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A+B=n

.Another method :find tow points on lsuch as A and B

D is equal to the length of the component

of the vector

that is or thogonal to the vector

ex : find the distance from the point s (4-4)to the line x+3y =6

sol. Method 1:

v==4i+zj = 1 + 9 = 10 = . = 4 + . +310 D=4+610 = 10

Method 2

Find another point : x+3y=6 , y=0, x=6 , Q(6,0)

V== 4 + 2 = = 6 2

proj = .2 . , = = .2 .

v.b =24-4=20

2=36+4=40N=4i+2j-

20

40.(6i-2j) =i+3j

D= = 1 + 9 = 10Cross products

Definition

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N: unit vector perpendicular to the plane of A and B, determined by the right hand

rule

A X B =0 1) if=00 or 1800(A//B)2)if A=0 or B=0 or both A and B is zero

Ex : Apply the definition of the cross product to the basic vectors I,j, and K.

Ixj = n sin = = = 1(unit vectors)Sin=1 ( = 900 ,sin900 = 1)N=k (since n unit vector normal to xy plane )

= = = ()Jxk= -k = ()Kxi=j= ()

= = = 1 ()

= The area of a parallelogram

Area = =

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THE Determinant formula of AXB

Let A=a,i+azj+a3k and B= b,I +b2j +b3k

AXB= (a,I,+azj+a3k)x (b,i+bzj+b3k)

The result can be tained by using the distributive laws and the rules for multiplying

I,j, and k the sameresult can be obtained by expanding the symbolic determinant

AXB= 1 2 3

1

2

3

Ex1 find AXB and AXB if

A=2i +j+k ,B=-4i+3j+k

So1 AXB= 2 1 14 3 1=1 13 1

2 14 1 + 2 14 3 =-2i-6j+6j-10k

Ex 2 let A=i+zjk 9 B=-2i-4j+zk,c=-2i+zj+2k.

Which vectors , if any ,are (a) perpendicular .(b) parallel?

So1 :a)

A.B =(1)(-2)+(2)(-4)+(-1)(2)=-2-8-2=-12

B-C =(-2)(-2)(-4)(2)+(2)(2)=4-8-4+0

A.C= (1)(-2)+(2)(2)+(-1)(2)=-2+4-2=0

A and C and Band C are perpendicular

B) AXB = 2 1 14 3 1 = 2 14 2 2 14 2 + 1 12 2 1 22 4 =

+ + = 0

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BXC= 2 4 22 2 2=- 12i -12kAXC=

1 4 12 2 2 = 6 + 6

A and B are parallel.

Note: the cross product is defined only for vectors in 3-space where as the dot

product is difined for vectors in 2-space and 3-space

Ex3 a) find a vector perpendicular to the plan of

P(1,-1,0), Q(2,1,-1),and R(-1,1,2).

b) find the area of the triangle with vertices p,Q, and R

c) find a unit vector perpendicular to the plane of p,Q, and R

Sol . a) the vector x is perpendicular to the plan because it isperpendicular to both vectors =i+zj-k = -2i +zj +2k

= 1 2 1

2 2 2= 6 + 6

b) the area A of the triangle is half the area of the parallelogram determined

by the vectors and A=

1

2 = 12 6 + 6 = 32

c) = n ()

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, =

=

6+662 = 12 + 12 (from a and b )

The triple scalar or Box product

IF A= (a1, a2,a3 ) B= (b1,b2,b3) and C =(c1,c2,c3)then the triple scalar product of A,B

and C is A.(BXC)

It is not necessary to compute the dot product and cross product to evaluate a

scalar triple product . the value can be obtained directly from the formula

. =1

2

3

1 2 31 2 3 .9And it can be proved as follows :

BXC= 1 2 31 2 3 = 2 33 3 1 31 3 1 21 2

A.(BXC)=

1

2

3

2 3 2

1

3

1 3+

3

1

2

1 3=

1 2 3

1

2

3

1 2 3

Ex 1 calculate the scalar triple product A.(BXC) of the vectors A=3i-2j-, B=i+4j-

4k, C=3i+2k

Sol. A(BXC)=3 2 51 4 40 3 2

= 49Ex2 prove that

A.(BXC)=C. (AXB)=B. (CXA)

Sol//

C.(AXB)=(C1i+C2J+C3K). 1 2 31 2 3 =

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1 = 2 32 3 2 1 31 3 + 3 1 21 2 = 1 2 31 2 31 2 3 =

1 2 3

1

2

3

1 2 3

(inter change the rows twice ) (inter changing tow rows of a determinant

multiplies its value by -1)

Note ; rewrite the first equality as

A.(BXC)= (AXB).C

This formula states that the dot and cross may be inter changed in a triple

scalar product without altering its value.

Geometric properties of the scalar triple product

A.(BXC)= 10 (dot product de finition) from this formula itcan be concluded that the triple scalar product is the volume of the

parallelepiped (parallel ogram - sided box)that has A,B, and C as adjacent

edges.

V=. ()(note that A.(BXC)= V)

The(+) occurs when A makes an a cute

Angle with BXC and (-) occurs when it makes an obtuse angle)

V= (area of base) (height)

= cos = Or :h = = .()2 = .() V= (area of base)(height)= = . ()Note: from the figure we can see that

A.(BXC)= C.(AXB)= B.(CXA) = volume of the parallelepiped which is proved

earlier in ex2.

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Ex: find the volume of the box (parallelepiped)determined by A=i-2j-k, B=-

2i+3k,

and C =7J-4K.

SOL. A. (BXC)= 1 2 12 0 30 7 1 = 23

V=. () = 23 = 23Parametric Eqs. Of lines

1- The line in 2-space that passes through the point po (xo,yo) and isparallel to the non zero vector v=(a,b) =ai+bi has parametric

equations . x=xo+at ,y=yo+bt11 <

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Is parallel to the line L and point

(-3,2,-3) lies on L , so it follows

That L has parametric equations

X=-3+4t ,y=2-3t ,z=-3+7t. 1

If we used Q (1,-1,4)as the point on L rather than P (-3,2,-3), we would have

obtained the eqs.

X=1+4t ,y=-1-3t ,z=4+7t 2

(1)(2)

LT