Vectors

**Tangera** · January 10th 2009, 04:59 AM

Q1: If a = i + 2j - k and b = j + k, find a unit vector perpendicular to both a and b.

===

Q2: The points P and Q have position vectors a + b and 3a - 2b respectively when referred from the origin O. Given that OPQR is a parallelogram, express the vectors PQ and PR in terms of a and b. [I have found PQ = 2a - 3b; PR = a - 4b] By evaluating 2 scalar products, show that if OPQR is a square, then|a| $^2$ = 2|b| $^2$ .

Thank you for helping!

**Prove It** · January 10th 2009, 05:03 AM

Originally Posted by Tangera

Q1: If a = i + 2j - k and b = j + k, find a unit vector perpendicular to both a and b.

===

Q2: The points P and Q have position vectors a + b and 3a - 2b respectively when referred from the origin O. Given that OPQR is a parallelogram, express the vectors PQ and PR in terms of a and b. [I have found PQ = 2a - 3b; PR = a - 4b] By evaluating 2 scalar products, show that if OPQR is a square, then|a| $^2$ = 2|b| $^2$ .

Thank you for helping!

1. Take the cross product of $\mathbf{a}$ and $\mathbf{b}$ and then divide this vector by its length.

**Prove It** · January 10th 2009, 05:05 AM

Originally Posted by Tangera

Q1: If a = i + 2j - k and b = j + k, find a unit vector perpendicular to both a and b.

===

Q2: The points P and Q have position vectors a + b and 3a - 2b respectively when referred from the origin O. Given that OPQR is a parallelogram, express the vectors PQ and PR in terms of a and b. [I have found PQ = 2a - 3b; PR = a - 4b] By evaluating 2 scalar products, show that if OPQR is a square, then|a| $^2$ = 2|b| $^2$ .

Thank you for helping!

2. Drawing a picture always helps.

Since it's a parallelogram, notice that OR is parallel and of equal magnitude to PQ and that QR is parallel and of equal magnitude to OP.

What do you have when vectors are parallel and of equal length? They are EQUAL.

**Tangera** · January 10th 2009, 05:21 AM

^ Thank you for your suggestions! I got stuck at part that requires the evaluation of the scalar product...

Do I evaluate PR.OQ = OP.PQ = 0?

**Prove It** · January 10th 2009, 05:26 AM

Originally Posted by Tangera

^ Thank you for your suggestions! I got stuck at part that requires the evaluation of the scalar product...

Do I evaluate PR.OQ = OP.PQ = 0?

What sides touch?

OP touches PQ

PQ touches QR

QR touches OR

OP touches OR.

If it's a square, then the angles should be right angles. Evaluating any of the dot products of touching sides should give 0 if this is the case.

Then show that the lengths are equal, and you've got a square.

See how you go from there.

**Tangera** · January 10th 2009, 05:37 AM

^ Um...sorry I am still confused...I understand your method, but the question wanted me to show |a|

= 2|b|

if OPQR is a square...so do I have to equate 2 scalar products?

**Prove It** · January 10th 2009, 05:44 AM

Originally Posted by Tangera

^ Um...sorry I am still confused...I understand your method, but the question wanted me to show |a|

= 2|b|

if OPQR is a square...so do I have to equate 2 scalar products?

Is that $|\mathbf{a}|^2 = 2|\mathbf{b}|^2$ ?

**Tangera** · January 10th 2009, 06:00 AM

^ Yup the question wanted

...

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