Recitation 2

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.