Vectors flash cards

1. 1.Give the condition for three vectors $\vec{a}, \vec{b}, \vec{c}$ to be coplanar in terms of the scalar triple product.

   They are coplanar iff $[\vec{a}\ \vec{b}\ \vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) = 0$.

   Hint: Think of the box product as a volume.

2. 2.If $\vec{a}$ and $\vec{b}$ are two vectors with angle $\theta$ between them, what do $\vec{a}\cdot\vec{b}$ and $|\vec{a}\times\vec{b}|$ represent, and how are they related?

   $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$ (scalar, projection-based) and $|\vec{a}\times\vec{b}| = |\vec{a}||\vec{b}|\sin\theta$ (vector magnitude, area-based); together, provided $\vec{a}\cdot\vec{b}\neq 0$, $\tan\theta = \dfrac{|\vec{a}\times\vec{b}|}{\vec{a}\cdot\vec{b}}$.

   Hint: Dot gives cosine, cross gives sine.

3. 3.State the formula for the position vector of a point dividing the join of $A(\vec{a})$ and $B(\vec{b})$ in the ratio $m:n$ internally.

   Internal division: $\vec{r} = \dfrac{m\vec{b}+n\vec{a}}{m+n}$; external division: $\vec{r} = \dfrac{m\vec{b}-n\vec{a}}{m-n}$.

   Hint: Section formula, vector form.

4. 4.What is the vector (cross) product's direction rule and its use in finding the area of a triangle with vertices $\vec{a}, \vec{b}, \vec{c}$?

   Direction is given by the right-hand rule (perpendicular to both vectors, forming a right-handed system); area of triangle $=\dfrac{1}{2}|(\vec{b}-\vec{a})\times(\vec{c}-\vec{a})|$.

   Hint: Half the magnitude of a cross product.

5. 5.Define the scalar projection and vector projection of $\vec{a}$ on $\vec{b}$.

   Scalar projection $=\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$; vector projection $=\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\right)\vec{b}$.

   Hint: Divide dot product by magnitude of $\vec{b}$, once or twice.

6. 6.Prove that $[\vec a+\vec b,\ \vec b+\vec c,\ \vec c+\vec a]=2[\vec a\ \vec b\ \vec c]$ for any vectors $\vec a,\vec b,\vec c$.

   Expand $(\vec b+\vec c)\times(\vec c+\vec a)=\vec b\times\vec c+\vec b\times\vec a+\vec c\times\vec a$ (since $\vec c\times\vec c=\vec 0$). Dotting with $(\vec a+\vec b)$, every term containing a repeated vector vanishes, leaving $\vec a\cdot(\vec b\times\vec c)+\vec b\cdot(\vec c\times\vec a)=[\vec a\vec b\vec c]+[\vec a\vec b\vec c]$. Hence $[\vec a+\vec b,\vec b+\vec c,\vec c+\vec a]=2[\vec a\vec b\vec c]$.

   Hint: Expand the cross product first, then dot; terms with a repeated vector vanish.

7. 7.For non-coplanar $\vec a,\vec b,\vec c$, express $[\vec a\times\vec b,\ \vec b\times\vec c,\ \vec c\times\vec a]$ in terms of $[\vec a\ \vec b\ \vec c]$.

   Using $(\vec b\times\vec c)\times(\vec c\times\vec a)=[\vec b\vec c\vec a]\vec c-[\vec b\vec c\vec c]\vec a=[\vec a\vec b\vec c]\vec c$, we get $(\vec a\times\vec b)\cdot\big([\vec a\vec b\vec c]\vec c\big)=[\vec a\vec b\vec c]\big[(\vec a\times\vec b)\cdot\vec c\big]=[\vec a\vec b\vec c]^2$. So $[\vec a\times\vec b,\vec b\times\vec c,\vec c\times\vec a]=[\vec a\ \vec b\ \vec c]^2$.

   Hint: Reduce the vector-vector cross product using the four-vector identity first.

8. 8.Without expanding fully, show that $[\vec a-\vec b,\ \vec b-\vec c,\ \vec c-\vec a]=0$ for every choice of vectors $\vec a,\vec b,\vec c$.

   Note $(\vec a-\vec b)+(\vec b-\vec c)+(\vec c-\vec a)=\vec 0$, so the three vectors are always linearly dependent (one is minus the sum of the other two) and hence coplanar, making their scalar triple product identically zero regardless of $\vec a,\vec b,\vec c$.

   Hint: Check if the three difference-vectors themselves sum to something special.

9. 9.If $\vec a,\vec b,\vec c$ are non-coplanar with $[\vec a\ \vec b\ \vec c]=5$, evaluate $[\vec a+2\vec b,\ \vec b+2\vec c,\ \vec c+2\vec a]$.

   By trilinearity of the scalar triple product, the new box product equals $\det\begin{pmatrix}1&2&0\\0&1&2\\2&0&1\end{pmatrix}[\vec a\vec b\vec c]$. This determinant is $1(1)-2(0-4)+0=1+8=9$. Hence the value is $9\times5=45$.

   Hint: Write the coefficient matrix of the three combinations and take its determinant.

10. 10.Points $A(1,0,-1)$, $B(2,3,1)$, $C(0,1,2)$, $D(4,2,-1)$ are given. Determine whether they are coplanar.

    Take $\vec{AB}=(1,3,2)$, $\vec{AC}=(-1,1,3)$, $\vec{AD}=(3,2,0)$. Then $\vec{AC}\times\vec{AD}=(1\cdot0-3\cdot2,\ -((-1)\cdot0-3\cdot3),\ (-1)\cdot2-1\cdot3)=(-6,9,-5)$, and $\vec{AB}\cdot(-6,9,-5)=-6+27-10=11\neq0$. Since the scalar triple product is nonzero, the four points are not coplanar — they form a genuine tetrahedron of volume $11/6$.

    Hint: Coplanarity of 4 points means the box product of any two edge-vectors and a third from a common vertex is zero.

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