Question

The vectors a-5b and a-b are perpendicular. If a and b are unit vectors, then determine a (dot) b.

Answer 1

You have vectors
\[v_1=\vec{a}-5\vec{b}\]
and \[v_2=\vec{a}-\vec{b}\]
they are perpendicular, so their dot product is...?

Answer 2

0

Answer 3

but the answer is 1

Answer 4

v1.v2=0
\[a ^{2}=1,b ^{2}=1,find a.b\]

Answer 5

@surjithayer I don't get it...sorry

Answer 6

@amistre64 please help

Answer 7

Does this question have an answer? If you take the dot product of the two vectors which should be perpendicular, you have an expression for the ratio of the length of the unit vectors a and b. They are not 1 however...am I missing something?

Answer 8

since they are being very general; lets go ahead and define some parts that would be useful to us: a = (1,0), and let b = (x,y)

a-5b

u = (1-5x,0-5y)
v = ( 1-x , 0-y)

then we can establish the dot product as zero

Answer 9

@TimSmit the answer is 1, but I don't understand how they get that

Answer 10

the dot product should give us the equation of a circle ...

Answer 11

or an ellipse, maybe a hyperbola ..... something conic fer sure :)

u = (1-5x,0-5y)
v = ( 1-x , 0-y)
---------------
(5x^2-6x+1) + (5y^2-6y) = 0

5(x^2-6x/5)+1 + 5(y^2 - 6y/5) = 0

(x^2-6x/5) +1/5 + (y^2 - 6y/5) = 0

(x^2-6x/5+36/100) - 36/100 +1/5 + (y^2 - 6y/5 + 36/100) -36/100 = 0

(x-6/10)^2 + (y-6/10)^2 + (20 - 72)/100 = 0

(x-6/10)^2 + (y-6/10)^2 = 27/50

yay!! its a circle

Answer 12

where was i going with this ?.....

Answer 13

oh right .... sqrt(x^2+y^2) = 1; therefore say: y = sqrt(1-x^2) and solve for x

Answer 14

(1,0) dot (x,y) = x

Answer 15

I can't seem to solve it :) the dot product of the two perpendicular vectors is:
\[||a||^2+5||b||^2=0\]
If a and b are unit vectors then their lengths are 1, so 6=0?

Answer 16

Is there any simpler explanation?

Answer 17

probably, but i tend to have to go off of what i can piece together from years gone by :)

Answer 18

Could you check your vectors eLg? are they really a-5b and a-b?

Answer 19

yes they are right

Answer 20

\[(x_a-5x_b,y_a-5y_b)\cdot (x_a-x_b,y_a-y_b)=0\]
i do get a solution close to 1, approx 0.99986...

Answer 21

Ah I was thinking too perpendicular for unit vectors while taking the dot product.

Answer 22

since the question is general, i dont see a reason why we couldnt define a or b as 1,0; lets use b instead and see how that goes

\[(x_a-5,y_a)\cdot (x_a-1,y_a)=0\]
\[(x^2-6x+5)+(y^2)=0\]
\[(x^2-6x+9)+(y^2)=4\]
since sqrt(x^2+y^2)=1, let y = sqrt(1-x^2)
\[(x^2-6x+9)+(1-x^2)=4\]
\[-6x+9+1=4\]
\[-6x=-6\]
\[x=1\]

Answer 23

therefore:

(x,y) dot (1,0) = x = 1

Answer 24

i most likely did some bad mathing the first go around :)

Answer 25

oh ok,

Answer 26

Working out your dot product amistre:
\[(x_a-5x_b,y_a-5y_b)\cdot(x_a-x_b,y_a-y_b)=0\]
\[(x_a-5x_b)*(x_a-x_b)+(y_a-5y_b)*(y_a-y_b)=0\]
\[x_a^2-6x_ax_b+5x_b^2+y_a^2-6y_ay_b+5y_b^2=0\]
As a and b are unit vectors:
\[\sqrt{x_a^2+y_a^2}=1=x_a^2+y_a^2\]
Then
\[-6x_ax_b-6y_ay_b+6=0\]
Now, choosing for a=(1,0) gives:
\[-6x_b+6=0\]
\[x_b=1\]
As b is a unit vector, b=(1,0)
So the dot product of a and b is 1 as a=b.
Correct?

Answer 27

id have to take a few aspirins to be sure :)

Answer 28

because they are perpendicular so dot product is 0
\[\left( vectora-vectorb \right).\left( vectora-vector5b \right)=0\]
\[or\left( vectora \right)^{2}-5\left( vectora \right).\left( vector5b \right)-\left( vectorb \right).\left( vectora \right)+5\left( vectorb \right)^{2}= 0\]
\[1-6 \left( vectora \right).\left( vectorb \right)+5\left( 1 \right)= 0 \]
\[6\left( vectora \right).\left( vectorb \right) =6\]
\[( vector a).\left( vectorb \right) =\frac{ 6 }{6 }=1 \]

Answer 29

\[(x^2_a-6x_ax_b+5x^2_b)+(y^2_a-6y_ay_b+5y^2_b)=0\]
\[y_a=\sqrt{1-x_a^2~}~:~y_b=\sqrt{1-x_b^2~}\]
\[x^2_a-6x_ax_b+5x^2_b+(\sqrt{1-x_a^2~})^2-6\sqrt{(1-x_a^2)(1-x_b^2)}+5(\sqrt{1-x_b^2~})^2=0\]
\[x^2_a-6x_ax_b+5x^2_b+1-x_a^2-6\sqrt{(1-x_a^2)(1-x_b^2)}+5-5x_b^2=0\]
\[-6x_ax_b-6\sqrt{(1-x_a^2)(1-x_b^2)}+6=0\]
\[1=x_ax_b+\sqrt{(x_ax_b)^2-x_a^2-x^2_b+1}\]
and as long as i dont mess it up along the way :)

Answer 30

pfft, i think i should have kept the ys in there

Answer 31

surji showed how do it easily
\[ (a-5b)\cdot ( a-b) =0 \\ a^2 -5(a\cdot b) - a\cdot b +5 b^2 =0\]
a, and b are unit length. simplify and combine like terms
\[ 1 -6(a\cdot b) + 5 =0 \\ a\cdot b= 1\]

Answer 32

\[-6x_ax_b-6\sqrt{(1-x_a^2)(1-x_b^2)}+6=0\]
\[-6x_ax_b-6\sqrt{(1-x_a^2)}\sqrt{(1-x_b^2)}+6=0\]
\[x_ax_b+\sqrt{(1-x_a^2)}\sqrt{(1-x_b^2)}=1\]
\[x_ax_b+y_ay_b=1\]
and therefore the the defintion of a.b

Answer 33

correction i typed 5 extra in the third line.