MHTCET 22 Apr 2025 Shift 2 Paper

Concept:Each line is the intersection of two planes. The direction ratios of the line are obtained by the cross product of the normals of its two planes.Explanation:For the first line, the planes are $x-3y-4=0$ and $4y-z+5=0$.Their normals are $\vec{n}_1=(1,-3,0)$ and $\vec{n}_2=(0,4,-1)$.So the direction ratios of the first line are:$$(l_1,m_1,n_1)=\vec{n}_1\times\vec{n}_2=(3,1,4).$$For the second line, the planes are $x+3y-11=0$ and $2y-z+6=0$.Their normals are $\vec{n}_3=(1,3,0)$ and $\vec{n}_4=(0,2,-1)$.Thus the direction ratios of the second line are:$$(l_2,m_2,n_2)=\vec{n}_3\times\vec{n}_4=(-3,1,2).$$Let $\theta$ be the angle between the lines. Then:$$\cos\theta=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_2^2+m_2^2+n_2^2}}.$$Substitute the values:$$\cos\theta=\frac{3(-3)+1(1)+4(2)}{\sqrt{9+1+16}\sqrt{9+1+4}}=\frac{-9+1+8}{\sqrt{26}\sqrt{14}}=0.$$Therefore, $\theta=\frac{\pi}{2}$.Answer:$\frac{\pi}{2}$, i.e., option A.