Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The direction ratios of the two lines are $(3,5,7)$ and $(1,4,7)$.

The normal vector $\vec{n}$ to the plane containing these two lines is the cross product of their direction vectors:

$\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 5& 7\\ 1& 4& 7\end{array}\right|$

$\vec{n}=\hat{i}(35-28)-\hat{j}(21-7)+\hat{k}(12-5)$

$\vec{n}=7\hat{i}-14\hat{j}+7\hat{k}$

Dividing by $7$, the direction ratios of the normal to the plane are $(1,-2,1)$.

The plane passes through the point $(-2,2,-5)$ which lies on the first line. The equation of the plane is:

$1(x+2)-2(y-2)+1(z+5)=0$

$x+2-2y+4+z+5=0$

$x-2y+z+11=0$

The perpendicular distance from the origin $(0,0,0)$ to this plane is:

$d=\frac{|1(0)-2(0)+1(0)+11|}{\sqrt{1^2+(-2{)}^2+1^2}}$

$d=\frac{11}{\sqrt{1+4+1}}=\frac{11}{\sqrt{6}}$ units

Answer: $\frac{11}{\sqrt{6}}$ units