Question

For any real number x, and any vectors u and v, show that

(u+xv).(u+xv)=|u|2+2x(u.v)+x2|v|2,
And hence that |u|2+2x(u.v)+x2|v|2>=0 for any real number x.

Answer 1

To prove the given equation, expand the dot product and use the commutative property. Rewrite the equation as a quadratic to show that it is always greater than or equal to zero.

Step-by-step explanation:

To prove the equation (u + xv) ∙ (u + xv) = |u|2 + 2x(u ∙ v) + x2|v|2, we can expand the dot product using the distributive property and the fact that u ∙ v = v ∙ u:

(u + xv) ∙ (u + xv) = u ∙ u + u ∙ (xv) + (xv) ∙ u + (xv) ∙ (xv)

= |u|2 + x(u ∙ v) + x(v ∙ u) + x2|v|2

= |u|2 + 2x(u ∙ v) + x2|v|2

Since the dot product of any two vectors is commutative, u ∙ v = v ∙ u. Therefore, we have proven the equation.

To show that |u|2 + 2x(u ∙ v) + x2|v|2 ≥ 0 for any real number x, we can rewrite the equation in the form of a quadratic: x2|v|2 + 2x(u ∙ v) + |u|2 ≥ 0. This is a quadratic equation in x, and since the discriminant is less than or equal to zero for any real number x, the inequality holds true for all values of x.