91Ó°ÊÓ

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

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

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

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,$P(\mathrm{Î\pm },0,0),Q(0,\mathrm{β},0),$and$R(0,0,\mathrm{λ})$. Then points satisfy the equation of the plane. Thus, the obtained results are:-

$2â‹\dots \mathrm{Î\pm }+3â‹\dots 0+6â‹\dots 0=6⇒2â‹\dots \mathrm{Î\pm }=6⇒\mathrm{Î\pm }=3,$, i.e., coordinates of point P are$(3,0,0)$

$2â‹\dots 0+3â‹\dots \mathrm{β}+6â‹\dots 0=6⇒3â‹\dots \mathrm{β}=6⇒\mathrm{β}=2,i.e.,$, i.e., coordinates of point are$(0,2,0)$

$2â‹\dots 0+3â‹\dots \mathrm{β}+6â‹\dots 0=6⇒3â‹\dots \mathrm{β}=6⇒\mathrm{β}=2,$, i.e., coordinates of point Rare$(0,0,1)$

.

Then the vector

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

$P=(0i+2j+0k)-(3i+0j+0k)=-3i+2j$and the vector $\stackrel{→}{PR}=P.V$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" $△PQR=\frac{1}{2}|\stackrel{→}{PQ}×\stackrel{→}{PR}|.$.

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

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