Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The plane contains the point $Q(1,0,-1)$ and the line $\vec{r}=(\hat{i}-3\hat{j}+\hat{k})+\lambda (2\hat{i}-2\hat{j}+\hat{k})$.

The line passes through the point $A(1,-3,1)$ and has the direction vector $\vec{b}=2\hat{i}-2\hat{j}+\hat{k}$.

Since the plane contains both the point $Q$ and the line, it contains the vector $\overrightarrow{QA}$ joining $Q$ and $A$.

$\overrightarrow{QA}=(1-1)\hat{i}+(-3-0)\hat{j}+(1-(-1))\hat{k}=-3\hat{j}+2\hat{k}$

The normal vector $\vec{n}$ to the plane is perpendicular to both $\vec{b}$ and $\overrightarrow{QA}$.

$\vec{n}=\vec{b}\times \overrightarrow{QA}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& -2& 1\\ 0& -3& 2\end{array}\right|$

$\vec{n}=\hat{i}(-4-(-3))-\hat{j}(4-0)+\hat{k}(-6-0)=-\hat{i}-4\hat{j}-6\hat{k}$

Taking the normal vector as $\hat{i}+4\hat{j}+6\hat{k}$, the equation of the plane passing through $Q(1,0,-1)$ is:

$1(x-1)+4(y-0)+6(z+1)=0$

$x+4y+6z+5=0$

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

$d=\frac{|1(0)+4(0)+6(0)+5|}{\sqrt{1^2+4^2+6^2}}=\frac{5}{\sqrt{1+16+36}}=\frac{5}{\sqrt{53}}$

Comparing this with the given distance $\frac{p}{\sqrt{53}}$, we get $p=5$.

Answer: $5$