Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the equation of the plane be $ax+by+cz+d=0$.

Since the plane passes through $(1,0,0)$, we have $a+d=0\Rightarrow a=-d$.

Since the plane passes through $(0,1,0)$, we have $b+d=0\Rightarrow b=-d$.

Substituting $a$ and $b$ into the equation of the plane, we get:

$-dx-dy+cz+d=0$

Dividing by $-d$, the equation becomes:

$x+y-\frac{c}{d}z-1=0$

Let $k=-\frac{c}{d}$. The equation of the plane is $x+y+kz-1=0$.

The normal vector to this plane is $\overrightarrow{n_1}=\hat{i}+\hat{j}+k\hat{k}$.

The normal vector to the given plane $x+y-3=0$ is $\overrightarrow{n_2}=\hat{i}+\hat{j}$.

The angle between the planes is $45^{\circ }$, so:

$\cos 45^{\circ }=\frac{|\overrightarrow{n_1}\cdot \overrightarrow{n_2}|}{|\overrightarrow{n_1}||\overrightarrow{n_2}|}$

$\frac{1}{\sqrt{2}}=\frac{|(1)(1)+(1)(1)+(k)(0)|}{\sqrt{1^2+1^2+k^2}\sqrt{1^2+1^2+0^2}}$

$\frac{1}{\sqrt{2}}=\frac{2}{\sqrt{2+k^2}\sqrt{2}}$

$\sqrt{2+k^2}=2$

Squaring both sides, we get:

$2+k^2=4$

$k^2=2\Rightarrow k=\pm \sqrt{2}$

Substituting the value of $k$ back into the equation of the plane, we get:

$x+y\pm \sqrt{2}z-1=0$

Answer: $x+y\pm \sqrt{2}z-1=0$