To solve this exercise regarding the tetrahedron $ABCD$ and the point $M$, let’s break down the problem step by step.

Part a) Express vector $\overrightarrow{BM}$ as a function of vectors $\overrightarrow{BC}$ and $\overrightarrow{BD}$

We are given: $\overrightarrow{AM} = 3\overrightarrow{AB} - \overrightarrow{AC} - \overrightarrow{AD}.$ We need to express $\overrightarrow{BM}$ in terms of $\overrightarrow{BC}$ and $\overrightarrow{BD}$.

First, using the relationship $\overrightarrow{BM} = \overrightarrow{BA} + \overrightarrow{AM}$, and knowing that: $\overrightarrow{BA} = -\overrightarrow{AB},$ we substitute $\overrightarrow{AM}$ into this equation: $\overrightarrow{BM} = -\overrightarrow{AB} + (3\overrightarrow{AB} - \overrightarrow{AC} - \overrightarrow{AD}).$ Simplifying this expression: $\overrightarrow{BM} = (3\overrightarrow{AB} - \overrightarrow{AB}) - \overrightarrow{AC} - \overrightarrow{AD},$ $\overrightarrow{BM} = 2\overrightarrow{AB} - \overrightarrow{AC} - \overrightarrow{AD}.$

Next, we express $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$ in terms of $\overrightarrow{BC}$ and $\overrightarrow{BD}$:

• $\overrightarrow{AB} = -\overrightarrow{BA} = \overrightarrow{BC} + \overrightarrow{BD}$,
• $\overrightarrow{AC} = \overrightarrow{BC}$,
• $\overrightarrow{AD} = \overrightarrow{BD}$.

Substituting these into the equation for $\overrightarrow{BM}$: $\overrightarrow{BM} = 2(\overrightarrow{BC} + \overrightarrow{BD}) - \overrightarrow{BC} - \overrightarrow{BD}.$ Simplifying this expression: $\overrightarrow{BM} = 2\overrightarrow{BC} + 2\overrightarrow{BD} - \overrightarrow{BC} - \overrightarrow{BD},$ $\overrightarrow{BM} = \overrightarrow{BC} + \overrightarrow{BD}.$

Thus, the vector $\overrightarrow{BM}$ is expressed as: $\overrightarrow{BM} = \overrightarrow{BC} + \overrightarrow{BD}.$

Part b) Determine which plane containing a face of the tetrahedron the point $M$ lies in

We are given that: $\overrightarrow{AM} = 3\overrightarrow{AB} - \overrightarrow{AC} - \overrightarrow{AD}.$ By analyzing this equation, note that it expresses $M$ as a linear combination of the vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$.

This means that point $M$ lies on the plane formed by points $A$, $C$, and $D$, which is the face of the tetrahedron $ACD$.

Final Answers:

• Part a): $\overrightarrow{BM} = \overrightarrow{BC} + \overrightarrow{BD}$.
• Part b): The point $M$ lies on the plane containing the face $ACD$ of the tetrahedron.

Let me know if you need more details or have any questions!

Here are 5 related questions:

1. How can we express $\overrightarrow{AM}$ in terms of vectors $\overrightarrow{BC}$ and $\overrightarrow{BD}$?
2. How do we find the equation of the plane passing through points $A$, $C$, and $D$?
3. Can we express $\overrightarrow{BM}$ in terms of different vector bases of the tetrahedron?
4. How does the position of point $M$ change if the coefficients in $\overrightarrow{AM}$ are altered?
5. How would the solution change if the given point $M$ was defined differently, say as $AM = 2\overrightarrow{AB} + \overrightarrow{AC} - \overrightarrow{AD}$?

Tip: When dealing with tetrahedrons, a useful strategy is to express vectors in terms of multiple bases (like $\overrightarrow{BC}$ and $\overrightarrow{BD}$) to simplify the problem geometry.