**Section 10.3, Exercise 33**: Show that the vector \mathrm{orth}_{\mathbf{a}}\mathbf{b}=\mathbf{b}-\mathrm{proj}_{\mathbf{a}}\mathbf{b} is orthogonal to \mathbf{a}. (it is called an orthogonal projection of \mathbf{b}).

Comment: The idea is to verify the dot product of \mathrm{orth}_{\mathbf{a}}\mathbf{b} and a is 0. You are going to use the definition of \mathrm{proj}_{\mathbf{a}}\mathbf{b}.

**Section 10.3, Exercise 34**: For the vectors \mathbf{a}=\langle 1,4\rangle, \mathbf{b}=\langle 2,3\rangle, find \mathrm{orth}_{\mathbf{a}}\mathbf{b} and illustrate by drawing the vectors \mathbf{a}, \mathbf{b}, \mathrm{proj}_{\mathbf{a}}\mathbf{b}, \mathrm{orth}_{\mathbf{a}}\mathbf{b}.

Comment: Use the definition of \mathrm{orth}_{\mathbf{a}}\mathbf{b} from the previous exercise to give the answer.

**Section 10.3, Exercise 43**: Find the angle between a diagonal of a cube and a diagonal of one of its faces.

Solution: Let the cube be the unit cube whose vertices are (0,0,0),(0,0,1),\ldots,(1,1,0),(1,1,1). Then the diagonal of the cube is given by the vector \mathbf{d}=(1,1,1). There are in total 12 diagonals on its faces. However, by symmetry, there are only two that matter. One is the diagonal from (0,0,0) to (1,1,0) associated to the vector \mathbf{a}=(1,1,0) and the other one is from (1,0,0) to (0,1,0) associated to the vector \mathbf{a}=(-1,1,0). Use the dot product to see the cosine of the angle between the diagonal and the normal vector is \sqrt{\frac{2}{3}} and 0 respectively. Hence the angle we want is \arccos{\sqrt{\frac{2}{3}}} and \frac{\pi}{2}.

**Section 10.3, Exercise 51**: The Parallelogram Law states that |\mathbf{a}+\mathbf{b}|^2+|\mathbf{a}-\mathbf{b}|^2=2|\mathbf{a}|^2+2|\mathbf{b}|^2. (a) Give a geometric interpretation of the Parallelogram Law. (b) Prove the Parallelogram Law.

Comment: (a) The geometric interpretation comes from the parallelogram spanned by vectors \mathbf{a} and \mathbf{b}. (b) Use dot product to express the square of the magnitude of a vector.

**Section 10.4, Exercise 12**: Find the vector (\mathbf{i}+\mathbf{j})\times(\mathbf{i}-\mathbf{j}), not with determinants, but by using properties of cross products.

Comment: Use distributivity of the cross product.

**Section 10.4, Exercise 22**: Show that (\mathbf{a}\times\mathbf{b})\cdot \mathbf{b}=0 for all vectors \mathbf{a} and \mathbf{b} in V_3.

Comment: Intuitively, the cross product of \mathbf{a} and \mathbf{b} is always perpendicular to \mathbf{b}. Thus their dot product is 0. One the other hand, this identity can be also derived from the property of the triple product that (\mathbf{a}\times\mathbf{b})\cdot \mathbf{c}=\mathbf{a}\cdot (\mathbf{b})\times\mathbf{c}).

**Section 10.4, Exercise 33**: Find the volume of the parallelepiped determined by the vectors \mathbf{a}=\langle 1,2,3\rangle, \mathbf{b}=\langle -1,1,2\rangle and \mathbf{c}\langle 2,1,4\rangle.

Comment: The magnitude of the triple product gives the volume of the parallelepiped.

**Section 10.4, Exercise 46**: (a) Let P be a point not on the plane that passes through the points Q, R and S. Show that the distance d from P to the plane is d=\frac{|\mathbf{a}\cdot(\mathbf{b}\times \mathbf{c})|}{|\mathbf{a}\times\mathbf{b}|}, where \mathbf{a}=\vec{QR}, \mathbf{b}=\vec{QS} and \mathbf{c}=\vec{QP}. (b) Use the formula in part (a) to find the distance from the point P(2,1,4) to the plane through the points Q(1,0,0), R(0,2,0), and S(0,0,3).

Comment: The geometric intuition is to see the numerator is the volume of the parallelepiped spanned by \mathbf{a}, \mathbf{b} and \mathbf{c} and the denominator is the area of the parallelogram spanned by \mathbf{a} and \mathbf{b}. Thus the formula gives the height of the parallelepiped which is just the distance we want. Another way to wrap your mind around it is to see that \mathbf{a}\times\mathbf{b} gives the normal vector of the plane. Thus the distance is the magnitude of the projection of \mathbf{c} onto the normal vector. Then use the formula for the projection vector to get the distance.

**Section 10.4, Exercise 49**: Prove that (\mathbf{a}-\mathbf{b})\times(\mathbf{a}+\mathbf{b})=2(\mathbf{a}\times\mathbf{b}).

Comment: Distributivity.