b,\qquad\vec v=\vec c,$$

so that $$\vec a=\vec u-\vec v.$$

The square of the magnitude is

$$|\vec a|^{2}=|\vec u-\vec v|^{2}=|\vec u|^{2}+|\vec v|^{2}-2\,\vec u\cdot\vec v.$$

We evaluate each term separately:

$$|\vec u|^{2}=(2\cos\theta)^{2}|\vec b|^{2}=4\cos^{2}\theta\cdot1=4\cos^{2}\theta,$$

$$|\vec v|^{2}=|\vec c|^{2}=2^{2}=4,$$

$$\vec u\cdot\vec v=(2\cos\theta\,\vec b)\cdot\vec c=2\cos\theta(\vec b\cdot\vec c)=2\cos\theta(2\cos\theta)=4\cos^{2}\theta.$$

Putting these into the formula for $$|\vec a|^{2}$$ we obtain

$$|\vec a|^{2}=4\cos^{2}\theta+4-2\cdot4\cos^{2}\theta=4\cos^{2}\theta+4-8\cos^{2}\theta=4-4\cos^{2}\theta.$$

Simplifying we have

$$|\vec a|^{2}=4(1-\cos^{2}\theta)=4\sin^{2}\theta.$$

But the problem states that $$|\vec a|=\sqrt{2},$$ so $$|\vec a|^{2}=2.$$ Therefore

$$4\sin^{2}\theta=2\quad\Longrightarrow\quad\sin^{2}\theta=\frac12.$$

Since $$0<\theta<\tfrac{\pi}{2},$$ the sine and cosine are both positive, giving

$$\sin\theta=\frac1{\sqrt2},\qquad\cos\theta=\frac1{\sqrt2}.$$

Hence

$$\tan\theta=\frac{\sin\theta}{\cos\theta}=1.$$

Finally,

$$1+\tan\theta=1+1=2.$$

Hence, the correct answer is Option B.