Question

The plane $2x+3y+6z=6$ intersects the coordinate axes at points $P,Q,R,$ forming a triangle. Find the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$. Write a vector formula for the area of the triangle $PQR,$ and find the area.

Short answer

The area of triangle PQR is 3.5 square units.

Step-by-step solution

- Find intersection points

Determine where the plane intersects the coordinate axes by setting two of the variables to zero at a time. For the x-axis (y=0, z=0), solve for x: $2x+3(0)+6(0)=6$ $x=3$ Point P is (3,0,0). For the y-axis (x=0, z=0), solve for y: $2(0)+3y+6(0)=6$ $y=2$ Point Q is (0,2,0). For the z-axis (x=0, y=0), solve for z: $2(0)+3(0)+6z=6$ $z=1$ Point R is (0,0,1).

- Find vectors PQ and PR

Find the vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ using the coordinates of P, Q, and R. $\overrightarrow{PQ}=Q-P=(0,2,0)-(3,0,0)=(-3,2,0)$ $\overrightarrow{PR}=R-P=(0,0,1)-(3,0,0)=(-3,0,1)$

- Vector formula for the area of triangle

The area of triangle PQR can be found using the cross product formula for vectors. The area is half the magnitude of the cross product of $\overrightarrow{PQ}$ and $\overrightarrow{PR}$. Find the cross product: $\overrightarrow{PQ}\times \overrightarrow{PR}=(-3,2,0)\times (-3,0,1)$ $=\left|\begin{array}{ccccccc}\mathbf{i}& \mathbf{j}& \mathbf{k}-3& 2& 0-3& 0& 1\end{array}\right|=\mathbf{i}(2\cdot 1-0\cdot 0)-\mathbf{j}(-3\cdot 1-0\cdot -3)+\mathbf{k}(-3\cdot 0-(-3\cdot 2))$ $=2\mathbf{i}+3\mathbf{j}+6\mathbf{k}$ The magnitude of $2\mathbf{i}+3\mathbf{j}+6\mathbf{k}$ is: $\sqrt{2^2+3^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7$ The area of the triangle is half the magnitude: $\text{Area}=\frac{1}{2}\times 7=\frac{7}{2}=3.5$

Key concepts

plane intersection

To determine where a plane intersects the coordinate axes, we need to find points where two of the coordinates are zero, while the third one is non-zero. This method helps in simplifying the equation. For example, for the plane equation given as $2x+3y+6z=6$:
* For the x-intercept, set $y=0$ and $z=0$. Solving $2x=6$ gives $x=3$. Hence, the point is $(3,0,0)$, known as point P.
* For the y-intercept, set $x=0$ and $z=0$. Solving $3y=6$ gives $y=2$. Hence, the point is $(0,2,0)$, known as point Q.
* For the z-intercept, set $x=0$ and $y=0$. Solving $6z=6$ gives $z=1$. Hence, the point is $(0,0,1)$, known as point R. These points form the vertices of the triangle on the plane. Understanding these intersections is crucial for forming the vectors needed for further calculations.

cross product

The cross product is a vector operation that takes two vectors and returns a third vector that is perpendicular to both input vectors. This is particularly useful in finding the area of a triangle formed by vectors. For vectors $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ calculated previously:
$$\overrightarrow{PQ}=(-3,2,0)$$ and
$$\overrightarrow{PR}=(-3,0,1)$$
You perform the cross product as follows:
$$\overrightarrow{PQ}\times \overrightarrow{PR}=\left|\begin{array}{ccccccc}\mathbf{i}& \mathbf{j}& \mathbf{k}-3& 2& 0-3& 0& 1\end{array}\right|$$
This results in the vector $2\mathbf{i}+3\mathbf{j}+6\mathbf{k}$. The resulting vector is perpendicular to both $\overrightarrow{PQ}$ and $\overrightarrow{PR}$, providing a direction that is normal to the plane formed by these vectors.

magnitude of vectors

The magnitude of a vector represents its length. To find the magnitude of a vector $\overrightarrow{A}=a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$, you use the formula:
$$\Vert \overrightarrow{A}\Vert =\sqrt{a^2+b^2+c^2}$$
From our cross product result, the vector was $2\mathbf{i}+3\mathbf{j}+6\mathbf{k}$. Therefore, its magnitude is:
$$\sqrt{2^2+3^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7$$
Understanding the magnitude helps in quantifying the 'size' of the vector, which is important when calculating areas or other quantities involving vectors.

triangle area formula

To find the area of a triangle using vectors, we use the magnitude of the cross product of two of its side vectors. The area $A$ of the triangle is given by:
$$A=\frac{1}{2}\times \Vert \overrightarrow{PQ}\times \overrightarrow{PR}\Vert$$
From the previous step, we found the cross product's magnitude to be 7. Therefore, the area of the triangle PQR is:
$$A=\frac{1}{2}\times 7=3.5$$
This method leverages the properties of the cross product and its magnitude, making it a powerful way to compute the area for triangles in vector analysis.