Let $$\vec{a}$$ and $$\vec{b}$$ be two vectors such that $$|2\vec{a} + 3\vec{b}| = |3\vec{a} + \vec{b}|$$ and the angle between $$\vec{a}$$ and $$\vec{b}$$ is 60°. If $$\frac{1}{8}\vec{a}$$ is a unit vector, then $$|\vec{b}|$$ is equal to:

We are given that the magnitudes of the two vectors

$$\;2\vec a+3\vec b\quad\text{and}\quad 3\vec a+\vec b$$

are equal, that the angle between $$\vec a$$ and $$\vec b$$ is $$60^\circ,$$ and that $$\displaystyle \frac18\vec a$$ is a unit vector. Our task is to find $$|\vec b|.$$

First, from “$$\dfrac18\vec a$$ is a unit vector” we have

$$\Bigl|\dfrac18\vec a\Bigr|=1 \;\;\Longrightarrow\;\; \dfrac1{8}\,|\vec a|=1 \;\;\Longrightarrow\;\; |\vec a|=8.$$ Hence

$$|\vec a|^2 = 8^2 = 64.$$

Next we use the condition

$$|\,2\vec a+3\vec b\,| \;=\; |\,3\vec a+\vec b\,|.$$

Squaring both sides (because both magnitudes are non-negative) and applying the dot-product identity $$|\vec u|^2=\vec u\!\cdot\!\vec u,$$ we get

$$ (2\vec a+3\vec b)\cdot(2\vec a+3\vec b) \;=\; (3\vec a+\vec b)\cdot(3\vec a+\vec b). $$

Now we expand each side completely, remembering that $$\vec a\!\cdot\!\vec b=\vec b\!\cdot\!\vec a.$$ For the left-hand side:

$$ (2\vec a+3\vec b)\cdot(2\vec a+3\vec b) \;=\; (2\vec a)\cdot(2\vec a) \;+\; 2\,(2\vec a)\cdot(3\vec b) \;+\; (3\vec b)\cdot(3\vec b). $$ Evaluating term by term, we obtain

$$ 4|\vec a|^2 \;+\; 12\,\vec a\!\cdot\!\vec b \;+\; 9|\vec b|^2. $$

For the right-hand side:

$$ (3\vec a+\vec b)\cdot(3\vec a+\vec b) \;=\; (3\vec a)\cdot(3\vec a) \;+\; 2\,(3\vec a)\cdot\vec b \;+\; \vec b\cdot\vec b, $$ which evaluates to

$$ 9|\vec a|^2 \;+\; 6\,\vec a\!\cdot\!\vec b \;+\; |\vec b|^2. $$

Because the two magnitudes are equal, the two expansions themselves must be equal. Thus

$$ 4|\vec a|^2 + 12\,\vec a\!\cdot\!\vec b + 9|\vec b|^2 \;=\; 9|\vec a|^2 + 6\,\vec a\!\cdot\!\vec b + |\vec b|^2. $$

We now move every term to one side to set the entire expression to zero:

$$ 4|\vec a|^2 - 9|\vec a|^2 \;+\; 12\,\vec a\!\cdot\!\vec b - 6\,\vec a\!\cdot\!\vec b \;+\; 9|\vec b|^2 - |\vec b|^2 \;=\; 0. $$ Combining like terms gives

$$ -5|\vec a|^2 \;+\; 6\,\vec a\!\cdot\!\vec b \;+\; 8|\vec b|^2 = 0. $$

Next we write the scalar product $$\vec a\!\cdot\!\vec b$$ in terms of the magnitudes and the included angle. The standard formula is

$$ \vec a\!\cdot\!\vec b = |\vec a|\,|\vec b|\,\cos\theta, $$ where $$\theta$$ is the angle between the two vectors. Here $$\theta=60^\circ,$$ so $$\cos60^\circ=\dfrac12.$$ Therefore

$$ \vec a\!\cdot\!\vec b = |\vec a|\,|\vec b|\,\dfrac12 = 8\,|\vec b|\cdot\dfrac12 = 4\,|\vec b|. $$

Let us denote $$|\vec b|=x$$ to simplify writing. Then

$$ \vec a\!\cdot\!\vec b = 4x, \qquad |\vec b|^2 = x^2. $$

Substituting these expressions, together with $$|\vec a|^2=64,$$ into the earlier equation $$-5|\vec a|^2 + 6\,\vec a\!\cdot\!\vec b + 8|\vec b|^2 = 0,$$ we get

$$ -5(64) \;+\; 6(4x) \;+\; 8(x^2) \;=\; 0. $$ Now we evaluate each term:

$$ -320 \;+\; 24x \;+\; 8x^2 = 0. $$

For convenience we multiply the whole equation by $$-1$$ to make the leading coefficient of $$x^2$$ positive:

$$ 320 \;-\; 24x \;-\; 8x^2 = 0. $$

Next we divide every term by $$8$$ to reduce the coefficients:

$$ 40 \;-\; 3x \;-\; x^2 = 0. $$

Re-ordering the terms in the usual descending powers of $$x$$ yields the quadratic equation

$$ x^2 + 3x - 40 = 0. $$

We now solve this quadratic. The quadratic formula states that for $$ax^2+bx+c=0,$$ $$ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}. $$ Here $$a=1,\; b=3,\; c=-40,$$ so

$$ x = \dfrac{-3 \pm \sqrt{\,3^2 - 4(1)(-40)\,}}{2(1)} = \dfrac{-3 \pm \sqrt{9 + 160}}{2} = \dfrac{-3 \pm \sqrt{169}}{2} = \dfrac{-3 \pm 13}{2}. $$

This gives the two solutions

$$ x = \dfrac{-3 + 13}{2} = \dfrac{10}{2} = 5, \qquad x = \dfrac{-3 - 13}{2} = \dfrac{-16}{2} = -8. $$

Because a magnitude can never be negative, we discard $$x=-8$$ and keep

$$ x = 5. $$

Recalling that $$x = |\vec b|,$$ we conclude

$$ |\vec b| = 5. $$

Hence, the correct answer is Option D.