|p + q| = |p - q|

|p + q|² = |p - q|²

(p + q)•(p + q) = (p - q)•(p - q)

p•p + 2p•q + q•q = p•p - 2p•q + q•q

4p•q = 0

p•q = 0

Since this is the case, they are perpendicular.


Or,


Applying law of cosines,

( |p + q|² = |p|² + |q|² + 2|p||q|cos θ )


|p|² + |q|² + 2|p||q|cos θ = |p|² + |q|² + 2|p||q|cos (π - θ)

2|p||q|cos θ = -2|p||q|cosθ

4|p||q| cos θ = 0

|p||q| cos θ = 0

p•q = 0

|(p + q) x (p - q)|

= |p x p - p x q + q x p - q x q|

= |0 + q x p + q x p - 0|

= |2 q x p|

= 2|q||p|sin θ

2 x 1 x 2 x sin(π/3)

= 4 sin(π/3)

= 4(√3/2)

= 2√3

iii) the cross product of p and q i.e p x q is perpendicular to both p and q.

So for p.(p x q), you are performing a dot product of p and a vector that is perpendicular to it.

The dot product of 2 perpendicular vectors is 0

p.(p x q) = |p||p x q|cos (π/2)

= |p||p x q| x 0 (since cos (π/2) = 0)

= 0

ii)

|p.q| = | |p||q|cos(π/3) |

= |2 x 1 x ½|

= |1|

= 1

It means that the magnitude/absolute value of the product of the magnitude of a vector q (2 units) and the length of projection of another vector p onto q equals to 1.

Omg i understand it now ! Thank you so much for your help ! :))) appreciate it

Welcome. Hope the answers are correct