Scalar products(dot product) in terms of $\lambda$

Question

After spending a solid day or two on vectors, particularly Scalar/Dot products. I come across one of the questions that doesn't quite match the same format I find online. Before continuing, I'll post the actual question:

---

With respect to origin O, the position vectors of points A, B and C are:

$$a = \lambda i + 2j -k, b = 3i + 4j -2k, c = 5i - j +2k$$ Respectively, $\lambda$ is a constant.

• (a) Find the scaler product $\vec{BA} . \vec{BC}$ in terms of $\lambda $

---

There's a whole(much larger) question proceeding that, however I would like to atleast try tackling it myself first before asking.

The first issue I've personally had was, $\lambda$. I've never come across it in a vector before, none of the vectors have had a variable so I would always find something like $b$ and $c$ however I'm positive it's just me overthinking all of it.

I've worked around with dot product and most of the time used equations: $$\cos(\theta) = \frac{\vec{BA}.\vec{BC}}{\left|\vec{BA}\right|\left|\vec{BC}\right|}$$

Or

$$\vec{BA}.\vec{BC} = \left | \vec{BA} \right |\left | \vec{BC} \right |\cos(\theta)$$

Amongst a few other situational ones

---

The question itself holds little to no points so I presume I'm overdoing what I need to do, spending 10 minutes each revision and not coming with a concrete answer is definitely not the way to go.

---

The final small question I would like to ask in relation to this is how $a = \lambda i + 2j -k, b = 3i + 4j -2k, c = 5i - j +2k$

Is related to $\vec{BA}.\vec{BC}$

---

Attempts so far:

(a)I first found all the values for $\vec{OA}$, $\vec{OB}$, $\vec{OC}$ which were:

• $\vec{OA} = ( \lambda , 2, -1)$
• $\vec{OB} = ( 3, 4, -2)$
• $\vec{OC} = ( 5, -2, 2)$

I then using one of the provided answers used those to find values for $\vec{BA}$ and $\vec{BC}$ like so: $$\vec{BA} = \vec{OA} - \vec{OB} $$ Therefore: $$(\lambda, 2, -1) - (3, 4, -2) = (\lambda - 3, -2, 1)$$

And did the same for $\vec{BC}$. The answers boiled down to: $$(\lambda-3, -2, 1).(2, -5, 4)$$

Thus: $(2\lambda + 8)$ seems to be the answer

---

I do not wish to ask for a final answer, but more of a guideline on how to achieve it

Answer 1

$\vec{BA}=\vec{A}-\vec{B}=(\lambda,2,-1)-(3,4,-2)=(\lambda-3,-2,1)$

Answer 2

Hints; $$\vec{OA}=(\lambda, 2, -1), \vec{OB}=(3, 4, -2), \vec{OC}=(5, -1, 2)$$ You can find $\vec{BA}$ and $\vec{BC}$ from here. When you have coordinates, it's convenient to use the vector representation I used.

Answer 3

Do you know the other definition of "standard scalar product" in this context?

I mean other than the $\cos (\theta)$-stuff?

The $i$, $j$, and $k$ are the base vectors, I suppose? Then you should be able to write the scalar product as $(a-b) \cdot (c-b)$ where the $\cdot$ refers to the scalar product an $a$, $b$, and $c$ are as in your text.

Hope this helps.