Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The direction ratios of the given lines are ${\vec{b}}_1=(2,3,4)$ and ${\vec{b}}_2=(3,4,5)$.

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

$\vec{n}={\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|$

$\vec{n}=\hat{i}(15-16)-\hat{j}(10-12)+\hat{k}(8-9)=-\hat{i}+2\hat{j}-\hat{k}$

The direction ratios of the normal are $(-1,2,-1)$, which can be written as $(1,-2,1)$.

The plane contains the point $(1,2,3)$ from the first line. The equation of the plane is:

$1(x-1)-2(y-2)+1(z-3)=0$

$x-1-2y+4+z-3=0$

$x-2y+z=0$

The given plane is $Ax-2y+z=d$. Since the distance between the two planes is a non-zero constant ($\sqrt{6}$ units), the planes must be parallel. Comparing the normal vectors, we get $A=1$.

The distance between the parallel planes $x-2y+z=0$ and $x-2y+z-d=0$ is given by:

$\text{Distance}=\frac{|d-0|}{\sqrt{1^2+(-2{)}^2+1^2}}=\frac{|d|}{\sqrt{6}}$

Given that the distance is $\sqrt{6}$ units:

$\frac{|d|}{\sqrt{6}}=\sqrt{6}$

$|d|=6$

Answer: 6