Finding Parallel and Perpendicular Vectors with a Given Value of x and y

[BilloRani2012]

Homework Statement

Consider the vectors p = xi + 5j + yk and q = 2i - 4j + 3k

a) for what values of x and y are the vectors p and q parallel?

b) If x = -7 for what value of y are the vectors p and q perpendicular? Given that x and y take these values find a third vector that is perpendicular to both p and q.

Homework Equations

The Attempt at a Solution

This is what i have done:

(2, -4, 3)

so, -4c = 5
therefore c = -5/4

so i checked for x and y:
x = -5/4 * 2 = 2
y = -5/4 * 3 = 3

so the values of x and y that i got were: x = 2 and y = 3

 

[I like Serena]

Hi BilloRani2012, welcome to PF!

so i checked for x and y:
x = -5/4 * 2 = 2
y = -5/4 * 3 = 3

so the values of x and y that i got were: x = 2 and y = 3

You say you "checked" for x and y.
I'd rather say that you "calculated" x and y.

However, it's not true that for instance -5/4 * 2 = 2.
Can you see the mistake there?

If you calculate this properly, you will indeed have the values for x and y.

 

[BilloRani2012]

oh i see..sorry, so my values for x and y would actually be

x = -5/4*2 = -2.5
y= -5/4*3 = -3.75

Is that right?

 

[I like Serena]

oh i see..sorry, so my values for x and y would actually be

x = -5/4*2 = -2.5
y= -5/4*3 = -3.75

Is that right?

Yep!

Cheers!

 

[BilloRani2012]

okay thanks :)

could you please help me with the other question i put up?
it is called "vector question"

 

[BilloRani2012]

so for part b) would i just sub -7 in for x and take the dot product?

 

[BilloRani2012]

Perpendicular vectors

Homework Statement

Consider the vectors p = xi + 5j + yk and q = 2i - 4j + 3k

a) If x = -7 for what value of y are the vectors p and q perpendicular? Given that x and y take these values find a third vector that is perpendicular to both p and q.

Homework Equations

The Attempt at a Solution

This is what i got for a)

p = -7i + 5j + yk abd q = 2i - 4j + 3k

2c = -7
c = -2/7

I then multiplied -2/7 by all the numbers in the q equation
so,

-7/2*2 = -7
-7/2*-4 = 14
-7/2*3= -10.5

therfore, i got that y = -10.5

is that right?? and how would i find another vector that is perpendicular to p and q?

 

[mege]

You should be able to use the cross product and dot product to find what you are looking for.

Dot Product = 0 = Perp.
Crossproduct generates a perp. vector given any 2.

 

[BilloRani2012]

okay, so i did the cross product of p and q and got this:

p = -7i + 5j + yk abd q = 2i - 4j + 3k

so,

p*q = (-7*2) + (5*-4) + (y*3)
= -14 - 20 + 37
= -34 + 3y = 0
3y = 34
y = 11.33

is that right?

 

[mege]

That is the dot product which you used, but that looks correct.

You can take your newly formed vectors and find a third vector orthogonal to both by using the cross product (explained well by the physorg link that automatically populates these threads).

 

[BilloRani2012]

okay so to find the third vector do i just multiply p and q together:

so, (-7i + 5j + 11.33k) * (2i - 4j + 3k)
and expand and solve right??

i've tried it and i got --> -30.32i + 1.66j + 18k

is that right?

 

[I like Serena]

so for part b) would i just sub -7 in for x and take the dot product?

Correct!

The dot product must come out zero, meaning the vectors are perpendicular.

 

[Dazz4C]

okay so to find the third vector do i just multiply p and q together:

so, (-7i + 5j + 11.33k) * (2i - 4j + 3k)
and expand and solve right??

i've tried it and i got --> -30.32i + 1.66j + 18k

is that right?

Are you expanding using determinants?...you will get a different answer.

 

[BilloRani2012]

by determinants do u mean, if i multiply i by j it gives k, and if a multiply i by k it gives me k...and so on...

cuz that's what i did...is it wrong?

 

[Dazz4C]

by determinants do u mean, if i multiply i by j it gives k, and if a multiply i by k it gives me k...and so on...

cuz that's what i did...is it wrong?

Check page 204 in your lecture notes...;) and yes, that is incorrect.

Hint: a * b = i|ad-bc|-j|ad-bc|+k|ad-bc|

 

[BilloRani2012]

how did you know that it's on pg 204?? lol...but yea i did it..thanks :)

but my value that i calcualtd for y is right isn't it??

so i did the dot product of p and q and got this:

p = -7i + 5j + yk abd q = 2i - 4j + 3k

so,

p*q = (-7*2) + (5*-4) + (y*3)
= -14 - 20 + 37
= -34 + 3y = 0
3y = 34
y = 11.33

 

[Dazz4C]

how did you know that it's on pg 204?? lol...but yea i did it..thanks :)

but my value that i calcualtd for y is right isn't it??

so i did the dot product of p and q and got this:

p = -7i + 5j + yk abd q = 2i - 4j + 3k

so,

p*q = (-7*2) + (5*-4) + (y*3)
= -14 - 20 + 37
= -34 + 3y = 0
3y = 34
y = 11.33

Your value for y is correct, but I don't understand how you went from (y*3) to 37. Unless it was a typo and was meant to be 3y. You should also be stating why you have it equate to 0.

 

[berkeman]

(two threads merged)

 

[shahumair12]

i didnt understand the answer please give it in detail