Quiz 1 Solution

Problem 1. Let $\vec{a}=\left(6,4,2\right),\quad \vec{b}=\left(-9,6,3\right),\quad \vec{c}=\left(-3,6,3\right)$.

(a) Compute $2\vec{a}-\vec{b}+\vec{c}$ and $-3\vec{a}+2\vec{b}+4\vec{c}$.

Solution.

\begin{align*} 2\vec{a}-\vec{b}+\vec{c} &= \left(12+9-3, 8-6+6, 4-3+3\right)=\left(18, 8, 4\right),\\ -3\vec{a}+2\vec{b}+4\vec{c} &= \left(-18-18-12, -12+12+24, -6+6+12\right)=\left(-48, 24, 12\right). \end{align*}

(b) Do $\vec{a},\vec{b},\vec{c}$ lie on the same plane? Show your computation.

Solution. Yes. Check the volume of the parallelepiped spanned by $\vec{a},\vec{b},\vec{c}$:

\begin{align*} \vec{a}\times\vec{b}\cdot\vec{c} =\begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \notag \end{vmatrix} =\begin{vmatrix} 6 & 4 & 2 \\ -9 & 6 & 3 \\ -3 & 6 & 3 \notag \end{vmatrix}=0. \end{align*}

(If you notice that the second column is a scalar multiple of the third column, you can deduce that the determinant must be $0$ — without any computation.)

Problem 2. Find the distance from the point $\left(0,2,1\right)$ to the plane $2x-3y+5z-1=0$. (Hint: The only formula you need is the dot product. Keep $\sqrt{38}$ as it is.)

Solution. Let $P$ be an arbitrary point on the plane, say $P=\left(-1,-1,0\right)$. The distance from $A=\left(0,2,1\right)$ equals to the absolute value of the component of the vector $\overrightarrow{AP}$ along the normal vector of the plane.

A normal vector of the plane can be chosen as $\vec{n}=\left(2,-3,5\right)$. Moreover, $\overrightarrow{AP}=\overrightarrow{OP}-\overrightarrow{OA}=\left(-1,-3,-1\right)$. Thus the desired distance is

$$\left|\mathrm{comp}_{\vec{n}}\,\overrightarrow{AP}\right|=\frac{\left|\overrightarrow{AP}\cdot\vec{n}\right|}{\left|\vec{n}\right|}=\frac{\left|-2+9-5\right|}{\sqrt{38}}=\frac{2}{\sqrt{38}}.$$

Problem 3. Consider a point $M$ in the same plane as the triangle $ABC$. Show that there exist real numbers $\lambda,\mu,\nu$ such that

$$\overrightarrow{OM}=\lambda \overrightarrow{OA}+\mu\overrightarrow{OB}+\nu\overrightarrow{OC}$$

and

$$\lambda + \mu + \nu = 1.$$

(Hint: $\overrightarrow{AM}$ lies in the plane spanned by $\overrightarrow{AB}$ and $\overrightarrow{AC}$.)

triangle

Proof. Since $M$ in the same plane as the triangle $ABC$, $\overrightarrow{AM}$ lies in the plane spanned by $\overrightarrow{AB}$ and $\overrightarrow{AC}$. Thus there exists real numbers $s,t$ such that

\begin{equation}\label{eq:linearDependent} \overrightarrow{AM} = s\,\overrightarrow{AB} + t\,\overrightarrow{AC}. \end{equation}

Note that

\begin{align*} \overrightarrow{AM} &= \overrightarrow{OM}-\overrightarrow{OA},\\ \overrightarrow{AB} &= \overrightarrow{OB}-\overrightarrow{OA},\\ \overrightarrow{AC} &= \overrightarrow{OC}-\overrightarrow{OA}, \end{align*}

thus \eqref{eq:linearDependent} reduces to

$$\left(\overrightarrow{OM}-\overrightarrow{OA}\right) = s\left(\overrightarrow{OB}-\overrightarrow{OA}\right) + t\left(\overrightarrow{OC}-\overrightarrow{OA}\right).$$

Re-grouping terms involving $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$, one has

$$\overrightarrow{OM}=\left(1-s-t\right) \overrightarrow{OA}+s\,\overrightarrow{OB}+t\,\overrightarrow{OC}.$$

Set

$$\lambda = 1-s-t,\quad \mu=s,\quad \nu=t,$$

then

$$\overrightarrow{OM}=\lambda \overrightarrow{OA}+\mu\overrightarrow{OB}+\nu\overrightarrow{OC}$$

and

$$\lambda + \mu + \nu = 1.$$

$\blacksquare$

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