Question

If e=l i+mj+nk is a unit vector, the maximum value of lm+mn+nl is\

Answer 1

It is given that, e = l i+m j+n k , is a unit vector.

As ,$|e|=1$, where e is a unit vector.

Therefore, l²+m²+n²=1   ........................(1)

As ,→ (l+m+n)²= l²+m²+n²+ 2 (l m+m n+n l)

   → (l+m+n)²= 1+ 2 (l m+m n+n l)→[Using (1)]

→ As square of any number is always greater than zero, i.e a²≥0

So, (l+m+n)²≥ 0

∴ →1 +2 (l m+m n+n l)≥0

→2 (l m+m n+n l)≥ -1

→ l m+m n+n l≥ -1/2

Answer 2

Step-by-step explanation:

e=li^+mj^+nk^

∣e∣=l2+m2+n2

1=l2+m2+n2....(1)

Now the expression

(l−m)2+(m−n)2+(n−l)2≥0

⇒2∑l2−2∑lm≥0

Using (1) we get ∑lm≤1