Chapter 6 Vectors

Exercise 6.1

Q1

Show that the equation represents a sphere, and find its center and radius.

\[ (a) x^2+y^2+z^2-2x-4y+8z=15 \\ (b) 2x^2+2y^2+2z^2=8x-24z+1 \]

Q2

Find a+b,4a+2b,|a|, and |a-b|.(a, b, i, j are vectors)

\[ (a) a=(8,1,-4),\ b=(5,-2,1)\\ (b) a=5i+3j,\ b=-i-2j \]

Q3

Find the angle between the vectors a=(2,2,-1) and b=(5,-3,2).

Q4

If a=(1,3,4) and b=(2,7,-5), find the cross product a × b.

Solutions to Exercise 6.1

Q1

Show that the equation represents a sphere, and find its center and radius.

1. \[ x^2+y^2+z^2-2x-4y+8z=15 \\ (x^2-2x+1)+(y^2-4y+4)+(z^2+8z+16)=15+1+4+16 \\ (x-1)^2+(y-2)^2+(z+4)^2=36 \\ the\ center\ is\ (1,2,-4),\ and \ radius \ is \sqrt{36}=6. \]
2. \[2x^2+2y^2+2z^2=8x-24z+1 \\ x^2+y^2+z^2=4x-12z+\frac{1}{2} \\ (x^2-4x+4)+y^2+(z^2+12z+36)=\frac{1}{2}+4+36 \\ (x-2)^2+(y-0)^2+(z+6)^2=\frac{81}{2}\\ the\ center\ is\ (2,0,-6),\ and \ radius \ is \sqrt{\frac{81}{2}}= \frac{9}{\sqrt{2}}. \]

Q2

Find a+b,4a+2b,|a|, and |a-b|.(a, b, i, j are vectors)

1. \[ \vec a=(8,1,-4),\ \vec b=(5,-2,1)\\ \]

\[ \vec a +\vec b = (8+5,1-2,-4+1) = (13,-1,-3)\\ \vec 4a +\vec 2b = (4\times 8+2\times5,4\times1+2\times(-2),4\times(-4)+2\times1) = (42,0,-14)\\ \mid \vec a \mid = \sqrt{8^2+1^2+(-4)^2} = \sqrt{81} = 9 \\ \vec a - \vec b = (8-5,1-(-2),-4-1) = (3,3,-5)\\ \mid \vec a - \vec b \mid = \sqrt{3^2+3^2+(-5)^2} = \sqrt{43} \\ \]

2. \[ \vec a=5\vec i+3\vec j,\ \vec b=-\vec i-2\vec j\]

\[ \vec a +\vec b =(5\vec i+3\vec j)+( -\vec i-2\vec j)=(5-1)\vec i + (3-2) \vec j = 4\vec i +\vec j \\ \vec 4a +\vec 2b = 4(5\vec i+3\vec j)+2(-\vec i-2\vec j) = 18\vec i+8\vec j \\ \mid \vec a \mid = \sqrt{5^2+3^2}=\sqrt{34}\\ \vec a - \vec b = (5\vec i+3\vec j)-( -\vec i-2\vec j)=6\vec i +5\vec j \\ \mid \vec a - \vec b \mid = \sqrt{6^2+5^2} = \sqrt{61} \]

Q3

Find the angle between the vectors a=(2,2,-1) and b=(5,-3,2).

\[ \mid \vec a \mid = \sqrt{2^2+2^2+(-1)^2}=3 \\ \mid \vec b \mid = \sqrt{5^2+(-3)^2+2^2}=\sqrt{38} \\ \vec a \times \vec b = 2(5)+2(-3)+(-1)2=2 \\ \cos \theta = \frac{\vec a \vec b}{\mid \vec a \mid \mid \vec b \mid}=\frac{2}{3\sqrt{38}} \\ Thus\ the\ angle\ between\ \vec a\ and\ \vec b\ is \\ \theta = \cos^{-1}{(\frac{2}{3\sqrt{38}})} \]

Q4

If a=(1,3,4) and b=(2,7,-5), find the cross product a × b.

\[\vec a \times \vec b = \begin{Vmatrix} \vec i& \vec j & \vec k\\1&3&4\\2&7&-5\\\end{Vmatrix} \\ = \begin{Vmatrix}3&4\\7&-5\\\end{Vmatrix} \vec i - \begin{Vmatrix}1&4\\2&-5\\\end{Vmatrix} \vec j + \begin{Vmatrix}1&3\\2&7\\\end{Vmatrix} \vec k \\ = (-15-28)\vec i - (-5-8)\vec j + (7-6)\vec k \\ = -43 \vec i +13\vec j+\vec k \]