**Section 10.1, Exercise 15**: Show that the equation 2x^2+2y^2+2z^2=8x-24z+1 represents a sphere, and find its center and radius.

Comment: The idea is to complete the squares. Once the equation is written in the canonical form (x-a)^2+(y-b)^2+(z-c)^2=r^2, the center is just (a,b,c) and the radius is r.

**Section 10.1, Exercise 37**: Find the distance between the spheres x^2+y^2+z^2=4 and x^2+y^2+z^2=4x+4y+4z-11

Comment: If two spheres whose centers are O_1, O_2 and radii r_1, r_2 are apart, the distance between them is given by |O_1O_2|-(r_1+r_2). Note that this formula doesn’t work if the spheres intersect or one is contained in the other.

**Section 10.2, Exercise 2**: Write each combination of vectors as a single vector. \vec{AB}+\vec{BC}, \vec{CD}+\vec{DB}, \vec{DB}-\vec{AB}, \vec{DC}+\vec{CA}+\vec{AB}.

Solution: \vec{AB}+\vec{BC}=\vec{AC}, \vec{CD}+\vec{DB}=\vec{CB}, \vec{DB}-\vec{AB}=\vec{DA}, \vec{DC}+\vec{CA}+\vec{AB}=\vec{DB}.

**Section 10.2, Exercise 18**: Find a vector that has the same direction as \langle 2, 4, 2\rangle but has length 6.

Comment: The quickest way to find the vector is to find the scale factor. The length of the original vector is \sqrt{24}. Thus the scale factor is 6/\sqrt{24}. Multiplying the scale factor to the vector (2,4,2) gives the answer.

**Section 10.2, Exercise 25**: Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (see p.550 for picture)

Solution: Let’s call the force pointing to the left F and the other one G. With some effort, F=(-300, 0) and G=(100,100\sqrt{3}). Now F+G=(-200, 100\sqrt{3}). The magnitude is \sqrt{200^2+100^2\times 3}=100\sqrt{7} and the angle it makes with the positive x-axis is 2\pi - \arctan \sqrt{3}/2.