The vertices B and C of a Δ ABC lie on the line, x+2 3= y-1 0= z 4 such that BC=5 units. Then the area (in sq. units) of this triangle, given the…

The points $B$B and $C$C lie on the line $\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}.$3x+2​=0y−1​=4z​.

Draw perpendicular $AD$AD on the line $BC.$BC.

Clearly area of $\Delta ABC=\frac{1}{2}\cdot AD\cdot BC$ΔABC=21​⋅AD⋅BC

To find a point on the line, let $\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}=r$3x+2​=0y−1​=4z​=r

$$\Rightarrow x+2=3r,y-1=0,z=4r$$⇒x+2=3r,y−1=0,z=4r

$$\Rightarrow x=3r-2,y=1,z=4r$$⇒x=3r−2,y=1,z=4r

Thus, the point $D\equiv (3r-2,1,4r)$D≡(3r−2,1,4r)

The direction ratios of a line joining two points $({x}_{1},{y}_{1},{z}_{1})$(x1​,y1​,z1​) and $({x}_{2},{y}_{2},{z}_{2})$(x2​,y2​,z2​) are $<{x}_{2}-{x}_{1},{y}_{2}-{y}_{1},{z}_{2}-{z}_{1}>$<x2​−x1​,y2​−y1​,z2​−z1​>

Thus, the direction ratios of $AD$AD are $<3r-2-1, 1+1, 4r-2>=<3r-3, 2, 4r-2>$<3r−2−1,1+1,4r−2>=<3r−3,2,4r−2>

Since $AD$AD is perpendicular to given line, hence the dot product of their direction ratios is zero.

$$\Rightarrow 3\cdot (3r-3)+0\cdot 2+4\cdot (4r-2)=0$$⇒3⋅(3r−3)+0⋅2+4⋅(4r−2)=0

$$\Rightarrow 9r-9+16r-8=0$$⇒9r−9+16r−8=0

$$\Rightarrow 25r-17=0$$⇒25r−17=0

$$\Rightarrow r=\frac{17}{25}$$⇒r=2517​

$$\Rightarrow D\equiv (3\times \frac{17}{25}-2,1,4\times \frac{17}{25})=(\frac{1}{25},1,\frac{68}{25})$$⇒D≡(3×2517​−2,1,4×2517​)=(251​,1,2568​)

The distance between the points ({x}_{1},{y}_{1},{z}_{1})&({x}_{2},{y}_{2},{z}_{2}) is $\sqrt{{({x}_{1}-{x}_{2})}^{2}+{({y}_{1}-{y}_{2})}^{2}+{({z}_{1}-{z}_{2})}^{2}}$(x1​−x2​)2+(y1​−y2​)2+(z1​−z2​)2​

$$\Rightarrow AD=\sqrt{{(\frac{1}{25}-1)}^{2}+{(1+1)}^{2}+{(\frac{68}{25}-2)}^{2}}$$⇒AD=(251​−1)2+(1+1)2+(2568​−2)2​

$$\Rightarrow AD=\sqrt{\frac{576}{625}+4+\frac{324}{625}}$$⇒AD=625576​+4+625324​​

$$\Rightarrow AD=\sqrt{\frac{136}{25}}$$⇒AD=25136​​

$\Rightarrow AD=\frac{2}{5}\sqrt{34}$⇒AD=52​34​ units

Hence, the area of $\Delta ABC=\frac{1}{2}\cdot (\frac{2}{5}\sqrt{34})\cdot 5=\sqrt{34}$ΔABC=21​⋅(52​34​)⋅5=34​ sq units.