# Need help with algebra questions

I need help understanding how to do these questions if the numbers are different etc. Thank you :)

I just need to make sure i understand these if i am faced with them and not just following routine.

4)

A line through point r0 = (a, b, c) parallel to vector u = (u, v, w) is given b

(x, y, z) = (a, b, c) + t·(u, v, w),

where t is any real number. In the vector form, we have

r = r0 + t· u,

where r = (x, y, z).

5)

Conversely, the vector equation for the line x=(-1, 3, 1) + t(2, -1, 3) is the same as the parametric equation x= -1+ 2t, y= 3- t, z= 1+3t.

Point (3, 1, -5).

Match the x, y and z coordinates to get:
x:
-1+ 2t = 3
2t = 3+1 =4
t =2

y:
3- t =1
t =2

z:

1+3t = -5

3t = 6

t= 2

Therefore: (3, 1, -5) lies on the plane.
------------------------

6)

The plane is 2x-y+z=9

Direction cosine vector  (2, -1, 1)

c) (1, 1, -1)

Perpendicular means dot product should be zero.

(2*1)+(-1*1)+(1*-1) = 2-1-1 =0

So optionc  c) is correct

selected Aug 18, 2014 by zoe

1)

a ⋅ b is a scaler product

if a =<x1,y1,z1>   b= <x2,y2,z2>

a ⋅ b =x1x2+y1y2+z1z2

a ⋅ b = |a||b|cosθ

|a|=magnitude of a

|b|=magnitude of b

--------------------------

2)

If nonzero vectors x and y are perpendicular then x ⋅y =0

Option d is correct.

x ⋅y =|x||y|cosθ

x and y are perpendicular means θ =90

x ⋅y =|x||y|cos90

=0

If nonzero vectors x and y are parallel then x =ky

Option b is correct.

"If  x is a unit vector same with same direction and sence as y, then"

Option a, c are correct.

--------------------------------

3)

The vectors a = < 1, 2,4 >; b = <4, -2, 1 >

Calculate the angle between the vectors by using following formula

cos θ =(u.v)/|u||v|

a . b = < 1, 2,4 > . <4, -2, 1 >

= 4(1) + (2)(-2)+(4)(1)

= 4

Calculate the magnitude of vectors.

Magnitude of a vector |a| = √(x2 + y2+ y2 )

|a| = √(12 +22 +42) = √21

|b| = √(42 +(-2)2 +12) = √21

cos θ =  4/(√21)(√21)

cos θ = 4/21

θ =  cos^-1(4/21)

= 79.02°

edited Aug 18, 2014 by bradely

7)

The point (1, 4, 2)

The plane, 5x+ y- 3z =k

The point contains on the plane.

5(1)+ 4- 3(2) =k

k =5+4-6

=3

------------------

8)

Conversely, the vector equation for the line x=(-1, 2, 4) + t(5, 1, 0) is the same as the parametric equation x= -1+ 5t, y= 2+ t, z= 4+0t.

Put y =0

2+ t = 0

t = -2

Calculate the x, z coordinates.

x= -1+ 5t

= -1+ 5(-2)

= -1- 10

= -11

z= 4+0t =4

The point is (-11, 0, 4)

-------------------

9)

The projection of the vector a onto the direction b is (a ⋅ b)/|b|

a . b = <4, 0, 7 > . <-1, -2, 2 >

= 4(-1) + (0)(-2)+(7)(2)

= 10

Calculate the magnitude of vector.

Magnitude of a vector |b| = √(x2 + y2+ y2 )

|b| = √((-1)2 +(-2)2 +22) = √9 =3

Projection of the vector a onto the direction b is

= (10)/3

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10)

Calculate the distance from point to the plane by using formula:

d =|Ax1+By1+zc1+D|/√(A2 + B2+ C2 )

Here, (A, B, C) = (2, 1, -2), D = -8 and (x1, y1, z1) = (1, 2, -5)

d =|Ax1+By1+zc1+D|/√(A2 + B2+ C2 )

=|2*1+ 1*2 +(-2)*(-5)-8|/√(22 + 12+(-2)2 )

=|2+ 2 +10 - 8|/√(4 + 1 + 4 )

=6/3

= 2