Need help with 10 algebra questions. Desperate

Question

I understand that not all of these have answers but i just need a simplifed way of working them out if i am faced with these questions. Thank you :) and i get how to find most of these, i just want to be 100% sure

 

Answer 1

4) a < 6, 4, -8 >, b < - 3, -2, 4 >, c < 9, 6, -12 > , d = < 3, -2, -4>

Since,cross product of a and b is zero they are parallel.

Since,cross product of b and c is zero they are parallel.

Since,cross product of c and d is not zero they are not parallel.

The cross product of d and a is not zero, they are not parallel.

The cross product of d and b is not zero, they are not parallel.

Therefore, d is not parallel to other vectors.

 

Answer 2

1)

The position vector of X, OX =<2,5,-1>

The position vector of Y, OY =<0,8,4>

Find the vector XY :

XY= OY-OX

    =<0,8,4>-<2,5,-1>

    =<0-2,8-5,4-(-1)>

    =<-2,3,5>

Therefore ,the vector XY is <-2,3,5>

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

2)

4x+<1,-1,3>= <1,3,-5>

4x =<1,3,-5>-<1,-1,3>

    =<1-1,3-(-1),-5-3)>

    =<0,4,-8>

Divide each side by 4.

4x /4=<0,4,-8>/4

x =<0/4,4/4,-8/4>

    =<0,1,-2>

Therefore ,the vector X is <0,1,-2>

Answer 3

3)

Given vector , v =<11,10, -2>

If v = <a, b, c> is a vector , then magnitude of the vector is |v| = √((a)² + (b)² +(c)²).

Here, a = 11, b = 10 and c = -2

|v| = √((11)² + (10)² + (-2)²)

     =√(121+100+4)

      =√225

      = 15 units.

The magnitude of the given vector is 15 units.

5)

Given vector , d =<6, -3, 2>

Formula for unit vector: â = a/|a|

Find magnitude of vector:

|d| = √((6)² + (-3)² + (2)²)

     =√(36+9+4)

      =√49

      = 7 units.

Unit vector of the given vector:

d cap  = d/|d|

           = (1/7)(<6, -3, 2>)

The unit vector of the given vector is (1/7)(<6, -3, 2>)

Answer 4

6) (3, 6, 1) ( 9, -2, - 7)

Mid point is [ (3 + 9)/2 , (6 - 2) /2, (1 - 7)/ 2]

= [ 6, 2, -3]

7) PX : XQ = 3 : 8

∝ = 3 , β = 8

Formula x = (∝q + βp) /( ∝ + β)

x = (3q + 8p)/(3 + 8)

x = (8p + 3q)/ 11

8) a < 2, -1, 3 >,  b < 4,1,-2 >

a . b = 2(4) + (-1)(1) + 3(-2)

         = 8 - 1 - 6

a.b    = 1

a , b are not perpendicular since dot product of a and b is not equal to zero.

The vectors are not perpendicular.

9) If PXQY is a parallelogram, then PX = YQ

And PX parallel to YQ

Parallel vectors have same magnitude and direction.

So vector (PX) = vector(YQ)

x - p = q - y

y = q +  p - x

10) u, v , w are vectors

u .v is scalar.

Then (u . v) w is a vector.

Optin B is correct choice.

(u . v) . w is a scalar.

Option a is correct choce.