Question

Equation of a plane: Find the equation of the plane determined by the vectors (1, −3, 5)and (2, 4, −1) and containing the point(0, 3, −2).

Short answer

The equation of the plane is$-17x+11y+10z-13=0.$

Step-by-step solution

Step 1. Given Information.

The given vectors of the plane are (1, −3, 5)and (2, 4, −1)and the given containing point is(0, 3, −2).

Step 2. Finding the equation of a plane.

To find the equation of a plane determined by the vectors and containing the point, we will use the formula: $\left(\overrightarrow{r}-\stackrel{\rightharpoonup }{a}\right)·\left(\overrightarrow{\alpha }\times \overrightarrow{\beta }\right)=0.$

Let's find the cross product,

role="math" localid="1649641070012" $$\begin{gathered} \left(\overrightarrow{\alpha }\times \overrightarrow{\beta }\right) \\ =\left|\begin{array}{ccc}i& j& k\\ 1& -3& 5\\ 2& 4& -1\end{array}\right| \\ =i\left(\left(-3\right)\left(-1\right)-\left(4\right)\left(5\right)\right)-j\left(\left(1\right)\left(-1\right)-\left(2\right)\left(5\right)\right)+k\left(\left(1\right)\left(4\right)-\left(2\right)\left(-3\right)\right) \\ =-17i+11j+10k \end{gathered}$$

Now, let's do the difference,

role="math" localid="1649641452589" $$\begin{gathered} \left(\overrightarrow{r}-\overrightarrow{a}\right) \\ =\left(xi+yj+zk\right)-\left(0i+3j-2k\right) \\ =i\left(x-0\right)+j\left(y-3\right)+k\left(z-\left(-2\right)\right) \\ =xi+(y-3)j+(z+2)k \end{gathered}$$

Step 3. Calculate.

Now, let's do the dot product,

$$\begin{gathered} \left[xi+(y-3)j+(z+2)k\right]·\left(-17i+11j+10k\right)=0 \\ \left(x\right)\left(-17\right)+\left(y-3\right)\left(11\right)+\left(z+2\right)\left(10\right)=0 \\ -17x+11y-33+10z+20=0 \\ -17x+11y+10z-13=0 \end{gathered}$$