Find Scalar 'a' for Perpendicular Vectors L and K | Vector Product Homework

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To find the scalar 'a' such that the vector L - aK is perpendicular to L, the dot product must equal zero. The dot product was calculated incorrectly, leading to confusion about obtaining three values for 'a'. Participants emphasized that the correct application of the dot product formula should yield a single linear equation in 'a'. Clarification was provided to ensure the calculation aligns with the standard dot product method. Properly executing this will help identify the correct scalar 'a' that satisfies the condition of perpendicularity.
xphloem
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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
 
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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
 
xphloem said:

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.
 
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