Solutions to Problem Set 4

Math 311-01/02

1-30-2018

[Cross product]

1. Compute the cross product of $(2, -1, 5)$ and $(3, 2, 6)$ .

$$(2, -1, 5) \times (3, 2, 6) = \left|\matrix{ \ihat & \jhat & \khat \cr 2 & -1 & 5 \cr 3 & 2 & 6 \cr}\right| = (-16, 3, 7).\quad\halmos$$


2. Compute the cross product of $(2, -5, 0)$ and $(3, -4, 0)$ .

$$(2, -5, 0) \times (3, -4, 0) = \left|\matrix{ \ihat & \jhat & \khat \cr 2 & -5 & 0 \cr 3 & -4 & 0 \cr}\right| = (0, 0, 7).\quad\halmos$$


3. Compute the cross product of $(2, -3, 1)$ and $(-6, 9, -3)$ .

$$(2, -3, 1) \times (-6, 9, -3) = \left|\matrix{ \ihat & \jhat & \khat \cr 2 & -3 & 1 \cr -6 & 9 & -3 \cr}\right| = (0, 0, 0).\quad\halmos$$


4. Compute the cross product of $3 \ihat - \jhat + 4 \khat$ and $2 \ihat + 11 \khat$ .

$$(3 \ihat - \jhat + 4 \khat) \times (2 \ihat + 11 \khat) = \left|\matrix{ \ihat & \jhat & \khat \cr 3 & -1 & 4 \cr 2 & 0 & 11 \cr}\right| = -11 \ihat - 25 \jhat + 2 \khat.\quad\halmos$$


5. Find two unit vectors perpendicular to both of the vectors $(2, 3, 1)$ and $(4, 0, 1)$ .

$$(2, 3, 1) \times (4, 0, 1) = \left|\matrix{ \ihat & \jhat & \khat \cr 2 & 3 & 1 \cr 4 & 0 & 1 \cr}\right| = (3, 2, -12).$$

$$\|(3, 2, -12)\| = \sqrt{9 + 4 + 144} = \sqrt{157}.$$

The two unit vectors are $\pm
   \dfrac{1}{\sqrt{157}} (3, 2, -12)$ .


6. The vectors $(2, 1, 1)$ and $(-3, 4, 1)$ form adjacent sides of a parallelogram. Find the area of the parallelogram.

$$(2, 1, 1) \times (-3, 4, 1) = \left|\matrix{ \ihat & \jhat & \khat \cr 2 & 1 & 1 \cr -3 & 4 & 1 \cr}\right| = (-3, -5, 11).$$

The area is

$$\|(-3, -5, 11)\| = \sqrt{9 + 25 + 121} = \sqrt{155}.\quad\halmos$$


7. The vertices of a parallelogram, listed counterclockwise, are $A(1, 2, 4)$ , $B(-4, 3, 7)$ , $C(0, 4, 13)$ , and $D(5, 3, 10)$ . Find the area of the parallelogram.

The vectors for the sides adjacent to A are

$$\bvec{A B} = (-5, 1, 3) \quad\hbox{and}\quad \bvec{A D} = (4, 1, 6).$$

$$(-5, 1, 3) \times (4, 1, 6) = \left|\matrix{ \ihat & \jhat & \khat \cr -5 & 1 & 3 \cr 4 & 1 & 6 \cr}\right| = (3, 42, -9).$$

The area is

$$\|(3, 42, -9)\| = \sqrt{9 + 1764 + 81} = \sqrt{1854}.\quad\halmos$$


8. Find the volume of the parallelepiped with edges determined by the vectors $(4, -1, 5)$ , $(1,
   1, 1)$ , and $(2, 0, 3)$ .

$$\left|\matrix{ 4 & -1 & 5 \cr 1 & 1 & 1 \cr 2 & 0 & 3 \cr}\right| = 4 \cdot \left|\matrix{1 & 1 \cr 0 & 3 \cr}\right| - (-1) \cdot \left|\matrix{1 & 1 \cr 2 & 3 \cr}\right| + 5 \cdot \left|\matrix{1 & 1 \cr 2 & 0 \cr}\right| =$$

$$4 \cdot 3 - (-1) \cdot 1 + 5 \cdot (-2) = 3.$$

The volume is 3.


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