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 $(1,0,0)$ and $(0,1,0)$, we have:
$a+d=0\Rightarrow a=-d$
$b+d=0\Rightarrow b=-d$

The direction ratios of the normal to the plane are $(a,b,c)=(-d,-d,c)$.
Dividing by $-d$, the direction ratios can be written as $(1,1,k)$, where $k=-\frac{c}{d}$.

The normal to the given plane $2x+3y=7$ has direction ratios $(2,3,0)$.

The angle between the planes is $45^{\circ }$. Using the formula for the angle between two normals:
$\cos 45^{\circ }=\frac{|1(2)+1(3)+k(0)|}{\sqrt{1^2+1^2+k^2}\sqrt{2^2+3^2+0^2}}$

$\frac{1}{\sqrt{2}}=\frac{5}{\sqrt{2+k^2}\sqrt{13}}$

Squaring both sides:
$\frac{1}{2}=\frac{25}{13(2+k^2)}$

$13(2+k^2)=50$

$26+13k^2=50$

$13k^2=24$

$k^2=\frac{24}{13}\Rightarrow k=\pm \frac{2\sqrt{6}}{\sqrt{13}}$

The direction ratios of the normal are $\left(1,1,\pm \frac{2\sqrt{6}}{\sqrt{13}}\right)$.

Multiplying by $\sqrt{13}$, the direction ratios are $(\sqrt{13},\sqrt{13},\pm 2\sqrt{6})$.

Answer: $\sqrt{13},\sqrt{13},2\sqrt{6}$