Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let $P(x_1,y_1,z_1)$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z-3=0$.

Using the formula for the image of a point:

$\frac{x_1-3}{1}=\frac{y_1-1}{-1}=\frac{z_1-7}{1}=-2\frac{1(3)-1(1)+1(7)-3}{1^2+(-1{)}^2+1^2}$

$\frac{x_1-3}{1}=\frac{y_1-1}{-1}=\frac{z_1-7}{1}=-2\left(\frac{6}{3}\right)=-4$

Solving for $x_1,y_1,z_1$:

$x_1=3-4=-1$

$y_1=1+4=5$

$z_1=7-4=3$

Thus, the point $P$ is $(-1,5,3)$.

The required plane contains the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$, which passes through the origin $(0,0,0)$ and has direction ratios $(1,2,1)$.

Since the plane contains the origin, its equation is of the form $ax+by+cz=0$.

The normal vector to the plane is perpendicular to both the line's direction vector $\vec{d}=\hat{i}+2\hat{j}+\hat{k}$ and the position vector of $P$, $\overrightarrow{OP}=-\hat{i}+5\hat{j}+3\hat{k}$.

The normal vector $\vec{n}$ is given by the cross product:

$\vec{n}=\vec{d}\times \overrightarrow{OP}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ -1& 5& 3\end{array}\right|$

$\vec{n}=\hat{i}(6-5)-\hat{j}(3+1)+\hat{k}(5+2)$

$\vec{n}=\hat{i}-4\hat{j}+7\hat{k}$

The equation of the plane passing through the origin with normal vector $\hat{i}-4\hat{j}+7\hat{k}$ is:

$1(x-0)-4(y-0)+7(z-0)=0$

$x-4y+7z=0$

Answer: $x-4y+7z=0$