Finding scalar a where L - aK is perpendicular to L

[xphloem]

Homework Statement

Consider the two vectors L= i +2j+3K
K=4i+5j+6k
Find scalar 'a' such that:
L - aK is perpndicular to L.

Homework Equations

if two vectors are perpenicular dot product=0

The Attempt at a Solution

(i+2j+3k).{(1-4a)i+(2-5a)j+(3-6a)k}=0
I get three values of a here. but none satisfies th whole equations at the same time. Please help me

 

[Nick89]

I don't understand what you mean by the last line.

Can you show us how you calculated the dot product?
Surely it yields a linear equation in a? How can it possibly result in 3 values for a?
There is only one equation, how can you not be able to find one a that satisfies the whole equation at the same time?

You probably made a mistake with the dot product:

$$(a \hat{i} + b\hat{j} + c\hat{k} ) \cdot ( d\hat{i} + e\hat{j} + f\hat{k}) = ad + be + cf$$

 

[Redbelly98]

The Attempt at a Solution

(i+2j+3k).{(1-4a)i+(2-5a)j+(3-6a)k}=0
I get three values of a here. but none satisfies th whole equations at the same time. Please help me

Complete the dot product you wrote, using Nick89's formula if you didn't know it already. You'll get a linear equation in a.