Dot product with remarkable identities only

Question

I’m studying and doing an exercise from a school textbook. The wording is the following : $\triangle ABC$ is a triangle, $||AB|| = 3$, $||BC|| = 5$, and $||AC|| = 6.$ Calculate $AB ⋅ AC$. I couldn’t come up with a result using remarkable identities which seems the only method to use here. (By isolating the dot product on the right side of the equation). $$ ||AB + AC||^2 = ||AB||^2 + 2 AB ⋅ AC + ||AC||^2. $$ I’m stuck here. How will I replace $||AB+AC||^2$ knowing that I can’t just put $AB$ and $AC$ and then square it. I can’t use triangle inequalities since $-BA + AC$ is different than $-BC$? What can I do. The textbook correction says the result is $10$ but there’s no detail. Any answer appreciated.

Answer 1

The problem is

$\triangle ABC$ is a triangle, $||AB|| = 3$, $||BC|| = 5$, and $||AC|| = 6.$ Calculate $AB \cdot AC$.

You were close, but there is a sign difference to notice.

By definition of a Euclidean vector as directed line segment, $\vec{AB} := \vec{B}-\vec{A}$ is the vector from point $A$ to point $B$. Then, from the given side lengths get

$$ 3 = ||AB|| = ||\vec{AB}|| = ||\vec{B}-\vec{A}|| = ||\vec{A}-\vec{B}|| = ||\vec{BA}||. \tag1$$

Also, similarly given side lengths, get

$$ 5 = ||BC|| = ||\vec{BC}|| = ||\vec{B} -\vec{C}|| = ||\vec{B} -\vec{A} + \vec{A} - \vec{C}|| = ||\vec{BA} +\vec{AC}||. \tag2$$

Then use the dot product to get

$$ \vec{BC}\cdot\vec{BC} = ||\vec{BC}||^2 = ||\vec{BA} +\vec{AC}||^2 = ||\vec{BA}||^2 + 2 \vec{BA} \cdot \vec{AC} + ||\vec{AC}||^2. \tag3$$

Substitute the given side lengths in equation $(3)$ and move squares to the left side to get

$$ 5^2 - 3^2 - 6^2 = -20 = 2 \vec{BA} \cdot \vec{AC}. \tag4$$ But now, easily get

$$ \vec{BA}= \vec{A}-\vec{B}= -(\vec{B}-\vec{A})= -\vec{AB}. \tag5 $$

Combining equations $(4)$ and $(5)$, the answer is

$$ 10 = -\vec{BA} \cdot \vec{AC} = \vec{AB} \cdot \vec{AC}. \tag6 $$