If $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$ are unit vectors such that $$\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0}$$, then $$|\vec{a} \times \vec{c}|$$ is equal to:

We are told that $$\vec a,\;\vec b,\;\vec c$$ are unit vectors and they satisfy the vector equation

$$\vec a+2\vec b+2\vec c=\vec 0.$$

Because the right-hand side is the zero vector, its magnitude is zero. Hence, applying the definition of the square of a vector’s magnitude, we have

$$\bigl|\vec a+2\vec b+2\vec c\bigr|^{2}=0$$

and, by the dot-product expansion formula $$(\vec p+\vec q+\vec r)\cdot(\vec p+\vec q+\vec r)=\vec p\cdot\vec p+\vec q\cdot\vec q+\vec r\cdot\vec r+2\vec p\cdot\vec q+2\vec p\cdot\vec r+2\vec q\cdot\vec r,$$ we write

$$\bigl(\vec a+2\vec b+2\vec c\bigr)\cdot\bigl(\vec a+2\vec b+2\vec c\bigr)=0.$$

Carrying out the dot products term by term, we obtain

$$\vec a\cdot\vec a+4\,\vec b\cdot\vec b+4\,\vec c\cdot\vec c+2\bigl(\vec a\cdot2\vec b+\vec a\cdot2\vec c+2\vec b\cdot2\vec c\bigr)=0.$$

Because each of the three vectors is a unit vector, $$\vec a\cdot\vec a=\vec b\cdot\vec b=\vec c\cdot\vec c=1.$$ Introducing the notations

$$x=\vec a\cdot\vec b,\qquad y=\vec b\cdot\vec c,\qquad z=\vec a\cdot\vec c,$$

the above equation becomes

$$1+4+4+4x+4z+8y=0,$$

that is,

$$9+4x+4z+8y=0. \quad -(1)$$

Next, we exploit the original relation $$\vec a=-2\vec b-2\vec c.$$ Taking the dot product of this with each of the unit vectors in turn gives three more scalar equations.

First, dotting with $$\vec a:$$

$$\vec a\cdot\vec a=-2\bigl(\vec a\cdot\vec b+\vec a\cdot\vec c\bigr)\;\Longrightarrow\;1=-2(x+z),$$

so

$$x+z=-\dfrac12. \quad -(2)$$

Second, dotting with $$\vec b:$$

$$\vec b\cdot\vec a+2\,\vec b\cdot\vec b+2\,\vec b\cdot\vec c=0\;\Longrightarrow\;x+2+2y=0,$$

hence

$$x+2y=-2. \quad -(3)$$

Third, dotting with $$\vec c:$$

$$\vec c\cdot\vec a+2\,\vec c\cdot\vec b+2\,\vec c\cdot\vec c=0\;\Longrightarrow\;z+2y+2=0,$$

which simplifies to

$$z+2y=-2. \quad -(4)$$

We now solve the simultaneous linear equations (2), (3) and (4) for $$x,\;y,\;z.$$

Subtracting (4) from (3) gives

$$(x+2y)-(z+2y)=(-2)-(-2)\;\Longrightarrow\;x-z=0,$$

so

$$x=z. \quad -(5)$$

Substituting $$z=x$$ from (5) into (2) yields

$$x+x=-\dfrac12\;\Longrightarrow\;2x=-\dfrac12\;\Longrightarrow\;x=-\dfrac14.$$

Therefore, by (5),

$$z=-\dfrac14.$$

Finally, using $$x=-\dfrac14$$ in (3) gives

$$-\dfrac14+2y=-2\;\Longrightarrow\;2y=-2+\dfrac14=-\dfrac74\;\Longrightarrow\;y=-\dfrac78.$$

We now have the specific value we need:

$$\vec a\cdot\vec c=z=-\dfrac14.$$

Because both $$\vec a$$ and $$\vec c$$ are unit vectors, the magnitude of their cross product is determined by the fundamental formula

$$|\vec a\times\vec c|^{2}=|\vec a|^{2}\,|\vec c|^{2}-\bigl(\vec a\cdot\vec c\bigr)^{2}.$$

With $$|\vec a|=|\vec c|=1,$$ this gives

$$|\vec a\times\vec c|^{2}=1\cdot1-\left(-\dfrac14\right)^{2}=1-\dfrac1{16}=\dfrac{15}{16}.$$

Taking the positive square root,

$$|\vec a\times\vec c|=\sqrt{\dfrac{15}{16}}=\dfrac{\sqrt{15}}{4}.$$

Hence, the correct answer is Option B.