Question

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

Short answer

The formula for triangle is $△=\frac{1}{2}\left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|$, and the area of triangle calculated with this formula is $△PQR=\frac{7}{2}$units.

Step-by-step solution

Given information.

The given plane is $2x+3y+6z=6$that intersect the co-ordinate axes at points P,Q and R in order.

Area of triangle.

The areaof a triangle with base$△$and heightis given by $\mathbf{△}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{2}\mathbf{\times }\mathit{b}\mathbf{\times }\mathit{h}$

The vector in 3D-plane joining points Pand Qis $\overrightarrow{\mathbf{P}\mathbf{Q}}\mathbf{=}$Position vector of point Qposition vector of point P

Find the vectors representing the side and the area of a triangle made by the intersecting points of a plane with the co-ordinate axes.

The given plane is $2x+3y+6z=6$. Let the given plane intersects co-ordinate axes, localid="1659009881863" $OX,OY,andOZ$(O the origin) at points,$P(\alpha ,0,0),Q(0,\beta ,0),$and$R(0,0,\lambda )$. Then points satisfy the equation of the plane. Thus, the obtained results are:-

$2\cdot \alpha +3\cdot 0+6\cdot 0=6\Rightarrow 2\cdot \alpha =6\Rightarrow \alpha =3,$, i.e., coordinates of point P are$(3,0,0)$

$2\cdot 0+3\cdot \beta +6\cdot 0=6\Rightarrow 3\cdot \beta =6\Rightarrow \beta =2,i.e.,$, i.e., coordinates of point are$(0,2,0)$

$2\cdot 0+3\cdot \beta +6\cdot 0=6\Rightarrow 3\cdot \beta =6\Rightarrow \beta =2,$, i.e., coordinates of point Rare$(0,0,1)$

.

Then the vector

of point $\overrightarrow{PQ}=P.V$. of point $Q-P.V$,

$P=(0i+2j+0k)-(3i+0j+0k)=-3i+2j$and the vector $\overrightarrow{PR}=P.V$of point $R-P.V.$. of point $P=(0i+0j+1k)-(3i+0j+0k)=-3i+k.$.

The formula for the area ofrole="math" localid="1659010354799" $△PQR=\frac{1}{2}|\overrightarrow{PQ}\times \overrightarrow{PR}|.$.

role="math" localid="1659010827236" $$\begin{gathered} \overrightarrow{PQ}\times \overrightarrow{PR}=\left(-3i+2j\right)\times \left(-3i+k\right) \\ =\left(-3i\times 3i\right)+\left(-3i\times k\right)+\left(2j\times -3i\right)+\left(2j\times k\right) \\ =\left(9i\times i\right)-\left(3i\times k\right)-\left(6j\times i\right)+\left(2j\times k\right) \\ =3j+6k+2i\left(QI\times I=0,i\times k=-j,j\times i=-k,j\times k=i\right) \\ \left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|=\left|3j+6k+2i\right| \\ =\sqrt{3^2+6^2+2^2} \\ =\sqrt{49} \\ =7 \end{gathered}$$

Hence, area of the triangle PQR is $\frac{1}{2}\left|\overrightarrow{PQ}\times \overrightarrow{PR}\right|=\frac{7}{2}$units.