MHTCET 16 Apr 2026 Shift 1 Paper

Concept:The normal to a plane must be perpendicular to any line lying in the plane, and the angle between planes equals the angle between their normals.Explanation:Let the direction ratios of the required normal be $(a,b,c)$.The plane passes through $(1,0,0)$ and $(0,1,0)$, so the direction of the line joining these points is $(-1,1,0)$.Since this line lies in the plane, the normal is perpendicular to it:$$(a,b,c)\cdot(-1,1,0)=0 \Rightarrow -a+b=0 \Rightarrow a=b.$$So the normal can be taken as $(a,a,c)$.The normal to the given plane $2x+3y=7$ is $(2,3,0)$.The angle between the two planes is $45^\circ$, so the angle between their normals is also $45^\circ$.Using the dot product formula:$$\frac{|(a,a,c)\cdot(2,3,0)|}{\sqrt{a^2+a^2+c^2}\sqrt{2^2+3^2}}=\frac{1}{\sqrt2}.$$This simplifies to:$$\frac{|5a|}{\sqrt{2a^2+c^2}\sqrt{13}}=\frac{1}{\sqrt2}.$$Squaring both sides:$$\frac{25a^2}{13(2a^2+c^2)}=\frac12.$$Cross-multiplying:$$50a^2=26a^2+13c^2 \Rightarrow 24a^2=13c^2.$$Choose $a=\sqrt{13}$. Then:$$24(13)=13c^2 \Rightarrow c^2=24 \Rightarrow c=2\sqrt6.$$Hence the direction ratios are $\sqrt{13},\sqrt{13},2\sqrt6$.Answer:$$\sqrt{13},\ \sqrt{13},\ 2\sqrt6 \quad \text{(Option D)}$$